AP-C4-MJ

Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 1

This print-out should have 47 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

Weight of a Boxer001 (part 1 of 4) 10.0 points

A(n) 98.3 kg boxer has his first match inthe Canal Zone with gravitational acceler-ation 9.782 m/s2 and his second match atthe North Pole with gravitational accelera-tion 9.832 m/s2.

a) What is his mass in the Canal Zone?Correct answer: 98.3 kg.

Explanation:An object’s mass is constant, regardless of

the gravitational acceleration.

002 (part 2 of 4) 10.0 pointsb) What is his weight in the Canal Zone?Correct answer: 961.571 N.

Explanation:An object’s weight varies with gravitational

position and is given by

W = mg = (98.3 kg)(9.782 m/s2)

= 98.3 kg

003 (part 3 of 4) 10.0 pointsc) What is his mass at the North Pole?Correct answer: 98.3 kg.

Explanation:His mass doesn’t vary.

004 (part 4 of 4) 10.0 pointsd) What is his weight at the North Pole?Correct answer: 966.486 N.

Explanation:

W = mg = (98.3 kg)(9.832 m/s2)

= 98.3 kg

Hewitt CP9 04 E15005 (part 1 of 4) 10.0 points

Gravity on the surface of the moon is only1

6as strong as gravity on the Earth.

What is the weight of a 16 kg object onthe Earth? The acceleration of gravity is10 m/s2 .Correct answer: 160 N.

Explanation:On the Earth the weight is

W = mg = (16 kg)(10 m/s2) = 160 N .

006 (part 2 of 4) 10.0 pointsWhat is the weight on the moon?Correct answer: 26.6667 N.

Explanation:On the moon the weight is

W =1

6mg =

1

6(16 kg)(10 m/s2)

= 26.6667 N .

007 (part 3 of 4) 10.0 pointsWhat is the mass on the earth?Correct answer: 16 kg.

Explanation:The mass would be 16 kg everywhere.

008 (part 4 of 4) 10.0 pointsWhat is the mass on the moon?Correct answer: 16 kg.

Explanation:The mass would be 16 kg everywhere.

Dragster Acceleration009 (part 1 of 3) 10.0 points

A dragster and driver together have mass930.6 kg . The dragster, starting from rest,attains a speed of 25 m/s in 0.57 s .

Find the average acceleration of the drag-ster during this time interval.Correct answer: 43.8596 m/s2.

Explanation:The average acceleration during time t is

a =∆v

t


=v − 0

t

=v

t

=(25 m/s)

(0.57 s)

= 43.8596 m/s2 .

010 (part 2 of 3) 10.0 pointsWhat is the size of the average force on thedragster during this time interval?Correct answer: 40815.8 N.

Explanation:The average force on the dragster is

fd = md a

= (930.6 kg) (43.8596 m/s2)

= 40815.8 N .

011 (part 3 of 3) 10.0 pointsAssume: The driver has a mass of 78.6 kg .

What horizontal force does the seat exerton the driver?Correct answer: 3447.37 N.

Explanation:The force on the driver is

F = ma

= (78.6 kg) (43.8596 m/s2)

= 3447.37 N .

Force on a Bullet 02012 10.0 points

A 9.7 g bullet leaves the muzzle of a rifle witha speed of 520.5 m/s.

What constant force is exerted on the bulletwhile it is traveling down the 0.9 m length ofthe barrel of the rifle?Correct answer: 1459.96 N.

Explanation:Average acceleration can be found from

v2

f = v2

o + 2 a ℓ

Since vo = 0, we have

a =v2

2 ℓ

Thus

F = ma = mv2

2 ℓ

=(9.7 g)(520.5 m/s)2

2 (0.9 m)·

1 kg

1000 g

= 1459.96 N .

Force on a Coasting Car013 10.0 points

A 1698.7 kg car is traveling at 35 m/s whenthe driver takes his foot off the gas pedal. Ittakes 5.1 s for the car to slow down to 20 m/s.

How large is the net force slowing the car?Correct answer: 4996.18 N.

Explanation:The acceleration of the car is given by

a =v − v0

t

=20 m/s − 35 m/s

5.1 s= −2.94118 m/s2 .

Applying Newton’s second law yields the forceslowing the car:

‖~F‖ = |ma|

=∣

∣(1698.7 kg)(−2.94118 m/s2)∣

∣

= 4996.18 N .

Force on a Lawn Spreader014 10.0 points

Joe pushes down the length of the handle ofa 14.3 kg lawn spreader. The handle makesan angle of 45.6 ◦ with the horizontal. Joewishes to accelerate the spreader from rest to1.35 m/s in 1.3 s.

What force must Joe apply to the handle?Correct answer: 21.2245 N.

Explanation:


The horizontal component of the force is

Fh = F cos θ.

Let v be the final velocity of the spreader.According to Newton’s second law,

Fh = mah

so

F cos θ = m∆vh

∆t

F =m∆vh

∆t cos θ

=mv

t cos θ

=(14.3 kg) (1.35 m/s)

(1.3 s) cos 45.6◦

= 21.2245 N

keywords:

Hanging Weight 01015 (part 1 of 2) 10.0 points

Consider the 666 N weight held by two cablesshown below. The left-hand cable had tensionT and makes an angle of θ with the wall. Theright-hand cable had tension 740 N and makesan angle of 33◦ with the ceiling.

666 N

T

740 N33

◦

θ

a) What is the tension T in the left-handcable slanted at an angle of θ with respect tothe wall?Correct answer: 674.03 N.

Explanation:Observe the free-body diagram below.

F2 F1 θ1θ2

Wg

Note: The sum of the x- andy-components of F1 , F2 , andWg are equal to zero.

Let : F2 ≡ T ,

F1 = 740 N ,

Wg = 666 N ,

θ1 = 33◦ , and

θ2 = 90◦ − θ .

Basic Concept: Vertically and Horizontally,we have

F xnet = F x

1 − F x2 = 0

= F1 cos θ1 − F2 cos θ2 = 0 (1)

F ynet = F y

1+ F y

2−Wg = 0

= F1 sin θ1 + F2 sin θ2 −Wg = 0 (2)

Solution: Using Eqs. 1 and 2, we have

F x2 = F1 cos θ1 (3)

= (740 N) cos 33◦

= 620.616 N , and

F y2

= F3 − F1 sin θ1 (4)

= (666 N) − (740 N) sin 33◦

= (666 N) − (403.033 N)

= 262.967 N

So

F2 =√

(F x2)2 + (F y

2)2

=√

(620.616 N)2 + (262.967 N)2

= 674.03 N .

016 (part 2 of 2) 10.0 pointsb) What is the angle θ which the left-handcable makes with respect to the wall?Correct answer: 67.0367◦.

Explanation:


Using Eqs. 3 & 4, we have

θ2 = arctan

(

F y2

F x2

)

= arctan

(

262.967 N

620.616 N

)

= 22.9633◦ , so

θ = 90◦ − θ2

= 90◦ − 22.9633◦

= 67.0367◦ .

Hanging Weight 04017 (part 1 of 2) 10.0 points

Consider the 671 N weight held by two cablesshown below. The left-hand cable had tensionT2 and makes an angle of θ2 with the ceiling.The right-hand cable had tension 460 N andmakes an angle of 43◦ with the ceiling.

The right-hand cable makes an angle of 43◦

with the ceiling and has a tension of 460 N .

671 N

T2

460N

43◦θ

2

a) What is the tension T2 in the left-handcable slanted at an angle of θ2 with respect tothe wall?Correct answer: 490.744 N.


F2 F 1 θ1

θ2

Wg

Note: The sum of the x- andy-components of F1 , F2 , andWg are equal to zero.

Given : Wg = 671 N ,

F1 = 460 N ,

θ1 = 43◦ , and

θ2 = 90◦ − θ .

Basic Concept: Vertically and Horizontally,we have

F xnet = F x

1 − F x2 = 0

= F1 cos θ1 − F2 cos θ2 = 0 (1)

F ynet = F y

1+ F y

2−Wg = 0

= F1 sin θ1 + F2 sin θ2 −Wg = 0 (2)


F x2 = F1 cos θ1 (1)

= (460 N) cos 43◦

= 336.423 N , and

F y2

= F3 − F1 sin θ1 (2)

= 671 N − (460 N) sin 43◦

= 671 N − 313.719 N

= 357.281 N , so

F2 =√

(F x2)2 + (F y

2)2

=√

(336.423 N)2 + (357.281 N)2

= 490.744 N .

018 (part 2 of 2) 10.0 pointsb) What is the angle θ2 which the left-handcable makes with respect to the ceiling?Correct answer: 46.7222◦.

Explanation:Using Eq. 2, we have

θ2 = arctan

(

F y1

F x1

)

= arctan

(

313.719 N

336.423 N

)

= 46.7222◦ .


Hanging Weights 02019 (part 1 of 3) 10.0 points

In the figure below the left-hand cable has atension T1 and makes an angle of 55◦ withthe horizontal. The right-hand cable has atension T3 and makes an angle of θ3 with thehorizontal. A 61.4 N weight is on the left anda 53.1 N weight is on the right. The cableconnecting the two weights is horizontal.

61.4 N

53.1 N

T2

T1 T 3

θ355 ◦

a) Find the tension T1.Correct answer: 74.9556 N.

Explanation:

Given : W1 = M1 g ,

W2 = M2 g ,

θ1 = 55◦ ,

θ3 = 51.0044◦ , and

T2 = 42.9927 N .

T 3

T1 θ3

θ1

T3 cos θ3T1 cos θ1

W2W1

Note: T1 cos θ1 = T2 = T3 cos θ3

Basic Concepts:

∑

~F = m~a = 0

~W = m~g

Solution: Consider the point of attachmentof cable 1 and cable 2.

Vertically, W1 = M1 g acts down andT1 sin θ1 acts up, so

Fnet = W1 − T1 sin θ1 = 0

=⇒ T1 sin θ1 = W1 . (1)

Horizontally, T1 acts to the left and T1 cos θ1

acts to the right, so

Fnet = T1 − T1 cos θ1 = 0

=⇒ T1 cos θ1 = T2 . (2)

Using Eq. 1 we have

T1 =W1

sin θ1

=61.4 N

sin 55◦

= 74.9556 N .

020 (part 2 of 3) 10.0 pointsb) Find the tension T2.Correct answer: 42.9927 N.

Explanation:Dividing Eq. 1 by Eq. 2, we have

T2 =W1

tan θ1

=61.4 N

tan 55◦

= 42.9927 N .

021 (part 3 of 3) 10.0 pointsc) Find the angle θ3.Correct answer: 51.0044◦.

Explanation:Balancing forces at the point where the

right-hand weight is suspended, we have

T3 sin θ3 = W2 vertical (3)

T3 cos θ3 = T2 horizontal . (4)

dividing Eq. 3 by Eq. 4, we have

tan θ3 =W2

T2

θ = arctan

(

W2

T2

)

= arctan

(

53.1 N

42.9927 N

)

= 51.0044◦ .


The tension T3 =T2

cos θ3

= 68.3277 N is not

required in this problem.

Hanging Weights 01022 10.0 points

In the figure below the left-hand cable has atension T1 and makes an angle of 48◦ withthe horizontal. The right-hand cable has atension T3 and makes an angle of 54◦ withthe horizontal. A W1 weight is on the leftand a W2 weight is on the right. The cableconnecting the two weights has a tension 39 Nand is horizontal.

The acceleration of gravity is 9.8 m/s2 .

M1

M2

39 N

T1 T 3

54◦48 ◦

Determine the mass M2.Correct answer: 5.47744 kg.

Explanation:

Given : W1 = M1 g ,

W2 = M2 g ,

θ1 = 48◦ ,

θ3 = 54◦ , and

T2 = 39 N .

T 3T1 θ3θ

1

T3 cos θ3T1 cos θ1

W2

W1

Note: T1 cos θ1 = T2 = T3 cos θ3

Basic Concepts:∑

~F = m~a = 0

~W = m~g

Solution: Consider the point of attachmentof cable 2 and cable 3.

Vertically, W2 = M2 g acts down andT3 sin θ3 acts up, so

Fnet = W2 − T3 sin θ3 = 0

=⇒ T3 sin θ3 = W2 . (1)

Horizontally, T2 acts to the left and T3 cos θ3

acts to the right, so

Fnet = T2 − T3 cos θ3 = 0

=⇒ T3 cos θ3 = T2 . (2)

Dividing Eq. 1 by Eq. 2, we have

tan θ3 =W2

T2

.

W2 = T2 tan θ3

= (39 N) tan 54◦ = 66.3508 N

M2 =W2

g=

53.6789 N

9.8 m/s2= 5.47744 kg

and by symmetry, we have

T1 cos θ1 − T3 cos θ3 = 0 , so

W1 = T2 tan θ1

= (39 N) tan 48◦ = 58.2846 N

M1 =W1

g=

43.3139 N

9.8 m/s2= 4.41978 kg .

Holt SF 04Rev 68023 (part 1 of 2) 10.0 points

Consider the 44 N weight held by two cablesshown below. The left-hand cable is horizon-tal.

44 N

33◦

a) What is the tension in the cable slantedat an angle of 33◦?Correct answer: 80.7875 N.



67.7541 N67.7541 N

80.78

75N

44

N44

N

Scale: 10 N

33◦

Note: The sum of the x- andy-components of T1 , T2 , andWg are equal to zero.

Given : Wg = 44 N and

θ = 33◦ .

Basic Concept: Vertically, we have

Fy,net = F1 sin θ −Wg = 0

Solution:

F1(sin θ) = Wg

F1 =Wg

sin θ

=44 N

sin 33◦

= 80.7875 N

024 (part 2 of 2) 10.0 pointsb) What is the tension in the horizontal ca-ble?Correct answer: 67.7541 N.

Explanation:Basic Concept: Horizontally,

Fx,net = F1 cos θ − F2 = 0

Solution:

F2 = F1 cos θ

=Wg cos θ

sin θ

=(44 N) cos 33◦

sin 33◦

= 67.7541 N

Serway CP 04 16025 10.0 points

Consider the 90 N light fixture supported asin the figure.

35◦ 35◦

Find the tension in the supporting wires.Correct answer: 78.4551 N.

Explanation:

Given : W = 90 N and

θ = 35◦ .

W = 90 N

x

y

T2

θT1

θ

Horizontally,∑

Fx = 0 , so

T1 cos θ − T2 cos θ = 0

T1 = T2 .

Vertically,∑

Fy = 0 , so

2 (T1 sin θ) −W = 0

T1 = T2

=W

2 sin θ

=90 N

2 sin 35◦

= 78.4551 N .


Static Equilibrium 01026 (part 1 of 2) 10.0 points

The 7.1 N weight is in equilibrium under theinfluence of the three forces acting on it. TheF force acts from above on the left at an angleof α with the horizontal. The 5.8 N force actsfrom above on the right at an angle of 48◦ withthe horizontal. The force 7.1 N acts straightdown.

F5.8N

7.1 N

48◦α

Note: F, α are not to scale.What is the magnitude of the force F?

Correct answer: 4.7796 N.

Explanation:Note: Standard angular measurements are

from the positive x-axis in a counter-clockwisedirection.

Given : F1 = F ,

α1 = 180◦ − α ,

F2 = 5.8 N ,

α2 = 48◦ ,

F3 = 7.1 N , and

α3 = 270◦ ,

Basic Concepts:∑

F x = 0

F x1 + F x

2 + F x3 = 0

F x1 + F2 cos α2 + F3 cos α3 = 0 (1)

∑

F y = 0

F y1

+ F y2

+ F y3

= 0

F y1

+ F2 sin α2 + F3 sinα3 = 0 (2)


F x1 = −F2 cos α2 − F3 cos α3 (1)

= −(5.8 N) cos 48◦ − (7.1 N) cos 270◦

= −3.88096 N , and

F y1

= −F2 sin α2 − F3 sin α3 (2)

= −(5.8 N) sin 48◦ − (7.1 N) sin 270◦

= 2.78976 N , so

F1 =√

(F x1)2 + (F y

1)2

=√

(−3.88096 N)2 + (2.78976 N)2

= 4.7796 N .

027 (part 2 of 2) 10.0 pointsWhat is the angle α of the force F as shownin the figure above?Correct answer: 35.7098◦.

Explanation:

α1 = arctan

(

F y1

F x1

)

= arctan

(

2.78976 N

−3.88096 N

)

= 144.29◦ , from the positive x−axis

α = 180◦ − α1

= 180◦ − 144.29◦

= 35.7098◦ .

Observe the free-body diagram below.

4.78 N 5.8N

7.1 N

48◦

35.7◦

The vertical dashed vector is the sumof F1 sin α1 = 2.78976 N and F2 sinα1 =4.31024 N and is equal in length to F y

3=

7.1 N .

Static Equilibrium 04028 (part 1 of 2) 10.0 points

The knot at the junction is in equilibriumunder the influence of four forces acting onit. The F force acts from above on the left atan angle of α with the horizontal. The 5.6 Nforce acts from above on the right at an angle


of 17◦ with the horizontal. The 6.2 N forceacts from below on the left at an angle of 79◦

with the horizontal.

5.6 N

6.2N

F

knot17◦

79 ◦

α

Note: F, α are not to scale.What is the magnitude of the force F?

Correct answer: 6.09917 N.

Explanation:Note: Standard angular measurements are

from the positive x-axis in a counter-clockwisedirection.

Given : F1 = 5.6 N ,

α1 = 180◦ − 17◦ = 163◦ ,

F2 = 6.2 N ,

α2 = 180◦ + 79◦ = 281◦ ,

F3 = F , and

α3 = α .

Observe the free-body diagram belowwhere the vectors are decomposed into com-ponents along the x- and y-axes.

5.6 N

6.2N

6.1N

17◦

79 ◦

46.8

Note: The vectors along the x-and y-coordinates add to zero.

Basic Concepts:

∑

F x = 0

F x1 + F x

2 + F x3 = 0

F1 cos α1 + F2 cos α2 + F x3 = 0 (1)

∑

F y = 0

F y1

+ F y2

+ F y3

= 0

F1 sin α1 + F2 sinα2 + F y3

= 0 (2)


F x3 = −F1 cos α1 − F2 cos α2 (1)

= −(5.6 N) cos 163◦ − (6.2 N) cos 281◦

= 4.17229 N , and

F y3

= −F1 sin α1 − F2 sinα2 (2)

= −(5.6 N) sin 163◦ − (6.2 N) sin 281◦

= 4.44881 N , so

F3 =√

(F x3)2 + (F y

3)2

=√

(4.17229 N)2 + (4.44881 N)2

= 6.09917 N .

029 (part 2 of 2) 10.0 pointsWhat is the angle α of the force F as shownin the figure above?Correct answer: 46.8371◦.

Explanation:

α3 = arctan

(

F y3

F x3

)

= arctan

(

4.44881 N

4.17229 N

)

= 46.8371◦ , from the positive x−axis

α = 46.8371◦ .

Tipler PSE5 04 45030 (part 1 of 4) 10.0 points

A 0.62 kg block is suspended from the middleof a 1.68 m long string. The ends of the stringare attached to the ceiling at points separatedby 1 m, and the block can slip along the longstring.



1 m

0.84 m

T1

0.84 m

T2

0.62 kg

What angle does the string make with theceiling?Correct answer: 53.4704◦.

Explanation:

Given : ℓ = 1.68 m ,

d = 1 m .

Consider the physical distances

cos θ =

1

2d

1

2ℓ

=d

ℓ

θ = cos−1d

ℓ

= cos−11 m

1.68 m

= 53.4704◦ .

031 (part 2 of 4) 10.0 pointsWhat is the tension in the string?Correct answer: 3.78458 N.

Explanation:

Given : m = 0.62 kg .

Applying∑

Fy = may = 0 to the mass,

2 T sin θ − mg = 0

T =mg

2 sin θ

=(0.62 kg)

(

9.81 m/s2)

2 sin 53.4704◦

= 3.78458 N .

032 (part 3 of 4) 10.0 pointsThe 0.62 kg block is removed and two 0.31 kgblocks are attached to the string such that thelengths of the three string segments are equal.

1 m

0.56 mT1

0.56 m

T2

0.56 mT3

0.31 kg 0.31 kg

What is the tension in the string segmentattached to the ceiling on the right ?Correct answer: 3.30698 N.

Explanation:

Let m = 0.31 kg .

The length of each segment is

ℓ′ =ℓ

3=

1.68 m

3= 0.56 m .

Find the distance d′

d′ =d − ℓ′

2

=(1 m) − (0.56 m)

2= 0.22 m .

cos θ′ =d′

ℓ′

θ′ = cos−1

(

d′

ℓ′

)

= cos−1

(

0.22 m

0.56 m

)

= 66.8676◦ .


Applying∑

Fy = may = 0 to the 0.31 kg

block

T sin θ′ − mg = 0

T3 =mg

sin θ′

=(0.31 kg)

(

9.81 m/s2)

sin 66.8676◦

= 3.30698 N .

033 (part 4 of 4) 10.0 pointsWhat is the tension in the horizontal seg-ment?Correct answer: 1.29917 N.

Explanation:

Applying∑

Fx = max = 0 to the block,

T3 cos θ′ − T2 = 0

T2 = T3 cos θ′

= (3.30698 N) cos 66.8676◦

= 1.29917 N .

Tipler PSE5 04 53034 (part 1 of 2) 10.0 points

Your car is stuck in a mud hole. You are alone,but you have a long, strong rope. Havingstudied physics, you tie the rope tautly to atelephone pole and pull on it sideways at themidpoint, as shown.

404 N 3.3◦

22 m

Find the force exerted by the rope on thecar when the angle is 3.3◦ and you are pullingwith a force of 404 N but the car does notmove.Correct answer: 3.50914 kN.

Explanation:

Let : F = 404 N ,

θ = 3.3◦ , and

ℓ = 22 m .

Before the car starts moving,

∑

Fy = may ,

so

2 T sin θ − F = 0

T =F

2 sin θ

=404 N

2 sin 3.3◦·

kN

1000 N

= 3.50914 kN .

035 (part 2 of 2) 10.0 pointsHow strong must the rope be if it takes a forceof 599 N to move the car when θ is 3.1◦?Correct answer: 5.53821 kN.

Explanation:

Let : θ = 3.1◦ .

T =F

2 sin θ

=599 N

2 sin 3.1◦·

kN

1000 N

= 5.53821 kN .

Forces on a Sled036 (part 1 of 2) 10.0 points

A child holds a sled on a frictionless, snow-covered hill, inclined at an angle of 31◦.

62N

F

31◦

If the sled weighs 62 N, find the force ex-erted on the rope by the child.Correct answer: 31.9324 N.

Explanation:


Given : W = 62 N and

θ = 31◦ .

Consider the free body diagram for theblock

mg sin

θ N=

mg cos

θ

F

W

Basic Concepts: If we “tilt” our world,and consider the forces parallel to the hill,

Fnet =∑

Fup −∑

Fdown = 0

then the forces perpendicular to the hill,

Fnet =∑

Fout −∑

Fin = 0

Solution: Consider the free body diagramfor the sled: The weight of the sled has compo-nents W sin θ acting down the hill and W cos θacting straight into the hill.

The system is in equilibrium, so for forcesparallel to the hill,

Fnet = T −W sin θ = 0

=⇒ T = W sin θ

= (62 N) sin 31◦

= 31.9324 N

037 (part 2 of 2) 10.0 pointsWhat force is exerted on the sled by the hill?Correct answer: 53.1444 N.

Explanation:For forces perpendicular to the hill,

Fnet = N −W cos θ = 0

=⇒ N = W cos θ

= (62 N) cos 31◦

= 53.1444 N

Forces on a Skier038 (part 1 of 2) 10.0 points

A skier of mass 109 kg comes down a slope ofconstant angle 25◦ with the horizontal.

The acceleration of gravity is 9.8 m/s2 .What is the force on the skier parallel to

the slope?Correct answer: 451.441 N.

Explanation:

a

N W θ

The weight of the skier acting verticallydownward can be split into two components.The sine component acts along the slope.

Wparallel = W sin θ

= mg sin θ

= (109 kg) (9.8 m/s2) sin 25◦

= 451.441 N.

039 (part 2 of 2) 10.0 pointsWhat force normal to the slope is exerted bythe skis?Correct answer: 968.118 N.

Explanation:The cosine component acts perpendicular

to the slope.

Wnormal = W cos θ

= mg cos θ

= (109 kg) (9.8 m/s2) cos 25◦

= 968.118 N.

Tipler PSE5 04 61040 10.0 points

A 65 kg student weighs himself by standingon a scale mounted on a skateboard that isrolling down an incline, as shown. Assumethere is no friction so that the force exerted


by the incline on the skateboard is normal tothe incline.


28◦

What is the reading on the scale if the angleof the slope is 28◦?Correct answer: 563.012 N.

Explanation:

Let : θ = 28◦ ,

m = 65 kg , and

g = 9.81 m/s2 .

Consider the forces acting on the scale:

y x

N

W

28◦

Let the positive x-axis be parallel to anddown the plane, and the positive y-axis in thedirection of the normal force.

Applying∑

Fy = may = 0 ,

N −W cos θ = 0

N = W cos θ

= mg cos θ

= (65 kg)(

9.81 m/s2)

cos 28◦

= 563.012 N .

Up a Slope

041 (part 1 of 2) 10.0 pointsTwo blocks are connected by an extensiblemassless cord on an inclined plane as shownin the figure below.


41 kg

T 68 kg

F

3.8m/s

2

20◦

What is the force F pulling both theblocks?Correct answer: 779.546 N.

Explanation:

Let : θ = 20◦ ,

M1 = 41 kg ,

M2 = 68 kg ,

g = 9.8 m/s2 , and

a = 3.8 m/s2 .

F

N2

W2‖

W2⊥W2

N1

W1⊥

W1‖

W1

Tθ

The resulting acceleration is due to the ap-plied force acting against the component ofthe weights along the inclined surface. Thissystem is equivalent to a combined block ofmass M1 + M2 being accelerated up the slope


by a force F , and can be described by theequation

F − (M1 + M2) g sin θ = (M1 + M2) a . (1)

Solving for F we obtain

F = (M1 + M2)(a + g sin θ)

= (41 kg + 68 kg)

×(

3.8 m/s2 + 9.8 m/s2)

sin 20◦

= 779.546 N .

042 (part 2 of 2) 10.0 pointsWhat is the tension in the cord pulling thelower block?Correct answer: 293.224 N.

Explanation:Using Eq. (1) from Part 1 for the lower

block, we have

T − M1 g sin θ = M1 a .

Solving for T we obtain

T = M1 (a + g sin θ)

= (41 kg)

×(

3.8 m/s2 + 9.8 m/s2)

sin 20◦

= 293.224 N .

Pulleys 01043 (part 1 of 3) 10.0 points

Assume all pulleys are massless and fric-tionless, and the systems are in equilibrium.

The acceleration due to gravity is 9.8 m/s2.

44 N

T

Find the tension T .Correct answer: 22 N.

Explanation:

Let : W = 44 N .

A weight hanging at the end of a stringdefines the tension in that string, which is thesame throughout the string.

When a pulley system is in equilibrium, thesum of the upward forces will equal the sumof the downward forces.

2 T = W

T =W

2=

44 N

2= 22 N .

044 (part 2 of 3) 10.0 pointsThe suspended weight is 43 N.

43 N

T


Explanation:

Let : W2 = 43 N .

The weight W defines the tension:

T = W2 = 43 N .

045 (part 3 of 3) 10.0 pointsThe suspended weight is 35 N.

35 N

T

What is the tension T?Correct answer: 70 N.

Explanation:

Let : W3 = 35 N .

The pulley is in equilibrium, so

T = 2W3 = 2 (35 N) = 70 N .


Pulleys 06046 10.0 points

In the pulley system, all pulleys are mass-less and frictionless.


40 N

T


Explanation:

Let : W = 40 N .

W

1

2

3

T1 T1

T1 T1

T

T2 T3

At pulleys 1 and 3,

2 T1 = T2 and 2 T1 = T3 .

At the weight W,

W = T2 + T3 = 4 T1

T1 =W

4.

At pulley 2,

T = 2 T1 =W

2=

40 N

2= 20 N .

Pulleys 12047 10.0 points

The system is in equilibrium and the pulleysare weightless and frictionless. The suspendedweight is 19 N.


19 N

T

Find the tension T .Correct answer: 4.75 N.

Explanation:

Let : W = 19 N .

W

2

1

T1

T

T1

T

T

Consider equilibrium at pulley 1, we have

2 T1 = W

T1 =W

2.

At pulley 2,

2 T = T1 so

T =T1

2

=W

4

=19 N

4

= 4.75 N .

AP-C4-MJ

Documents

Transcript of AP-C4-MJ