Dynamics Laws of Motion Elevators, Pulleys, and...

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Dynamics Laws of Motion Elevators, Pulleys, and Friction Lana Sheridan De Anza College Oct 12, 2017

Transcript of Dynamics Laws of Motion Elevators, Pulleys, and...

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DynamicsLaws of Motion

Elevators, Pulleys, and Friction

Lana Sheridan

De Anza College

Oct 12, 2017

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Last time

• equilibrium

• nonequilibrium

• Problem solving with• tensions• inclines

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Overview

• Problem solving with more complex scenarios

• elevators• pulleys• friction

• Extending these ideas

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Separate Objects Pushed Together

Consider a force F that acts on two objects, masses m1 and m2,free to slide on a frictionless surface:

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

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Separate Objects Pushed Together

Question. What is the acceleration of object m2?

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

(A) Fm2

(B) m1Fm2

(C) Fm1+m2

(D) Fm1+m2

+ Fm2

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Separate Objects Pushed Together

Question. What is the acceleration of object m2?

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

(A) Fm2

(B) m1Fm2

(C) Fm1+m2

←(D) F

m1+m2+ F

m2

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Separate Objects Pushed Together

Question. What is the net force on object m2?

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

(A) 0

(B) m1Fm1+m2

(C) m2Fm1+m2

(D) F

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Separate Objects Pushed Together

Question. What is the net force on object m2?

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

(A) 0

(B) m1Fm1+m2

(C) m2Fm1+m2

←(D) F

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Separate Objects Pushed TogetherMain Idea: if objects are pushed or pulled together, then they mustall accelerate at the same rate.

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

That means that the individual net forces on each must bedifferent:

5.7 Analysis Models Using Newton’s Second Law 125

Example 5.7 One Block Pushes Another

Two blocks of masses m1 and m2, with m1 . m2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A constant horizontal force F

S is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are look-ing for the acceleration of the system.

AM

S O L U T I O N

Analyze Defining the initial position of the front bumper as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, xf 5 xi 1 vxi t 1 1

2ax t 2:

d 5 12ax t 2

Solve for t: (4) t 5 Å2dax

5 Å 2dg sin u

Use Equation 2.17, with vxi 5 0, to find the final velocity of the car:

vxf2 5 2axd

(5) vxf 5 "2axd 5 "2gd sin u

Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need.

What previously solved problem does this situation become if u 5 90°?

Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes

ax 5 g sin u 5 g sin 90° 5 g

which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.

WHAT IF ?

m2m1

1

m1m2

2

x

y

FS

FS

21PS

12PS

nSnS

gS gS

a

b c

Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m2. (b) The forces act-ing on m1. (c) The forces acting on m2.

continued

▸ 5.6 c o n t i n u e d

Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom.

Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration.

S O L U T I O N

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Separate Objects Pulled Along

m m2 1

FT

The accelerations of the two blocks are the same.

Imagining them as a single block gives the acceleration straightaway:

a =F

m1 +m2

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Separate Objects Pulled Along

m m2 1

FT

The accelerations of the two blocks are the same.

Imagining them as a single block gives the acceleration straightaway:

a =F

m1 +m2

m m2 1

FT

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Elevator Problems

a = 0

Elevator is at rest or moving withconstant velocity. You feel the sameas you normally do. Your weight andnormal force are both of magnitudemg .

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Elevator Problems

a = +a j (a is a positive number)

Elevator could be moving upwardincreasing speed or downwarddecreasing speed. You feel as if yourweight has increased.

Your weight is −mg j, but thenormal force is n = m(g + a) j.

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Elevator Problems

a = −a j (a is a positive number)

Elevator could be moving upwardand slowing down or movingdownward increasing speed. You feelas if your weight has decreased.

Your weight is −mg j, but thenormal force is n = m(g − a) j.

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Elevator Problems

An example with the normal force replaced by a tension instead:

5.7 Analysis Models Using Newton’s Second Law 127

Conceptualize The reading on the scale is related to the extension of the spring in the scale, which is related to the force on the end of the spring as in Figure 5.2. Imagine that the fish is hanging on a string attached to the end of the spring. In this case, the magnitude of the force exerted on the spring is equal to the tension T in the string. There-fore, we are looking for T. The force T

S pulls down on the

string and pulls up on the fish.

Categorize We can categorize this problem by identify-ing the fish as a particle in equilibrium if the elevator is not accelerating or as a particle under a net force if the elevator is accelerating.

Analyze Inspect the diagrams of the forces acting on the fish in Figure 5.13 and notice that the external forces acting on the fish are the downward gravitational force F

Sg 5 mgS

and the force TS

exerted by the string. If the elevator is either at rest or moving at constant velocity, the fish is a par-ticle in equilibrium, so o Fy 5 T 2 Fg 5 0 or T 5 Fg 5 mg. (Remember that the scalar mg is the weight of the fish.) Now suppose the elevator is moving with an acceleration aS relative to an observer standing outside the elevator in an inertial frame. The fish is now a particle under a net force.

S O L U T I O N

123

45678

9 0TS

mgS

123

45678

9 0

TS

mgS

a b

aSaS

When the elevator accelerates upward, the spring scale reads a value greater than theweight of the fish.

When the elevator accelerates downward, the spring scale reads a value less than theweight of the fish.

Figure 5.13 (Example 5.8) A fish is weighed on a spring scale in an accelerating elevator car.

Apply Newton’s second law to the fish: o Fy 5 T 2 mg 5 may

Solve for T : (1) T 5 may 1 mg 5 mg aay

g1 1b 5 Fg aay

g1 1b

where we have chosen upward as the positive y direction. We conclude from Equation (1) that the scale reading T is greater than the fish’s weight mg if aS is upward, so ay is positive (Fig. 5.13a), and that the reading is less than mg if aS is downward, so ay is negative (Fig. 5.13b).

(B) Evaluate the scale readings for a 40.0-N fish if the elevator moves with an acceleration ay 5 62.00 m/s2.

S O L U T I O N

Evaluate the scale reading from Equation (1) if aS is upward:

T 5 140.0 N 2 a2.00 m/s2

9.80 m/s2 1 1b 5 48.2 N

Evaluate the scale reading from Equation (1) if aS is downward:

T 5 140.0 N 2 a22.00 m/s2

9.80 m/s2 1 1b 5 31.8 N

Finalize Take this advice: if you buy a fish in an elevator, make sure the fish is weighed while the elevator is either at rest or accelerating downward! Furthermore, notice that from the information given here, one cannot determine the direction of the velocity of the elevator.

Suppose the elevator cable breaks and the elevator and its contents are in free fall. What happens to the reading on the scale?

Answer If the elevator falls freely, the fish’s acceleration is ay 5 2g. We see from Equation (1) that the scale reading T is zero in this case; that is, the fish appears to be weightless.

WHAT IF ?

▸ 5.8 c o n t i n u e d

Whether the reaction of the floor of the tension in the cable causethe upward force, the net force is still calculated in the same way.

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Elevator Problems

Question. You are hired to design an elevator that can lift peopleup to the top of a 70 story building in a short amount of time.The weight of the elevator car is 3500 N and the max load theelevator should be rated to carry is 2000 N and the weight of theentire cable used to lift the elevator car is 3000 N. You add thesenumbers together and decide that you can choose a low-cost cablerated to carry a load of 9000 N without breaking as the elevatorcable. Will you end up fired or with a commendation?

(A) Commendation!

(B) Fired!

(C) I’m not sure.

Page 17: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Elevator Problems

Question. You are hired to design an elevator that can lift peopleup to the top of a 70 story building in a short amount of time.The weight of the elevator car is 3500 N and the max load theelevator should be rated to carry is 2000 N and the weight of theentire cable used to lift the elevator car is 3000 N. You add thesenumbers together and decide that you can choose a low-cost cablerated to carry a load of 9000 N without breaking as the elevatorcable. Will you end up fired or with a commendation?

(A) Commendation!

(B) Fired! ←(C) I’m not sure.

Page 18: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

PulleysPulleys “turn tensions around a corner”.

128 Chapter 5 The Laws of Motion

Example 5.9 The Atwood Machine

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g. Deter-mine the magnitude of the acceleration of the two objects and the tension in the lightweight string.

Conceptualize Imagine the situation pictured in Figure 5.14a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude.

Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them. Therefore, we can categorize this problem as one involving two particles under a net force.

Analyze The free-body diagrams for the two objects are shown in Figure 5.14b. Two forces act on each object: the upward force T

S exerted by the string and

the downward gravitational force. In problems such as this one in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10. We must be very careful with signs in problems such as this one. In Figure 5.14a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign conven-tion, the y component of the net force exerted on object 1 is T 2 m1g, and the y component of the net force exerted on object 2 is m2g 2 T.

AM

S O L U T I O N

Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the two objects.

+

+

m1

m1

m2

m2

a b

TS

TS

gS

gS

m1m2

From the particle under a net force model, apply New-ton’s second law to object 1:

(1) o Fy 5 T 2 m1g 5 m1ay

Apply Newton’s second law to object 2: (2) o Fy 5 m2g 2 T 5 m2ay

Add Equation (2) to Equation (1), noticing that T cancels: 2 m1g 1 m2g 5 m1ay 1 m2ay

Solve for the acceleration: (3) ay 5 am2 2 m1

m1 1 m2bg

Substitute Equation (3) into Equation (1) to find T : (4) T 5 m1(g 1 ay) 5 a 2m1m2

m1 1 m2bg

Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m2 2 m1)g to the total mass of the system (m1 1 m2), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects.

Describe the motion of the system if the objects have equal masses, that is, m1 5 m2.

Answer If we have the same mass on both sides, the system is balanced and should not accelerate. Mathematically, we see that if m1 5 m2, Equation (3) gives us ay 5 0.

What if one of the masses is much larger than the other: m1 .. m2?

Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass. Therefore, the larger mass should simply fall as if the smaller mass were not there. We see that if m1 .. m2, Equation (3) gives us ay 5 2g.

WHAT IF ?

WHAT IF ?

For the moment, we are just considering massless, frictionlesspulleys. What does that mean?

• Massless: we do not have to worry about force needed toaccelerate each atom in the pulley

• Frictionless: the axle of the pulley has no friction to resist thewheel turning

Page 19: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

PulleysPulleys “turn tensions around a corner”.

128 Chapter 5 The Laws of Motion

Example 5.9 The Atwood Machine

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g. Deter-mine the magnitude of the acceleration of the two objects and the tension in the lightweight string.

Conceptualize Imagine the situation pictured in Figure 5.14a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude.

Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them. Therefore, we can categorize this problem as one involving two particles under a net force.

Analyze The free-body diagrams for the two objects are shown in Figure 5.14b. Two forces act on each object: the upward force T

S exerted by the string and

the downward gravitational force. In problems such as this one in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10. We must be very careful with signs in problems such as this one. In Figure 5.14a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign conven-tion, the y component of the net force exerted on object 1 is T 2 m1g, and the y component of the net force exerted on object 2 is m2g 2 T.

AM

S O L U T I O N

Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the two objects.

+

+

m1

m1

m2

m2

a b

TS

TS

gS

gS

m1m2

From the particle under a net force model, apply New-ton’s second law to object 1:

(1) o Fy 5 T 2 m1g 5 m1ay

Apply Newton’s second law to object 2: (2) o Fy 5 m2g 2 T 5 m2ay

Add Equation (2) to Equation (1), noticing that T cancels: 2 m1g 1 m2g 5 m1ay 1 m2ay

Solve for the acceleration: (3) ay 5 am2 2 m1

m1 1 m2bg

Substitute Equation (3) into Equation (1) to find T : (4) T 5 m1(g 1 ay) 5 a 2m1m2

m1 1 m2bg

Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m2 2 m1)g to the total mass of the system (m1 1 m2), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects.

Describe the motion of the system if the objects have equal masses, that is, m1 5 m2.

Answer If we have the same mass on both sides, the system is balanced and should not accelerate. Mathematically, we see that if m1 5 m2, Equation (3) gives us ay 5 0.

What if one of the masses is much larger than the other: m1 .. m2?

Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass. Therefore, the larger mass should simply fall as if the smaller mass were not there. We see that if m1 .. m2, Equation (3) gives us ay 5 2g.

WHAT IF ?

WHAT IF ?

For the moment, we are just considering massless, frictionlesspulleys. What does that mean?

• Massless: we do not have to worry about force needed toaccelerate each atom in the pulley

• Frictionless: the axle of the pulley has no friction to resist thewheel turning

Page 20: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

PulleysFree-body diagrams:

128 Chapter 5 The Laws of Motion

Example 5.9 The Atwood Machine

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g. Deter-mine the magnitude of the acceleration of the two objects and the tension in the lightweight string.

Conceptualize Imagine the situation pictured in Figure 5.14a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude.

Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them. Therefore, we can categorize this problem as one involving two particles under a net force.

Analyze The free-body diagrams for the two objects are shown in Figure 5.14b. Two forces act on each object: the upward force T

S exerted by the string and

the downward gravitational force. In problems such as this one in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10. We must be very careful with signs in problems such as this one. In Figure 5.14a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign conven-tion, the y component of the net force exerted on object 1 is T 2 m1g, and the y component of the net force exerted on object 2 is m2g 2 T.

AM

S O L U T I O N

Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the two objects.

+

+

m1

m1

m2

m2

a b

TS

TS

gS

gS

m1m2

From the particle under a net force model, apply New-ton’s second law to object 1:

(1) o Fy 5 T 2 m1g 5 m1ay

Apply Newton’s second law to object 2: (2) o Fy 5 m2g 2 T 5 m2ay

Add Equation (2) to Equation (1), noticing that T cancels: 2 m1g 1 m2g 5 m1ay 1 m2ay

Solve for the acceleration: (3) ay 5 am2 2 m1

m1 1 m2bg

Substitute Equation (3) into Equation (1) to find T : (4) T 5 m1(g 1 ay) 5 a 2m1m2

m1 1 m2bg

Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m2 2 m1)g to the total mass of the system (m1 1 m2), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects.

Describe the motion of the system if the objects have equal masses, that is, m1 5 m2.

Answer If we have the same mass on both sides, the system is balanced and should not accelerate. Mathematically, we see that if m1 5 m2, Equation (3) gives us ay 5 0.

What if one of the masses is much larger than the other: m1 .. m2?

Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass. Therefore, the larger mass should simply fall as if the smaller mass were not there. We see that if m1 .. m2, Equation (3) gives us ay 5 2g.

WHAT IF ?

WHAT IF ?

There is another way of representing this, that for these massless,frictionless rope & pulley systems that I find helpful. Remove thepulley and lay everything out flat on a frictionless surface:

m m1

T m gm g

2

1 2

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Pulleys and the Atwood Machine

The Atwood Machine can be used to make careful determinationsof g , as well as explore the behavior of forces and accelerations.

128 Chapter 5 The Laws of Motion

Example 5.9 The Atwood Machine

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g. Deter-mine the magnitude of the acceleration of the two objects and the tension in the lightweight string.

Conceptualize Imagine the situation pictured in Figure 5.14a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude.

Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them. Therefore, we can categorize this problem as one involving two particles under a net force.

Analyze The free-body diagrams for the two objects are shown in Figure 5.14b. Two forces act on each object: the upward force T

S exerted by the string and

the downward gravitational force. In problems such as this one in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10. We must be very careful with signs in problems such as this one. In Figure 5.14a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign conven-tion, the y component of the net force exerted on object 1 is T 2 m1g, and the y component of the net force exerted on object 2 is m2g 2 T.

AM

S O L U T I O N

Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the two objects.

+

+

m1

m1

m2

m2

a b

TS

TS

gS

gS

m1m2

From the particle under a net force model, apply New-ton’s second law to object 1:

(1) o Fy 5 T 2 m1g 5 m1ay

Apply Newton’s second law to object 2: (2) o Fy 5 m2g 2 T 5 m2ay

Add Equation (2) to Equation (1), noticing that T cancels: 2 m1g 1 m2g 5 m1ay 1 m2ay

Solve for the acceleration: (3) ay 5 am2 2 m1

m1 1 m2bg

Substitute Equation (3) into Equation (1) to find T : (4) T 5 m1(g 1 ay) 5 a 2m1m2

m1 1 m2bg

Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m2 2 m1)g to the total mass of the system (m1 1 m2), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects.

Describe the motion of the system if the objects have equal masses, that is, m1 5 m2.

Answer If we have the same mass on both sides, the system is balanced and should not accelerate. Mathematically, we see that if m1 5 m2, Equation (3) gives us ay 5 0.

What if one of the masses is much larger than the other: m1 .. m2?

Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass. Therefore, the larger mass should simply fall as if the smaller mass were not there. We see that if m1 .. m2, Equation (3) gives us ay 5 2g.

WHAT IF ?

WHAT IF ?

1http://en.wikipedia.org/wiki/Atwood machine

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Pulleys and the Atwood Machine

Notice that again, like the pushed blocks, the two objects mustaccelerate together at the same rate, because they are connectedthrough an inextensible rope.

(Is that realistic?)

The tension in all parts of the rope will be the same.

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Pulleys and the Atwood Machine

Notice that again, like the pushed blocks, the two objects mustaccelerate together at the same rate, because they are connectedthrough an inextensible rope. (Is that realistic?)

The tension in all parts of the rope will be the same.

Page 24: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Pulleys and the Atwood Machine

Notice that again, like the pushed blocks, the two objects mustaccelerate together at the same rate, because they are connectedthrough an inextensible rope. (Is that realistic?)

The tension in all parts of the rope will be the same.

Page 25: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Pulleys and the Atwood Machine

We can consider the motion for each mass separately:

Fnet,1 = (T −m1g)j = m1a j (1)

Fnet,2 = (T −m2g)j = −m2a j (2)

Be careful about the signs! Both masses must acceleratetogether - one up, one down.

Two equations, two unknowns. Solve as you like!

Page 26: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Pulleys and the Atwood Machine

We can consider the motion for each mass separately:

Fnet,1 = (T −m1g)j = m1a j (1)

Fnet,2 = (T −m2g)j = −m2a j (2)

Be careful about the signs! Both masses must acceleratetogether - one up, one down.

Two equations, two unknowns. Solve as you like!

Page 27: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Pulleys and the Atwood Machine

Fnet,1 = (T −m1g)j = m1a j (1)

Fnet,2 = (T −m2g)j = −m2a j (2)

Take eq (1) − eq (2):

m2g −m1g = m1a +m2a

a =(m2 −m1)g

m1 +m2

T =2m1m2g

m1 +m2

We get the same result if we use the “horizontal layout” picture.

Page 28: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Pulleys and the Atwood Machine

Fnet,1 = (T −m1g)j = m1a j (1)

Fnet,2 = (T −m2g)j = −m2a j (2)

Take eq (1) − eq (2):

m2g −m1g = m1a +m2a

a =(m2 −m1)g

m1 +m2

T =2m1m2g

m1 +m2

We get the same result if we use the “horizontal layout” picture.

Page 29: Dynamics Laws of Motion Elevators, Pulleys, and Frictionnebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture14.pdf · Laws of Motion Elevators, Pulleys, and Friction ... Categorize

Summary

• more problem solving

• pulleys

First Test this Friday, Oct 13.

(Uncollected) HomeworkSerway & Jewett,

• Ch 5, onward from page 136. Obj.Q 1; Problems: 21