Week3

7/18/2019 Week3

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Week 3 homework

IMPORTANT NOTE ABOUT WEBASSIGN:In the WebAssign versions of these problems, various details have been changed, so that

the answers will come out differently. The method to find the solution is the same, butyou will need to repeat part of the calculation to find out what your answer should have

been.

WebAssign Problem 1: An arrow, starting from rest, leaves the bow with a speed of

25.0 ms. If the average force e!erted on the arrow by the bow were doubled, all elseremaining the same, with what speed would the arrow leave the bow"

REASONING AND SOLUTION #rom $%uation 2.&,

2 2

0

2v v ax= +

'ince the arrow starts from rest, v0 ( 0 ms. In both cases x is the same so

2

) ) ) ) )

2

2 2 2 22

2 or

2

v a x a v a

a x a v av= = =

'ince F ( ma, it follows that a ( F m. The mass of the arrow is unchanged, and

) )

2 2

v F

v F =

or

( )2 )

2 ) )

) )

225.0 ms 2 *5.+ ms

F F v v v

F F = = = =

WebAssign Problem : nly two forces act on an ob-ect mass ( *.00 /g, as in the

drawing. #ind the magnitude and direction relative to the x a!is of the acceleration of

the ob-ect.

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REASONING According to 1ewtons second law m=∑ ! " , the acceleration of the

ob-ect is given by m= ∑" ! , where ∑ ! is the net force that acts on the ob-ect. We

must first find the net force that acts on the ob-ect, and then determine the acceleration

using 1ewtons second law.

SOLUTION The following table gives the x and y components of the two forcesthat act on the ob-ect. The third row of that table gives the components of the net force.

Force x-Component y-Component

!)

+0.0 1 0 1

!2

30.0 1 cos +5.04 ( +2.+ 1 30.0 1 sin +5.04 ( +2.+ 1

) 2= +∑ ! ! ! 2.+ 1 +2.+ 1

The magnitude of ∑ ! is given by the 6ythagorean theorem as

2 22.+ 1 +2.+ &2.7 1 F Σ = + =

The angle θ that ∑ ! ma/es with the 8 x a!is is

) +2.+ 1tan 27.2

2.+ 1

θ − = = °

2.+ 1

!Σθ +2.+ 1

According to 1ewtons second law, the magnitude of the acceleration of theob-ect is

2&2.7 1*0.& ms

*.00 /g

F a

m

∑= = =

'ince 1ewtons second law is a vector e%uation, we /now that the direction of theright hand side must be e%ual to the direction of the left hand side. In other words, the

direction of the acceleration a is the same as the direction of the net force ∑ ! .

Therefore, the direction of the acceleration of the ob-ect is 27.2 above the 8 a!is x° .

WebAssign Problem 3: The drawing shows three particles far away from any otherob-ects and located on a straight line. The masses of these particles are ,

, and . #ind the magnitude and direction of the net gravitationalforce acting on a particle A, b particle 9, and c particle :.

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REASONING $ach particle e!periences two gravitational forces, one due to each of the

remaining particles. To get the net gravitational force, we must add the twocontributions, ta/ing into account the directions. The magnitude of the gravitational

force that any one particle e!erts on another is given by 1ewton;s law of gravitation as2

) 2 F Gm m r = . Thus, for particle A, we need to apply this law to its interaction with

particle 9 and with particle :. #or particle 9, we need to apply the law to its interaction

with particle A and with particle :. <astly, for particle :, we must apply the law to its

interaction with particle A and with particle 9. In considering the directions, weremember that the gravitational force between two particles is always a force of

attraction.

SOLUTION We begin by calculating the magnitude of the gravitational force for

each pair of particles=

( ) ( ) ( )

( )

( ) ( ) ( )

( )

( ) ( ) ( )

( )

>)) 2 2

>5

2 2

>)) 2 2

>5

2 2

>)) 2 2

>32 2

3.37 )0 1 m /g *3* /g 5)7 /g5.007 )0 1

0.500 m

3.37 )0 1 m /g 5)7 /g )5+ /g.+&7 )0 1

0.500 m

3.37 )0 1 m /g *3* /g )5+ /g3.32& )0 1

0.500 m

A B AB

B C BC

A C AC

Gm m F

r

Gm m F

r

Gm m F r

×= = = ×

×= = = ×

×

= = = ×

⋅

⋅

⋅

In using these magnitudes we ta/e the direction to the right as positive.

a. 9oth particles 9 and : attract particle A to the right, the net force being

>5 >3 >55.007 )0 1 3.32& )0 1 5.37 )0 1, right A AB AC

F F F = + = × + × = ×

b. 6article : attracts particle 9 to the right, while particle A attracts particle

9 to the left, the net force being

>5 >5 >5 > .+&7 )0 1 > 5.007 )0 1 *.+& )0 1, right B BC AB

F F F = = × × = ×

c. 9oth particles A and 9 attract particle : to the left, the net force being

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>3 >5 >53.32& )0 1 .+&7 )0 1 &.)3 )0 1, leftC AC BC

F F F = + = × + × = ×

WebAssign Problem #: A bloc/ whose weight is +5.0 1 rests on a hori?ontal table. A

hori?ontal force of *3.0 1 is applied to the bloc/. The coefficients of static and /ineticfriction are 0.350 and 0.+20, respectively. Will the bloc/ move under the influence of the

force, and, if so, what will be the bloc/;s acceleration" $!plain your reasoning.

REASONING AND SOLUTION The bloc/ will move only if the applied force is greater

than the ma!imum static frictional force acting on the bloc/. That is, if

F > µsF N = µsmg = (0.650)(45.0 N) = 29.2 N

The applied force is given to be F ( *3.0 1 which is greater than the ma!imum

static frictional force, so the bloc/ will move .

The bloc/s acceleration is found from 1ewtons second law.

2 > >

*.72 msk k F f F mg F

am m m

µ Σ= = = =

WebAssign Problem $: 6art a of the drawing shows a buc/et of water suspended fromthe pulley of a well@ the tension in the rope is &2.0 1. 6art b shows the same buc/et of

water being pulled up from the well at a constant velocity. What is the tension in the rope

in part b"

REASONING At first glance there seems to be very little information given. owever,it is enough. In part a of the drawing the buc/et is hanging stationary and, therefore,is in e%uilibrium. The forces acting on it are its weight and the two tension forces

from the rope. There are two tension forces from the rope, because the rope is

attached to the buc/et handle at two places. These three forces must balance, whichwill allow us to determine the weight of the buc/et. In part b of the drawing, the

buc/et is again in e%uilibrium, since it is traveling at a constant velocity and,

therefore, has no acceleration. The forces acting on the buc/et now are its weight

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and a single tension force from the rope, and they again must balance. In part b,

there is only a single tension force, because the rope is attached to the buc/et handle

only at one place. This will allow us to determine the tension in part b, since theweight is /nown.

SOLUTION <et W be the weight of the buc/et, and let T be the tension in the ropeas the buc/et is being pulled up at a constant velocity. The freeBbody diagrams for

the buc/et in parts a and b of the drawing are as follows=

'ince the buc/et in part a is in e%uilibrium, the net force acting on it is ?ero. Ta/ingupward to be the positive direction, we have

&2.0 1 &2.0 1 0 or )+ 1 F W W Σ = + − = =

'imilarly, in part b we have

0 or )+ 1 F T W T W Σ = − = = =

WebAssign Problem %: In the drawing, the weight of the bloc/ on the table is +22 1 and

that of the hanging bloc/ is )5 1. Ignoring all frictional effects and assuming the pulley

to be massless, find a the acceleration of the two bloc/s and b the tension in the cord.

37. REASONING AND

SOLUTION 1ewtons second law

applied to ob-ect ) +22 1 gives

#reeBbody diagram for part a

W

&2.0 1 &2.0 1

#reeBbody diagram for

part b

W

T

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T ( m)a

)

'imilarly, for ob-ect 2 )5

1

T > m2 g ( m2a2

If the string is not to brea/or go slac/, both ob-ects must have

accelerations of the same

magnitude.

T

W2

Object 2

W1

T

N1

Object 1

Then a) ( a and a

2 ( > a. The above e%uations become

T ( m)a )

T > m2 g ( > m

2a 2

a. 'ubstituting $%uation ) into $%uation 2 and solving for a yields

2

) 2

m g a

m m=

+

The masses of ob-ects ) and 2 are

( ) ( )

( ) ( )

2

) )

2

2 2

+22 1 &.0 ms +*.) /g

)5 1 &.0 ms ).& /g

m W g

m W g

= = =

= = =

The acceleration is

( ) ( )2

22

) 2

).& /g &.0 ms2.&& ms

+*.) /g ).& /g

m g a

m m= = =

+ +

b. Csing this value in $%uation ) gives

( ) ( )2

)+*.) /g 2.&& ms )2& 1T m a= = =

WebAssign Problem &: In the drawing, the rope and the pulleys are massless, and thereis no friction. #ind a the tension in the rope and b the acceleration of the )0.0B/g

bloc/. (Hint: The larger ma move t!ice a far a the maller ma"#

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REASONING AND SOLUTION a. 1ewtons second law for bloc/ ) )0.0 /g is

T ( m)a

)

9loc/ 2 *.00 /g has two ropes attached each carrying a tension T . Also, bloc/ 2only travels half the distance that bloc/ ) travels in the same amount of time so its

acceleration is only half of bloc/ )s acceleration. 1ewtons second law for bloc/ 2 is

then

)

2 222T m g m a− = −

2

'olving $%uation ) for a, substituting into $%uation 2, and rearranging gives

( )

)22

)

2 )+

)*.7 1)

m g T

m m= =

+

b. Csing this result in $%uation ) yields

2

)

)*.7 1).*7 ms

)0.0 /g

T a

m= = =

WebAssign Problem ': As part a of the drawing shows, two bloc/s are connected by arope that passes over a set of pulleys. ne bloc/ has a weight of +)2 1, and the other has

a weight of &0 1. The rope and the pulleys are massless and there is no friction. a

What is the acceleration of the lighter bloc/" b 'uppose that the heavier bloc/ isremoved, and a downward force of &0 1 is provided by someone pulling on the rope, as

part b of the drawing shows. #ind the acceleration of the remaining bloc/. c $!plain

why the answers in a and b are different.

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REASONING AND SOLUTION

a. The rope e!erts a tension, T , acting upward on each bloc/. Applying 1ewtonssecond law to the lighter bloc/ bloc/ ) gives

T – m1g = m1a

'imilarly, for the heavier bloc/ bloc/ 2

T – m2g = – m2a

'ubtracting the second e%uation from the first and rearranging yields

22 )

2 )

> *.3 ms

m ma g

m m

= = +

b. The tension in the rope is now &0 1 since the tension is the reaction to the

applied force e!erted by the hand. 1ewtons second law applied to the bloc/ is

T – m1g = m1a

'olving for a gives

( ) 2 2

)

&0 1 > > &.0 ms )). ms

+2.0 /g

T a g

m= = =

c. In the first case, the inertia of 9T bloc/s affects the acceleration whereas,in the second case, only the lighter bloc/s inertia remains.

Pr"()i(e (on(e*)+"l *roblems:

1, Why do you lunge forward when your car suddenly comes to a halt" Why are you pressed bac/ward against the seat when your car rapidly accelerates" In your

e!planation, refer to the most appropriate one of 1ewton;s three laws of motion.

REASONING AND SOLUTION When the car comes to a sudden halt, the upper

part of the body continues forward as predicted by 1ewtons first law if the force

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e!erted by the lower bac/ muscles is not great enough to give the upper body the

same deceleration as the car. The lower portion of the body is held in place by the

force of friction e!erted by the car seat and the floor.When the car rapidly accelerates, the upper part of the body tries to remain at

a constant velocity again as predicted by 1ewtons first law. If the force provided

by the lower bac/ muscles is not great enough to give the upper body the sameacceleration as the car, the upper body appears to be pressed bac/ward against the

seat as the car moves forward.

1$, A )0B/g suitcase is placed on a scale that is in an elevator. Is the elevator

accelerating up or down when the scale reads a 75 1 and b )20 1" Dustify your

answers.

REASONING AND SOLUTION If the elevator were at rest, or moving with a constant

velocity, the scale would read the true weight of mg ( & 1. When the elevator is

accelerating, the scale reading will differ from & 1 and will display the apparent

weight, F 1, which is given by $%uation +.3= F

1 = mg + ma

where a, theacceleration of the elevator, is positive when the elevator accelerates upward and

negative when the elevator accelerates downward.

a. When the apparent weight is F 1 ( 75 1, the apparent weight is less than the

true weight mg ( & 1 so a must be negative. The elevator is accelerating

downward.

b. When the apparent weight is F 1 ( )20 1, the apparent weight is greater than

the true weight mg ( & 1 so a must be positive. The elevator is accelerating

upward.

1', A person has a choice of either pushing or pulling a sled at a constant velocity, asthe drawing illustrates. #riction is present. If the angle % is the same in both

cases, does it re%uire less force to push or to pull" Account for your answer.

REASONING AND SOLUTION 'ince the sled moves with constant velocity, the forceof /inetic friction is present. The magnitude of this force is given by $

/

F 1

, where $/

is the coefficient of /inetic friction and F 1

is the magnitude of the normal force that

acts on the sled. #urthermore, the hori?ontal component of the applied force must be

e%ual in magnitude to the force of /inetic friction, since there is no acceleration.

When the person pulls on the sled, the vertical component of the pulling forcetends to decrease the magnitude of the normal force relative to that when the sled is

not being pulled or pushed. n the other hand, when the person pushes on the sled,

the vertical component of the pushing force tends to increase the normal force

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relative to that when the sled is not being pulled or pushed. Therefore, when the sled

is pulled, the magnitude of the force of /inetic friction, and therefore the magnitude

of the applied force, is less than when the sled is pushed.

, A stone is thrown from the top of a cliff. As the stone falls, is it in e%uilibrium"

$!plain, ignoring air resistance.

REASONING AND SOLUTION An ob-ect is in e%uilibrium when its acceleration is

?ero. When a stone is thrown from the top of a cliff, its acceleration is theacceleration due to gravity@ therefore, the stone is not in e%uilibrium.

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