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Physics 130 General Physics Fall 2012

Midterm Exam 2

October 30, 2012

Name:

Instructions

1. This examination is closed book and closed notes. All your belongingsexcept a pen or pencil and a calculator should be put away and yourbookbag should be placed on the floor.

2. You will find one page of useful formulae on the last page of the exam.

3. Please show all your work in the space provided on each page. If youneed more space, feel free to use the back side of each page.

4. Academic dishonesty (i.e., copying or cheating in any way) willresult in a zero for the exam, and may cause you to fail theclass.

In order to receive maximum credit,

each solution should have:

1. A labeled picture or diagram, if appropriate.2. A list of given variables.3. A list of the unknown quantities (i.e., what you are being asked to find).4. A force-interaction diagram, if appropriate.5. One or more free-body diagrams, as appropriate, with labeled 1D or 2D

coordinate axes.6. Algebraic expression for the net force along each dimension, as appro-

priate.7. An algebraic solution of the unknown variables in terms of the

known variables.8. A final numerical solution, including units, with a box around it.9. An answer to additional questions posed in the problem, if any.

1


1. A heavy box is in the back of a truck. The truck is accelerating to the right. Draw amotion diagram, a force-interaction diagram, and a free-body diagram for the box.

Solution:

You can see from the motion diagram that the box accelerates to the right along withthe truck. According to Newton’s second law, ~F “ m~a, there must be a force to theright acting on the box. This is friction, but not kinetic friction, because the boxis not sliding against the truck. Instead, it is static friction, the force that preventsslipping. Were it not for static friction, the box would slip off the back of the truck.Static friction acts in the direction needed to prevent slipping. In this case, frictionmust act in the forward (toward the right) direction.

2


2. A bag of groceries is on the seat of your car as you stop at a stop light. The bag doesnot slide. Draw a motion diagram, a force-interaction diagram, and a free-body diagramfor the bag.

Solution:

You can see from the motion diagram that the bag accelerates to the left along withthe car as the car slows down. According to Newton’s second law, ~F “ m~a, theremust be a force to the left acting on the bag. This force must be static friction, theforce that prevents slipping, because the bag is not sliding across the seat. Were itnot for static friction, the bag would slide off the seat as the car stops. Static frictionacts in the direction needed to prevent slipping, and in this case, friction must actin the backward (toward the left) direction.

3


3. A football coach sits on a sled while two of his players build their strength by draggingthe sled across the field with ropes. Assume that the two players pull with equal strength.The friction force on the sled is 1000 N and the angle between the two ropes is 20˝. Howhard must each player pull to drag the coach at a steady 2.0 m{s?

Solution:

The pictorial representation and free-body diagram are shown below:

This is a 1D dynamics problem. The relevant forces are the force of kinetic frictionon the sled and the tension of each rope. Although there is also gravity, we are giventhe friction force, and therefore we do not need to know the weight of the sled.

Since the sled is not accelerating, it is in dynamic equilibrium along both axes, andNewton’s first law applies:

pFnetqy “ÿ

Fy “ T1y ` T2y ` fky “ 0 N (1)

pFnetqx “ÿ

Fx “ T1x ` T2x ` fkx “ 0 N (2)

Examining the free-body diagram, along the y-axis we have

T1y “ `T1 sinpθ{2q (3)

T2y “ ´T2 sinpθ{2q (4)

fky “ 0. (5)

Substituting into equation (1), we get

T1 sinpθ{2q ´ T2 sinpθ{2q “ 0 (6)

ñ T1 “ T2 “ T. (7)

4


In other words, the tension force of each rope is the same.

From the free-body diagram along the x-axis we have

T1x “ T1 cospθ{2q “ T cospθ{2q (8)

T2x “ T2 cospθ{2q “ T cospθ{2q. (9)

Substituting into equation (2) and solving for T we get

T cospθ{2q ` T cospθ{2q ´ fkx “ 0 (10)

2T cospθ{2q “ fkx (11)

ñ T “ fkx

2 cospθ{2q (12)

T “ 1000 N

2 ˆ cosp20˝{2q (13)

“ 508 N « 510 N. (14)

Note that in the last line we simplified our result to have just two significant figures.

5


4. You’re driving along at 25 m{s with your aunt’s valuable antiques in the back of yourpickup truck when suddenly you see a giant hole in the road 55 m ahead of you. For-tunately, your foot is right beside the brake and your reaction time is zero! Will theantiques be as fortunate? Assume that the coefficient of kinetic friction for rubber (i.e.,a tire) on concrete is 0.80.

(a) Can you stop the truck before it falls in the hole?

(b) If your answer to part (a) is yes, can you stop without the antiques sliding andbeing damaged? Their coefficients of friction are µs “ 0.60 and µk “ 0.30.

Solution:

To solve this problem we will treat the antiques (mass “ m) in the back of the pickup(mass “ M) both as particles. The antiques touch the truck’s steel bed, so only thesteel bed can exert contact forces on the antiques. The pickup-antiques system willalso be treated as a particle, and the contact force on this particle will be due to theroad.

The pictorial representation, motion diagram, force-interaction diagram, and free-body diagram for the combined pickup-antiques system are shown below:

6


The free-body diagram of just the box of antiques is:

(a) Our strategy for solving this problem is to find the smallest coefficient of frictionthat will allow the truck to stop in 55 m, then compare that to the known coefficientsfor rubber on concrete. For the pickup-antiques system, with mass m`M , Newton’ssecond law is

pFnetqx “ÿ

Fx “ ´f “ pm ` Mqax “ pm ` Mqa (1)

pFnetqy “ÿ

Fy “ N ´ pFGqPA “ 0 (2)

“ N ´ pm ` Mqg “ 0 (3)

ñ N “ pm ` Mqg. (4)

The model of static friction is f “ µN , where µ is the coefficient of friction betweenthe tires and the road. Substituting into the equations above, we get

´f “ ´µN “ ´µpm ` Mqg “ pm ` Mqa (5)

ñ a “ ´µg. (6)

The mass of the pickup-antiques system drops out! Next, from the constant-accelerationkinematics equation with v21 “ 0 and x0 “ 0, we have

v21 “ v20 ` 2apx1 ` x0q (7)

ñ a “ ´ v202x1

(8)

ñ µmin “ v202gx1

(9)

“ p25 m{sq22 ˆ p9.8 m{s2q ˆ p55 mq (10)

“ 0.58. (11)

7


Since this is smaller than the coefficient of static friction of rubber on concrete, 0.80,the truck can stop.

(b) The analysis of the pickup-antiques system applies to the antiques as well, andgives the same value of µmin “ 0.58. This value is smaller than the given coefficientof static friction (µs “ 0.60) between the antiques and the truck bed. Therefore, theantiques will not slide as the truck stops over a distance of 55 m.

8


5. You and your friend Peter are putting new shingles on a roof pitched at 25˝. You’resitting on the very top of the roof when Peter, who is at the edge of the roof directlybelow you, 5.0 m away, asks you for the box of nails. Rather than carry the 2.5 kg boxof nails down to Peter, you decide to give the box a push and have it slide down to him.If the coefficient of kinetic friction between the box and the roof is 0.55, with what speedshould you push the box to have it gently come to rest right at the edge of the roof?

Solution:

The pictorial representation and free-body diagram are shown below:

This is a 1D dynamics problem. The relevant forces are gravity, FG, the normalforce, n, and the kinetic friction force, fk. Note that we do not include the initialforce that was applied to the box of nails to get it moving, but we do include thefact that the box has some initial velocity.

The most natural coordinate system is one that is rotated by 25˝ and thereforealigned with the roof. The interaction and free-body diagrams are shown above.The net force along the x- and y-axis is

pFnetqx “ÿ

Fx “ FG sin 25˝ ´ fk “ ma (1)

pFnetqy “ÿ

Fy “ n ´ FG cos 25˝ “ 0 (2)

ñ n “ FG cos 25˝ (3)

In the second line we used the fact that the shingles are not leaping off the roof toset the acceleration in the y-direction equal to zero. The magnitude of the force ofgravity is FG “ mg, and the magnitude of the kinetic force of friction is

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fk “ µkn (4)

“ µkFG cos 25˝ (5)

“ µkmg cos 25˝. (6)

Substituting equation (6) into equation (1) and solving for the acceleration a we get

mg sin 25˝ ´ µkmg cos 25˝ “ ma (7)

ñ a “ psin 25˝ ´ µk cos 25˝qg (8)

“ psin 25˝ ´ 0.55 ˆ cos 25˝q ˆ 9.8 m{s2 (9)

“ ´0.743 m{s2 (10)

where the minus sign indicates the acceleration is directed up the incline. To find theinitial speed, vi, necessary to have the box come to rest (i.e., vf “ 0) after ∆x “ 5.0 mcan be found from the kinematic equation linking velocity and acceleration:

v2f “ v2i ` 2a∆x (11)

ñ vi “?

´2a∆x (12)

“b

´2p´0.743 m{s2qp5.0 mq (13)

“ 2.7 m{s. (14)

10


6. A 1.0 kg wood block is pressed against a vertical wood wall by the 12 N force shown inthe figure below. If the block is initially at rest, will it move upward, move downward, orstay at rest? Assume that the coefficient of static friction for wood on wood is µs “ 0.5.

Solution:

The block is initially at rest, so initially the friction force is static friction. If theforce of the push is too strong, the box will begin to move up the wall. If it is tooweak, the box will begin to slide down the wall. And if the pushing force is withinthe proper range, the box will remain stuck in place.

From the free-body diagram, the block is in static equilibrium along the x-axis.Therefore, according to Newton’s first law the net force is

pFnetqx “ÿ

Fx “ n ´ Fpush cos θ “ 0 (1)

ñ n “ Fpush cos θ. (2)

From our model for friction, the maximum static friction force is

pfsqmax “ µsn “ µsFpush cos θ (3)

“ 0.5 ˆ p12 Nq ˆ cosp30˝q (4)

“ 5.2 N. (5)

To determine if the block stays in place or moves up or down, we need to determineif the net force along the y-axis, pFnetqy, is greater or smaller than pfsqmax. The netforce excluding the force of friction is

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pFnetqy “ÿ

Fy “ Fpush sin θ ´ FG (6)

“ Fpush sin θ ´ mg (7)

“ p12 Nq ˆ sinp30˝q ´ p1.0 kgq ˆ p9.8 m{s2q (8)

“ ´3.8 N. (9)

In other words, a static friction force fs “ `3.8 N would prevent the block frommoving. Since this force is smaller than pfsqmax the box stays in place and does notmove.

12


7. The 1.0 kg block in the figure below is tied to the wall with a rope. It sits on top of the2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. Thecoefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block isµk “ 0.40.

(a) What is the tension in the rope holding the 1.0 kg block to the wall?

(b) What is the acceleration of the 2.0 kg block?

Solution:

The free-body diagram for the problem is shown below:

To solve this problem we need to use both Newton’s third and second laws. The sep-arate free-body diagrams for the two blocks show that there are two action/reaction

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pairs. Notice how the top block (block 1) both pushes down on the bottom block

(block 2) with force ~n1

1, and exerts a retarding friction force ~f2,top on the top surfaceof block 2.

(a) Block 1 is in static equilibrium (a1 “ 0 m{s2), but block 2 is accelerating to theright. Newton’s second law for block 1 is

pFnet on 1qx “ f1 ´ Trope “ 0 ñ Trope “ f1 (1)

pFnet on 1qy “ n1 ´ m1g “ 0 ñ n1 “ m1g. (2)

Although block 1 is stationary, there is a kinetic force of friction because there ismotion between block 1 and block 2. The friction model means

f1 “ µkn1 “ µkm1g. (3)

Substituting this result into equation (1) we get the tension of the rope:

Trope “ f1 “ µkm1g (4)

“ p0.40q ˆ p1.0 kgq ˆ p9.8 m{s2q (5)

“ 3.9 N. (6)

(b) Newton’s second law for block 2 is

ax ” a “ pFnet on 2qxm2

“ Tpull ´ f2 top ´ f2 bot

m2

(7)

ay “ pFnet on 2qym2

“ n2 ´ n1

1 ´ m2g

m2

“ 0 (8)

Forces ~n1 and ~n1

1 are an action/reaction pair, so ~n1

1 “ ~n1 “ m1g. Substituting intoequation (8) gives

n2 “ pm1 ` m2qg. (9)

This result is not surprising because the combined weight of both objects pressesdown on the surface. The kinetic friction on the bottom surface of block 2 is then

f2 bot “ µkn2 “ µkpm1 ` m2qg. (10)

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Next, we recognize that the forces ~f1 and ~f2 top are an action/reaction pair, so

f2 top “ f1 “ µkm1g. (11)

Finally inserting these friction results into equation (7) gives

a “ Tpull ´ µkm1g ´ µkpm1 ` m2qgm2

(12)

“ 20 N ´ p0.40qp1.0 kgqp9.8 m{s2q ´ p0.40qp1.0 kg ` 2.0 kgqp9.8 m{s2q2.0 kg

(13)

“ 2.2 m{s2. (14)

15


8. The lower block in the figure below is pulled on by a rope with a tension force of 20 N.The coefficient of kinetic friction between the lower block and the surface is 0.30. Thecoefficient of kinetic friction between the lower block and the upper block is also 0.30.What is the acceleration of the 2.0 kg block?

Solution:

The pictorial representation and free-body diagrams are shown below:

The blocks accelerate with the same magnitude but in opposite directions. Thus theacceleration constraint is a2 “ a “ ´a1, where a will have a positive value becauseof how we have defined the `x direction. There are two real action/reaction pairs.The two tension forces will act as if they are action/reaction pairs because we areassuming a massless rope and a frictionless pulley.

Make sure you understand why the friction forces point in the directions shown inthe free-body diagrams, especially force ~f 1

1 exerted on the bottom block (block 2) bythe top block (block 1).

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We have quite a few pieces of information to include. First, Newton’s second lawapplied to block 1 gives

pFnet on 1qx “ f1 ´ T1 (1)

“ µkn1 ´ T1 “ m1a1 “ ´m1a (2)

pFnet on 1qy “ n1 ´ m1g “ 0 (3)

ñ n1 “ m1g. (4)

For block 2 we have

pFnet on 2qx “ Tpull ´ f 1

1 ´ f2 ´ T2 (5)

“ Tpull ´ f 1

1 ´ µkn2 ´ T2 “ m2a2 “ m2a (6)

pFnet on 2qy “ n2 ´ n1

1 ´ m2g “ 0 (7)

ñ n2 “ n1

1 ` m2g. (8)

Note that to simplify the two x-equations above we have already used the kineticfriction model. Next, from Newton’s third law we have three additional constraints:

n1

1 “ n1 “ m1g (9)

f 1

1 “ f1 “ µkn1 “ µkm1g (10)

T1 “ T2 “ T. (11)

Knowing n1

1 we can now use the y-equation for block 2 to find n2. Substitutingall these pieces into the two x-equations, we end up with two equations with twounknowns:

µkm1g ´ T “ ´m1a (12)

Tpull ´ T ´ µkm1g ´ µkpm1 ` m2qg “ m2a. (13)

Subtracting the first equation from the second we get

Tpull ´ T ´ µkm1g ´ µkpm1 ` m2qg ´ µkm1g ` T “ m2a ` m1a (14)

Tpull ´ 3µkm1g ´ µkm2g “ pm2 ` m1qa (15)

Tpull ´ µkp3m1 ` m2qg “ pm2 ` m1qa. (16)

And finally solving for a we get

ñ a “ Tpull ´ µkp3m1 ` m2qgm1 ` m2

(17)

“ 20 N ´ p0.30qp3 ˆ 1.0 kg ` 2.0 kgqp9.8 m{s2q1.0 kg ` 2.0 kg

(18)

“ 1.8 m{s2. (19)

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9. The coefficient of kinetic friction between the 2.0 kg block and the table in the figurebelow is 0.30. What is the acceleration of the 2.0 kg block?

Solution:


Newton’s second law for m1 and m3 gives

T1 ´ pFGq1 “ T1 ´ m1g “ m1a1 (1)

T2 ´ pFGq3 “ T2 ´ m3g “ m3a3. (2)

For m2 Newton’s second law gives

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ÿ

pFon m2qy “ n2 ´ pFGq2 “ n2 ´ m2g “ 0 (3)

ñ n2 “ m2g (4)ÿ

pFon m2qx “ T2 ´ fk2 ´ T1 (5)

“ T2 ´ µkn2 ´ T1 (6)

“ T2 ´ µkm2g ´ T1 “ m2a2. (7)

Since all three blocks move together, the acceleration constraint gives

a1 “ a2 “ ´a3 ” a. (8)

Therefore, the equations for the three masses are:

T1 ´ m1g “ m1a (9)

T2 ´ µkm2g ´ T1 “ m2a (10)

T2 ´ m3g “ ´m3a. (11)

Subtracting the third equation from the sum of the first two equations yields:

pT1 ´ m1gq ` pT2 ´ µkm2g ´ T1q ´ pT2 ´ m3gq “ m1a ` m2a ` m3a (12)

p´m1 ´ µkm2 ` m3qg “ apm1 ` m2 ` m3q (13)

ñ a “ ´m1 ´ µkm2 ` m3qm1 ` m2 ` m3

g (14)

“ ´1.0 kg ´ 0.30 ˆ 2.0 kg ` 3.0 kg

1.0 kg ` 2.0 kg ` 3.0 kgˆ p9.8 m{s2q (15)

“ 2.3 m{s2. (16)

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10. The 1.0 kg physics book in the figure below is connected by a string to a 500 g coffeecup. The book is given a push up the slope and released with a speed of 3.0 m{s. Thecoefficients of friction are µs “ 0.50 and µk “ 0.20.

(a) How far does the book slide?

(b) At the highest point, does the book stick to the slope, or does it slide back down?

Solution:


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To solve this problem we use the particle model for the book (B) and the coffee cup(C), the models of kinetic and static friction, and the constant-acceleration kinematicequations.

(a) To find the distance x1 the book slides we must find its acceleration. Newton’ssecond law applied to the book gives

ÿ

pFon Bqy “ nB ´ pFGqB cos θ “ 0 (1)

ñ nB “ mBg cos θ (2)ÿ

pFon Bqx “ ´T ´ fk ´ pFGqB sin θ (3)

“ ´T ´ µknB ´ mBg sin θ (4)

“ ´T ´ µkmBg cos θ ´ mBg sin θ “ mBaB. (5)

Similarly, for the coffee cup we have

ÿ

pFon Cqy “ T ´ pFGqC “ T ´ mCg “ mCaC . (6)

Equations (5) and (6) can be rewritten using the acceleration constrain aC “ aB “ a

as

´T ´ µkmBg cos θ ´ mBg sin θ “ mBa (7)

T ´ mCg “ mCa. (8)

Adding these two equations and solving for the acceleration gives

´µkmBg cos θ ´ mBg sin θ ´ mCg “ pmB ` mCqa (9)

ñ a “ ´„

mBpµk cos θ ` sin θq ` mC

mB ` mC

g (10)

“ ´„p1.0 kgq ˆ p0.20 cos 20˝ ` sin 20˝q ` 0.5 kg

1.0 kg ` 0.5 kg

ˆ p9.8 m{s2q (11)

“ ´6.73 m{s2. (12)

Finally, to solve for the distance x1 we use the following kinematic equation withv1x “ 0, v0x “ 3.0 m{s, and x0 “ 0:

21


v21x “ v20x ` 2apx1 ´ x0q (13)

0 “ v20x ` 2ax1 (14)

ñ x1 “ ´v20x2a

(15)

“ ´p3.0 m{sq22 ˆ p´6.73 m{s2q (16)

“ 0.67 m. (17)

(b) In order to figure out if the book sticks to the slope or slides back down we haveto determine if the static friction force needed to keep the book in place, fs is largeror smaller than the maximum static friction force

pfsqmax “ µsnB “ µsmBgcosθ (18)

“ p0.50q ˆ p9.8 m{s2q ˆ cos 20˝ (19)

“ 4.60 N. (20)

When the cup is at rest, the string tension is T “ mCg. In this case, Newton’s firstlaw for the book becomes

ÿ

pFon Bqx “ fs ´ T ´ mBg sin θ (21)

“ fs ´ mCg ´ mBg sin θ “ 0 (22)

ñ fs “ pmC ` mB sin θqg (23)

“ p0.5 kg ` 1.0 kg sin 20˝q ˆ p9.8 m{s2q (24)

“ 8.25 N. (25)

Because fs ą pfsqmax, the book slides back down.

22


11. A house painter uses the chair and pully arrangement shown below to lift himself up theside of a house. The painter’s mass is 70 kg and the chair’s mass is 10 kg. With whatforce must he pull down on the rope in order to accelerate upward at 0.20 m{s2?

Solution:

The free-body diagram for the problem is shown below:

If the painter pulls down on the rope with force F , Newton’s third law requires therope to pull up on the painter with force F , which is just the tension in the rope.Based on the picture, the same tension force F also pulls up on the painter’s chair.

Newton’s second law for the painter and the chair gives

2F ´ FG “ pmP ` mCqa (1)

2F ´ pmP ` mCqg “ pmP ` mCqa (2)

ñ F “ pmP ` mCqa ` pmP ` mCqg2

(3)

“ pmP ` mCqpa ` gq2

(4)

“ p70 kg ` 10 kgq ˆ p0.20 m{s2 ` 9.8 m{s2q2

(5)

“ 400 N. (6)

23


12. Based on the figure below, find an expression for the acceleration of m1. Assume thetable is frictionless. Hint : Think carefully about the acceleration constraint.

Solution:

The pictorial representation and free-body diagram for the problem are shown below:

For every meter that block 1 moves forward, one meter is provided to block 2. Inother words, each rope on m2 has to be lengthened by one-half meter. Therefore,the acceleration constraint is

a2 “ ´1

2a1. (1)

Newton’s second law applied to block 1 and block 2, respectively, gives

T “ m1a1 (2)

2T ´ pFGq2 “ m2a2 “ ´m2a1

2. (3)

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Combining these two equations gives

2m1a1 ´ m2g “ ´m2a1

2(4)

4m1a1 ´ 2m2g “ ´m2a1 (5)

4m1a1 ` m2a1 “ 2m2g (6)

a1p4m1 ` m2q “ 2m2g (7)

ñ a1 “ 2m2g

p4m1 ` m2q (8)

Note: if m1 “ 0 then a2 “ ´g, which is what we would expect for a free-fallingobject.

25


13. What is the acceleration of the 2.0 kg block shown in the figure below across the fric-tionless table? Hint : Think carefully about the acceleration constraint.

Solution:

The pictorial representation and free-body diagram for the problem are shown below:

For every one meter that block 1 goes down, each rope on block 2 will be shortenedby one-half meter. Therefore, the acceleration constraint is

a1 “ ´2a2. (1)

Newton’s second law applied to block one gives

2T “ m2a2 (2)

ñ T “ m2a2

2. (3)

Similarly, applying Newton’s second law to block two gives

T ´ pFGq1 “ m1a1 “ ´2m1a2 (4)

T ´ m1g “ ´2m1a2 (5)m2a2

2´ m1g “ ´2m1a2 (6)

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m2a2 ` 4m1a2 “ 2m1g (7)

a2pm2 ` 4m1q “ 2m1g (8)

ñ a2 “ 2m1g

m2 ` 4m1

(9)

“ 2 ˆ p1.0 kgq ˆ p9.8 m{s2qp2.0 kg ` 4 ˆ 1.0 kgq (10)

“ 3.3 m{s2. (11)

Note that if m1 “ 0 then a2 “ 0, which is what we would expect.

27


14. As a science fair project, you want to launch an 800 g model rocket straight up and hit ahorizontally moving target as it passes 30 m above the launch point. The rocket engineprovides a constant thrust of 15 N. The target is approaching at a speed of 15 m/s. Atwhat horizontal distance between the target and the rocket should you launch?

Solution:

The pictorial representation and motion diagram for the problem are shown below:

This is a 2D dynamics/kinematics problem. For the rocket, Newton’s second lawalong the y-direction is

pFnetqy “ FR ´ mg “ maR (1)

ñ aR “ 1

mpFR ´ mgq (2)

“ 1

0.8 kgr15 N ´ p0.8 kgq ˆ p9.8 m{s2qs (3)

“ 8.95 m{s2. (4)

Next, we use the kinematic equation for the rocket to find the time, noting thaty0R “ pv0Rqy “ t0R “ 0:

y1R “ y0R ` pv0Rqypt1R ´ t0Rq ` 1

2aRpt1R ´ t0Rq2 (5)

“ 1

2aRt

21R (6)

ñ t1R “c

2y1RaR

(7)

“d

2 ˆ 30 m

8.95 m{s2 (8)

“ 2.589 s. (9)

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For the target, we use the same kinematic equation along the x-direction, notingthat t1T “ t1R (the times are the same for both the rocket and the target) and thatx0T “ aT “ 0:

x1T “ x0T ` pv0T qxpt1T ´ t0T q ` 1

2aT pt1T ´ t0T q2 (10)

“ pv0T qxt1T (11)

“ p15 m{sq ˆ p2.589 sq (12)

“ 39 m. (13)

In other words, the rocket should be launched when the target is 39 m away.

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15. A conical pendulum is formed by attaching a 500 g ball to a 1.0 m-long string, thenallowing the mass to move in a horizontal circle of radius 20 cm. The figure shows thatthe string traces out the surface of a cone, hence the name.

(a) What is the tension in the string?

(b) What is the ball’s angular speed, in rpm?

Solution:

(a) What is the tension in the string?

The forces in the z-direction in the free body diagram are the component of thetension in the z-direction and the force of gravity. To find the z-component ofthe tension,

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cos θ “ adjacent

hypotenuse“ Tz

T(1)

Tz “ T cos θ (2)

Apply Newton’s second law in the z-directionÿ

Fz “ T cos θ ´ mg “ 0 (3)

T “ mg

cos θ(4)

To calculate cos θ,

cos θ “ adjacent

hypotenuse“

?L2 ´ r2

L“

a

p1mq2 ´ p0.2mq21.0m

“ 0.98 (5)

ñ θ “ 11.5˝ (6)

Plugging this in for tension,

T “ mg

cos θ“ p0.500 kgq ˆ p9.80 m{s2q

0.98(7)

T “ 5.0 N (8)

(b) What is the ball’s angular speed, in rpm?Referring back to the tension triangle, the radial component of tension is

sin θ “ opposite

hypotenuse“ Tr

T(9)

Tr “ T sin θ (10)

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Referring back to the triangle for the length of the pendulum,

sin θ “ opposite

hypotenuse“ r

L(11)

r “ L sin θ (12)

Apply Newton’s second law in the r-direction

ÿ

Fr “ Tr “ T sin θ “ mω2r (13)

ñ ω “c

T sin θ

mr(14)

“d

5.0 N ˆ sin 11.5˝

p0.500 kgq ˆ p0.2 mq (15)

“ 3.16 rad{sec (16)

“ 3.16rad

secˆ

ˆ

1 rev

2π rad

˙

ˆˆ

60 sec

1 min

˙

(17)

“ 30 rpm (18)

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16. In an old-fashioned amusement park ride, passengers stand inside a 5.0 m diameter hollowsteel cylinder with their backs against the wall. The cylinder begins to rotate about avertical axis. Then the floor on which the passengers are standing suddenly drops away!If all goes well, the passengers will “stick” to the wall and not slide. Clothing has astatic coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficientin the range 0.40 to 0.70. A sign next to the entrance says, “No children under 30 kgallowed.” What is the minimum angular speed, in rpm, for which the ride is safe?

Solution:

The passengers stick to the wall if the static friction force is sufficient to support thegravitational force on them: fs “ Fg. The minimum angular velocity occurs whenstatic friction reaches its maximum possible value pfsqmax “ µsn. Although clothinghas a range of coefficients of friction, it is the clothing with the smallest coefficient(µs “ 0.60q that will slip first, so this is the case we need to examine. Assuming thatthe person is stuck to the wall, Newton’s second law gives

ÿ

Fr “ n “ mω2rÿ

Fz “ fs ´ Fg “ 0 (1)

ñ fs “ mg (2)

The minimum angular speed occurs when

fs “ pfsqmax “ µsn “ µsmrω2min (3)

Substituting this expression into the z-equation gives

fs “ µsmrω2min “ mg (4)

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Solving for ωmin

ωmin “c

g

µsr(5)

“d

9.80 m{s20.60 ˆ p2.5 mq (6)

“ 2.56 rad{s ˆˆ

1 rev

2π rad

˙

ˆˆ

60 sec

1 min

˙

(7)

“ 24 rpm (8)

Note that the angular velocity does not depend on the mass of the individual.

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17. You have a new job designing rides for an amusement park. In one ride, the rider’s chairis attached by a 9.0 m long chain to the top of a tall rotating tower. The tower spins thechair and rider around at the rate of 1.0 rev every 4.0 s. In your design, you’ve assumedthat the maximum possible combined weight of the chair and rider is 150 kg. You’vefound a great price for chain at the local discount store, but your supervisor wonders ifthe chain is strong enough. You contact the manufacturer and learn that the chain israted to withstand a tension of 3000 N. Will this chain be strong enough for the ride?

Solution:

For circular motion, the force is to the center of the circle. The only force in the r-direction in the free body diagram is the component of the tension in the r-direction.To find the r-component,

sin θ “ opposite

hypotenuse“ Tr

T(1)

Tr “ T sin θ (2)

35


Newton’s second law along the r-axis is

ÿ

Fr “ Tr “ mar “ mv2

r“ mω2r (3)

T sin θ “ mω2r (4)

The radius, r, is a component of the length of the chain, L.

sin θ “ opposite

adjacent“ r

L(5)

r “ L sin θ (6)

Plugging this in, we find

T sin θ “ mω2L sin θ (7)

T “ mω2L (8)

T “ p150 kgq ˆ„

2π

4.0 sec

2

ˆ p9.0 mq (9)

T “ 3300 N (10)

Thus, the 3000 N chain is not strong enough for the ride.

36


18. Mass m1 on the frictionless table shown below is connected by a string through a hole inthe table to a hanging mass m2. With what speed must m1 rotate in a circle of radiusr if m2 is to remain hanging at rest?

Solution:

This is a coupled problem where the two masses are connected by a string. m1 ismoving in circular motion andm2’s motion is linear. Notice in the free body diagramsthat the coordinate system for m1 is only defined for one direction. The coordinatesystem for m2 is the rtz-coordinate system. The tangential coordinate isn’t shownbecause it isn’t needed for this problem.

In a coupled two body problem, a free body diagram is needed for each mass.

Next we use Newton’s second law for each mass. Notice that m2 is hanging - it’s notmoving. So, the sum of the forces on m2 equals zero.

m2 :ÿ

pFon m2qy “ T ´ pFGq2 (1)

“ T ´ m2g (2)

“ 0 (3)

T “ m2g (4)

37


For m2, object isn’t moving in the z-direction and the normal force and the gravita-tional force are balanced. The two objects are connected by the string. The tensionof the string is in the radial direction and it’s providing the force for the centripetalforce that’s creating the circular motion of m1.

m1 :ÿ

pFon m1qr “ T “ m1a “ m1

v2

r(5)

The tension is the same for both masses and it’s what connects the objects in theequations. We can set the tensions equal and solve for the velocity of m1

m2g “ m1

v2

r(6)

ñ v2 “ m2gr

m1

(7)

ñ v “c

m2gr

m1

(8)

38


Kinematics and Mechanics

xf “ xi ` vxi∆t ` 1

2axp∆tq2

vxf “ vxi ` ax∆t

v2xf “ v2xi ` 2axpxf ´ xiq

yf “ yi ` vyit ` 1

2ayt

2

vyf “ vyi ` ayt

v2yf “ v2yi ` 2aypyf ´ yiq

θf “ θi ` ωi∆t ` 1

2α∆t2

ωf “ ωi ` α∆t

ωf2 “ ωi

2 ` 2α∆θ

s “ rθ

c “ 2πr

v “ 2πr

Tvt “ ωr

ar “ v2

r“ ω2r

at “ rα

Forces

~Fnet “ Σ~F “ m~a

~Fnet “ Σ~Fr “ m~a “ mv2

r

Fg “ mg

0 ă fs ă“ µsFN

fk “ µkFN

~FA on B “ ´~FB on A

Constants

g “ 9.8m

s2

39

midterm exam 2 - siena collegemmccolgan/gp130f12/exams/midterm2_all.pdf · midterm exam 2 october...

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