Topic 2: Mechanics 2.2 Forces and dynamics

Mechanics is the branch of physics which concerns itself with forces, and how they affect a body's motion.

Kinematics is the sub-branch of mechanics which studies only a body's motion without regard to causes.

Dynamics is the sub-branch of mechanics which studies the forces which cause a body's motion.

Topic 2: Mechanics2.2 Forces and dynamics

GalileoKinematics

NewtonDynamics

The two pillars of mechanics

Topic 2.1 Topic 2.2

2.2.1 Calculate the weight of a body using the expression W = mg.

2.2.2 Identify the forces acting on an object and draw free-body diagrams representing those forces.

2.2.3 Determine the resultant force in different situations.

2.2.4 State Newton’s first law.

2.2.5 Describe examples of Newton’s first law.

2.2.6 State the condition for translational equilibrium.

2.2.7 Solve problems involving translational equilibrium.


Calculate the weight of a body using the expression W = mg. Identify the forces acting on an object and draw free-body diagrams representing those forces.

A force is a push or a pull measured in newtons.One force we are very familiar with is the force of gravity, AKA the weight.

The very concepts of push and pull imply direction. The direction of the weight is down toward the center of the earth.

If you have a weight of 90 newtons (or 90 N), your weight can be expressed as a vector, 90 N, down.

We will show later that weight has the formula


W = mg weightwhere g = 10 m s-2

and m is the mass in kg

Free body diagram

Calculate the weight of a body using the expression W = mg. Identify the forces acting on an object and draw free-body diagrams representing those forces.


EXAMPLE: Calculate the weight of a 25-kg object.

SOLUTION:Since m = 25 kg and g = 10 m s-2,

W = mg = (25)(10) = 250 N (or 250 n).Note that W inherits its direction from the fact that g points downward.We sketch the mass as a dot, and the weight as a vector in a free body diagram:

mass

force

W

W = mg weightwhere g = 10 m s-2

and m is the mass in kg

Certainly there are other forces besides weight that you are familiar with.

For example, when you set a mass on a tabletop, even though it stops moving, it still has a weight. The implication is that the tabletop applies a counterforce to the weight, called a normal force.

Note that the weight and the normal forces are the same length – they balance.

The normal force is called a surface contact force.

Identify the forces acting on an object and draw free-body diagrams representing those forces.


WN


Tension can only be a pull and never a push.Friction tries to oppose the motion.Friction is parallel to the contact surface.Normal is perpendicular to the contact surface.Friction and normal are mutually perpendicular.Friction and normal are surface contact forces.


Tthe tension

W

N

fContact surface


Weight is drawn from the center of an object.Normal is always drawn from the contact surface.Friction is drawn along the contact surface.Tension is drawn at whatever angle is given.


T

W

N

f

EXAMPLE: An object has a tension acting on it at 30° as shown. Sketch in the forces, and draw a free body diagram.

SOLUTION:Weight from center, down.Normal from surface, up.Friction from surface,

parallel.

Free body diagram



T30°

W

N

f

T30°

W

N

f

Determine the resultant force in different situations.

The resultant (or net) force is just the vector sum of all of the forces acting on a body.


EXAMPLE: An object has mass of 25 kg. A tension of 50 n and a friction force of 30 n are acting on it as shown. What is the resultant force?SOLUTION:Since the weight and the normal forces cancel out in the y-direction, we only need to worry about the forces in the x-direction.The net force is thus 50 – 30 = 20 n (+x-dir)

T

W

N

f50 n

30 n




EXAMPLE: An object has exactly two forces F1 = 50 n and F2 = 30 n applied simultaneously to it. What is the resultant force’s magnitude?SOLUTION:Fnet = F = F1

+ F2 so we simply graphically add the two vectors:The magnitude is just given by Fnet

2 = 502 + 302 so that

Fnet = 58 n.F1

50 n

F2

30 n

Fnet = F net forceFx,net = Fx Fy,net = Fy

F net




EXAMPLE: An object has exactly two forces F1 = 50 n and F2 = 30 n applied simultaneously to it as shown. What is the resultant force’s direction?SOLUTION:Direction is measured from the x-axis traditionally.Opposite and adjacent are given directly, so use tangent.tan = opp/adj = 30/50 = 0.6 so that = tan-1(0.6) = 31°.

F150 n

F2

30 n


F net




EXAMPLE: An object has exactly two forces F1 = 50 n and F2 = 30 n applied simultaneously to it. What is the resultant force’s magnitude?SOLUTION:Begin by resolving F1 into its x- and y-components.Then Fnet,x = 44 n and

Fnet,y = 23 + 30 = 53 n.

Fnet2 = Fnet,x

2 + Fnet,y2 so that

Fnet2 = 442 + 532, Fnet = 70 n.

F 1

50 n

F2

30 n


28°

50cos2844 n

50sin2823 n

State Newton’s first law.

Newton’s first law is related to certain studies made by Galileo Galilee which contradicted Aristotelian tenets.

Aristotle basically said“The natural state of motion of all objects (except the heavenly ones) is one of rest.”

A child will learn that if you stop pushing a cart, the cart will eventually stop moving.

This simple observation will lead the child to come up with a force law that looks something like this:

“In order for a body to be in motion, there must be a force acting on it.”

As we will show on the next slide, both of these statements is false!


FALSE

FALSE

State Newton’s first law.

Here’s how Galileo (1564-1642) thought:If I give a cart a push on a smooth, level surface, it will eventually stop.

What can I do to increase the distance without pushing it harder?

If I can minimize the friction, it’ll go farther.In fact, he reasoned, if I eliminate the friction altogether the cart will roll forever!

Galileo called the tendency of an object to not change its state of motion inertia.


Inertia will only

change if there is a

force.

State Newton’s first law. State the condition for translational equilibrium.

Newton’s first law is drawn from his concept of net force and Galileo’s concept of inertia.

Essentially, Newton’s first law says that the velocity of an object will not change if there is no net force acting on it.

In his words...

In symbols...

The above equation is known as the condition for translational equilibrium.


Every body continues in its state of rest, or of uniform motion in a straight line, unless

it is compelled to change that state by forces impressed thereon.

v = 0

v = CONST F

If F = 0, Newton’s first lawthen v = CONST.

A body’s inertia will only change if there

is a net force applied to it.

Describe examples of Newton’s first law.

As a memorable demonstration of inertia – matter’s tendency to not change its state of motion (or its state of rest) - consider this:

A water balloon is cut very rapidly with a knife.For an instant the water remains at rest!Don’t try this at home, kids.


Solve translational equilibrium problems.


EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m.

SOLUTION:Give each tension a name to organize your effort.Draw a free body diagram of the mass and the knot.T3

is the easiest force to find. Why?

m is not moving so its FBD tells us thatFy = 0 or T3

– mg = 0 or T3 = mg .

knot

30° 45°T1

T2

T3

mg

T3

FBD, m

FBD, knot

T2T1

T3

30° 45°

m



EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m.

SOLUTION: T3 = mg

Now we break T1 and T2 down to components.

Looking at FBD knot we see thatT1x

= T1 cos 30° = 0.866T1

T1y = T1 sin 30° = 0.500T1

T2x = T2 cos 45° = 0.707T2

T2y = T2 sin 45° = 0.707T2

knot

30° 45°T1

T2

T3

mg

T3

FBD, m

FBD, knot

T2T1

T3

30° 45°



EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m.SOLUTION: T3

= mg

Putting this into the FBD of knot we get:

knot

30° 45°T1

T2

T3

mg

T3

FBD, m

FBD, knot

T2T1

T3

30° 45°

∑Fx = 0

T2 = 1.225T1

0.707T2 - 0.866T1 = 0

∑Fy = 00.707T2 + 0.500T1 - T3 = 0

0.707(1.225T1) + 0.500T1 = T3

T1 = mg/1.366

T2 = 1.225(mg/1.366)T2 = 0.897mg



EXAMPLE: A 25-kg mass is hanging via three cords as shown. Find the tension in each of the three cords, in newtons.

SOLUTION: Since all of the angles are the same use the formulas we just derived:T3

= mg = 25(10) = 250 n

T1 = mg/1.366 = 25(10)/1.366 = 180 n

T2 = 0.897mg = 0.897(25)(10) = 220 n

knot

30° 45°T1

T2

T3

FYI

This is an example of using Newton’s first law with v = 0. The next example shows how to use Newton’s first when v is constant, but not zero.



EXAMPLE: A 1000-kg airplane is flying at a constant velocity of 125 m s-1. Label and determine the value of the weight W, the lift L, the drag D and the thrust F if the drag is 25000 n.

SOLUTION: Since the velocity is constant, Newton’s first law applies. Thus Fx = 0 and Fy = 0.

W = mg = 1000(10) = 10000 n and points down.L = W = 10000 n and points up since Fy = 0.D tries to impede the aircraft and points left.F = D = 25000 n and points right since Fx = 0.

W

L

D F

2.2.8 State Newton’s second law of motion.

2.2.9 Solve problems using Newton’s second law.


State Newton’s second law of motion.

Newton reasoned: “If the sum of the forces is not zero, the velocity will change.”

But we know, and he also did, that a change in velocity is an acceleration.

So Newton then asked himself: “How is the sum of the forces related to the acceleration.”

Here is what Newton said: “The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass.”

In other words, the bigger the force the bigger the acceleration, and the bigger the mass the smaller the acceleration.

In formula form


Fnet = ma Newton’s second law(or F = ma )

a = F/m

State Newton’s second law of motion.

Looking at the form F = ma note that if a = 0 then F = 0. But if a = 0, v = CONST. Thus Newton’s first law is just a special case of his second.



Solve problems using Newton’s second law.



EXAMPLE: An object has mass of 25 kg. A tension of 50 n and a friction force of 30 n are acting on it as shown. What is its acceleration?

SOLUTION: The vertical forces W and N cancel out. The net force is thus 50 – 30 = 20 n (+x-dir).From Fnet = ma we get 20 = 25a so that

a = 20 / 25 = 0.8 m s-2.

T

W

N

f50 n

30 n




PRACTICE: Use F = ma to show that the formula for weight is correct.

F = ma.But F is the weight W.And a is the freefall acceleration g.Thus F = ma becomes W = mg.


EXAMPLE: A 1000-kg airplane is flying in perfectly level flight. The drag D is 25000 n and the thrust F is 40000 n. Find its acceleration.

SOLUTION: Since the flight is level, Fy = 0.

Fx = F – D = 40000 – 25000 = 15000 n = Fnet.

From Fnet = ma we get

15000 = 1000a

a = 15000 / 1000 = 15 m s-2.

W

L

D F






EXAMPLE: A 25-kg object has exactly two forces F1 = 40. n and F2 = 30. n applied simultaneously to it. What is the object’s acceleration?SOLUTION:Begin by resolving F1 into its x- and y-components.Then Fnet,x = 36 n and

Fnet,y = 17 + 30 = 47 n.

Fnet2 = Fnet,x

2 + Fnet,y2 so that

Fnet2 = 362 + 472, Fnet = 59 n.

Then from Fnet = ma we get

59 = 25a

a = 59 / 25 = 2.4 m s-2.

F 1

40 n

F2

30 n

25°

40cos2536 n

40sin2517 n




EXAMPLE: A 25-kg object resting on a frictionless incline is released, as shown. What is its acceleration?

SOLUTION:

Begin with a FBD.Break down the weight into its components.Since N and mg cos 30° are perpendicular to the path of the crate they do NOT contribute to its acceleration.

Fnet = ma

mg sin 30° = ma

a = 10 sin 30° = 5.0 m s-2.

30°

6.0 m

mg

N

6030

mg cos 30mg sin 30




EXAMPLE: A 25-kg object resting on a frictionless incline is released, as shown. What is its speed at the bottom?

SOLUTION:

We found that its acceleration is 5.0 m s-2.

We will use the timeless equation to find v so we need to know what s is.

From trigonometry, we have opposite and we want hypotenuse so we use sin = opp/hyp. Thuss = hyp = opp/sin = 6 / sin 30° = 12 m. v2 = u2 + 2as = 02 + 2(5)(12)

v = 11 m s-1.

30°

6.0 m

s

u = 0

v = ?

a = 5

2.2.10 Define linear momentum and impulse.

2.2.11 Determine the impulse due to a time-varying force by interpreting a force-time graph.

2.2.12 State the law of conservation of linear momentum.

2.2.13 Solve momentum and impulse problems.


Define linear momentum and impulse.

Linear momentum, p, is defined to be the product of an object’s mass m with its velocity v.

Its units are obtained directly from the formula and are kg m s-1.


p = mv linear momentum

EXAMPLE: What is the linear momentum of a 4.0-gram NATO SS 109 bullet traveling at 950 m/s?

SOLUTION: Convert grams to kg (jump 3 decimal places left) to get m = .004 kg. Then p = mv = (.004)(950) = 3.8 kg m s-1.


From Fnet = ma we can get

Fnet = m∆v/∆t

Fnet = ∆p/∆t

This last is just Newton’s second law in terms of change in momentum rather than mass and acceleration.


p = mv linear momentum

Fnet = ∆p/∆t Newton’s second law (p-form)

EXAMPLE: A 6-kg object has its speed increase from 5 m s-1 to 25 m s-1 in 30 s. What is the net force acting on it?

SOLUTION: Fnet = ∆p/∆t = m(v – u)/∆t

= 6(25 – 5)/30 = 4 n.


If we manipulate Newton’s second law (p-form) to isolate the change in momentum we get:

Fnet = ∆p/∆t

Fnet∆t = ∆p

We call the force times the time the impulse J.



EXAMPLE: A baseball has an average force of 12000 n applied to it for 25 ms. What is the impulse imparted to the ball from the bat?

SOLUTION: J = Fnet∆t

= 12000(2510-3)

= 300 n s.

J = Fnet∆t = ∆p impulse

FYI

The units for impulse are the units of force (n) times time (s).

Determine the impulse due to a time-varying force by interpreting a force-time graph.

The impulse is the area under an F vs. t graph.


EXAMPLE: A bat striking a ball imparts a force to it as shown in the graph. Find the impulse.

SOLUTION: Break the graph into simple areas of rectangles and triangles.A1 = (1/2)(3)(9) = 13.5 n s

A2 = (4)(9) = 36 n s

A3 = (1/2)(3)(9) = 13.5 n s

Atot = A1 + A2 + A3

Atot = 13.5 + 36 + 13.5 = 63 n s.

J = Fnet∆t = ∆p impulse

0

9

3

6

0 5 10

Force F/n

Time t/s

State the law of conservation of linear momentum.Recall Newton’s second law (p-form):

If the net force acting on an object is zero, we haveFnet = ∆p/∆t 0 = ∆p/∆t 0 = ∆pIn words, if the net force is zero, then the momentum does not change – p is constant.In symbols



If Fnet = 0 then p = CONST conservation of linear momentum

2.2.14 State Newton’s third law.

2.2.15 Discuss examples of Newton’s third law. Students should understand that when two bodies interact, the forces they exert on each other are equal and opposite.


State Newton’s third law. Discuss examples of Newton’s third law.

In words“For every action force there is an equal and opposite reaction force.”

In symbols

In the big picture, if every force in the universe has a reaction force that is equal and opposite, the net force in the whole universe is zero!

So why are there accelerations all around us?


FAB = -FBA

FAB is the force on body A by body B.FBA is the force on body B by body A.

Newton’s third law



EXAMPLE: Consider a door. When you push on a door with 10 n, because of Newton’s third law the door pushes on your hand with the same 10 n, but in the opposite direction. Why does the door move, and you don’t?

SOLUTION: Even though the forces are equaland opposite, they are acting on different bodies. Each body acts in response only to the force acting on it.The door can’t resist FAB, but you CAN resist FBA.

youraction

A

BFAB

A

FBA

the door’s reaction



EXAMPLE: Consider a baseball resting on a tabletop. Discuss all of the forces acting on the baseball, and their reactions.

SOLUTION: Acting on the ball is its weight FBE and the normal force on the ball caused by the table NBT.

The reactions are FEB and NTB.Note that FBE (the weight force) and NBT (the normal force) are acting on the ball. NBT (the normal force) acts on the table.

FEB (the weight force) acts on the earth.

FEB

FBE

NBT

NTB


We define a system as a collection of more than one body, mutually interacting with each other.


EXAMPLE: Three billiard balls interacting on a pool table constitute a system.

The action-reaction force pairs between the balls are called internal forces.

For any system all internal forces cancel!




EXAMPLE: Three billiard balls interacting on a pool table constitute a system. Label and discuss all of the internal forces.

The internal force pairs only exist while the balls are in contact with one another.

Note that a blue and a red force act on the white ball. It responds only to those two forces.

Note that a single white force acts on the red ball. It responds only to that single force.

Note that a single white force acts on the blue ball. It responds only to that single force.




EXAMPLE: Three billiard balls interacting on a pool table constitute a system. Describe the external forces.

External forces are the forces that the balls feel from external origins.

For billiard balls, these forces are the balls’ weights, the cushion forces, and the queue stick forces.

State the law of conservation of linear momentum.In light of Newton’s third law and a knowledge of systems, the conservation of linear momentum can be refined for a system of particles.

Since in a system all of the internal forces sum up to zero, Fnet can only be the sum of the external forces. Thus

In other words, internal forces cannot change the momentum of a system of particles – ever!


If Fnet = 0 then p = CONST conservation of linear momentum

If Fnet,ext = 0 then p = CONST conservation of linear momentum - system

Solve momentum and impulse problems.


EXAMPLE: A 12-kg block of ice at rest has a fire cracker inside a hole drilled in its center. When it explodes, the block breaks into 2 pieces, one of which travels at +16 m s-1 in the x-direction. What is the velocity of the other 8.0 kg piece?

SOLUTION: Make before and after sketches. The initial velocity of the two is 0.From conservation of momentum we havep = CONST which means p0 = pf. Since p = mv,

(8+4)(0) = 8v + 4(16) so that v = -8 m s-1.

8 4

8 4 16v


EXAMPLE: A 730-kg Smart Car traveling at 25 m s-1

(x-dir) collides with a stationary 1800-kg Dodge Charger. The two vehicles stick together. Find their velocity immediately after the collision.

SOLUTION: Make before and after sketches.p0 = pf so that (730)(25) = (730 + 1800)v

18250 = 2530v

v = 18250/2530 = 7.2 m s-1.

Solve momentum and impulse problems.



730 1800

25 0 730 +1800

v

Topic 2: Mechanics 2.2 Forces and dynamics

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