Physics 18 Spring 2011

Homework 3 - Solutions

Wednesday February 2, 2011

Make sure your name is on your homework, and please box your final answer. Becausewe will be giving partial credit, be sure to attempt all the problems, even if you don’tfinish them. The homework is due at the beginning of class on Wednesday, February9th. Because the solutions will be posted immediately after class, no late homeworks canbe accepted! You are welcome to ask questions during the discussion session or during officehours.

1. Astronauts in apparent weightlessness during their stay on the International SpaceStation must carefully monitor their masses because significant loss of body mass isknown to cause serious medical problems. Give an example of how you might designequipment to measure the mass of an astronaut on the orbiting space station.

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Solution

There are all sorts of ways that you might measure the mass of an astronaut. Oneway would be to put the astronaut on a spring with known spring constant, k, sethim oscillating, and measure the oscillation frequency, which depends on the mass.Another way might be to attach him to that spring, and spin him in a circle. Then youcan measure the expansion of the spring, which is also equal to the centripetal force.In this case, you can relate the astronaut’s mass to the expansion of the spring.

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2. A net force of (6.0 N) i − (3.0 N) j acts on a 1.5 kg object. Find the acceleration ~a.What is the magnitude of the acceleration, a?

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Solution

Newton tells us that ~F = m~a, where ~F is the applied force, m is the mass of the object,and ~a is the acceleration the mass experiences. So, the acceleration is just ~a = ~F/m.So, we just find

~a =~F

m=

6.0i− 3.0j

1.5=(

4.0i− 2.0j)

m/s2.

This is our acceleration. The magnitude is just a =√a2x + a2

y =√

4.02 + 2.02 =√

20 =4.47 m/s2.

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3. Seat belts and air bags save lives by reducing the forces exerted on the driver andpassengers in an automobile collision. Cars are designed with a “crumple zone” in thefront of the car. In the event of an impact, the passenger compartment decelerates overa distance of about 1 m as the front of the car crumples. An occupant restrained byseat belts and air bags decelerates with the car. By contrast, an unrestrained occupantkeeps moving forward with no loss of speed (Newton’s first law!) until hitting thedashboard or windshield. These are unyielding surfaces, and the unfortunate occupantthen decelerates over a distance of only about 5 mm.

(a) A 60 kg person is in a head-on collusion. The car’s speed at impact is 15 m/s.Estimate the net force on the person if he or she is wearing a seat belt and if theair bag deploys.

(b) Estimate the net force that ultimately stops the person if he or she is not restrainedby a seat belt or air bag.

(c) How do these two forces compare to the person’s weight?

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Solution

For a constant acceleration we can relate the initial and final velocities as v2f = v2

i +

2a∆x. Since the final velocity is zero, then the acceleration is a = − v2i2∆x

. Thus, the

force acting on the passenger is F = ma = −mv2i2∆x

. We just need to plug in the differentdistances for each case.

(a) When the air bags deploy and the seat belts hold, then the stopping distance is∆x = 1, and so

F = −mv2i

2∆x= −60× 152

2× 1= −6750 N,

where the negative just means that it’s pushing the passenger back into the seat.

(b) If there are no restraints, then ∆x = 5 mm, or ∆x = 0.005 m, and so

F = −mv2i

2∆x= − 60× 152

2× 0.005= −1.35× 106 N!

This is much bigger than the restrained case!

(c) The passenger weighs W = mg = 60 × 9.8 = 588 N, and so the result from part(a) is about 11.5 times bigger than their weight, while the result from part (b) isabout 2300 times their weight! It’s clear which is a more survivable impact!

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4. A 35.0-kg traffic light is supported bytwo wires as in the figure. (a) Draw thelight’s free-body diagram and use it toanswer the following question qualita-tively: Is the tension in wire 2 greaterthan or less than the tension in wire 1?(b) Verify your answer by applying New-ton’s laws and solving for the two tensions.

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Solution

(a) The free-body diagram is seen in the figureto the right. You’ll notice that the vectorsdo add to zero, which, as we know, is thecase if the light isn’t accelerating. That is,

~T1 + ~T2 + ~Fg = 0.

From the diagram, it seems pretty clearthat the tension ~T2 > ~T1.

(b) Now, let’s check the math. The sum ofthe forces in each direction has to be zero.Since gravity points down, it doesn’t con-tribute to the forces along the x direction.So,∑

Fx = T1 cos 30◦ − T2 cos 60◦ = 0.

The two forces cancel out in the horizontaldirection. So, we can express T2 in termsof T1, which gives

T2 =cos 30◦

cos 60◦T1 =

√3/2

1/2T1 =

√3T1,

and so, indeed, T2 > T1, as we guessed.

3060T

1

T2

Fg

x

y

m

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5. A 65-kg student weights himself bystanding on a force scale mounted on askateboard that is rolling down an incline,as shown in the figure. Assume there is nofriction so that the force exerted by theincline on the skateboard is normal to theincline. What is the reading on the scaleif θ = 30◦?

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Solution

The reading that the scale gives is equalto the normal force, the force that thescale exerts on the skater. We can drawa free-body diagram, choosing our coordi-nates such that the x axis points along theramp, as in the figure to the right. Then,the normal force points entirely along they direction. The sum of the forces in the ydirection has to be zero, since the skater isaccelerating along the ramp (i.e., in the xdirection). So,∑

Fy = Fn − Fg cos 30◦ = 0.

Since the gravitational force is Fg = mg,then

Fn = mg cos 30◦ =

√3

2mg =

√3

2(65)(9.8) = 552 N.

So, the scale reads 552 N.

30

Fg

x

y

m

Fn

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6. A block of mass m slides across a friction-less floor and then up a frictionless ramp.The angle of the ramp θ and the speed ofthe block before it starts up the ramp isv0. The block will slide up to some heighth above the floor before stopping. Showthat h is independent of m and θ by deriv-ing an expression for h in terms of v0 and g.

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Solution

Since the floor is frictionless, the block slides alongthe floor at a constant speed, v0. Once it startsgoing up the ramp, then it’s slowed by gravitypulling it back down. We can draw a force dia-gram for the block, orienting our coordinate sys-tem along the ramp,. In this case, the normalforce points entirely along y. Suppose that theblock moves along the ramp a distance d, as inthe figure. Then, from the geometry, it is up aheight h = d sin θ. So, if we can figure out howfar it goes along the ramp, then we’ll know ouranswer.

θ

Fg

x

y

m

Fn

θ

d h

We can figure out how far along the ramp it goes using the kinematic equations. Sincewe don’t have the time it takes to travel up the ramp, we can use the velocity-distanceequation,

v2xf = v2

xi + 2ax∆x,

where ax is the acceleration along the x direction (which isn’t g!). When the blockreaches its maximum height, it stops moving, and so vf = 0. So, if it started with aninitial velocity v0, and travels a distance ∆x = d, then solving for d gives

d = − v20

2ax.

What’s ax? We can determine this from Newton’s laws. Since the block is acceleratingalong x, and the normal force points entirely along y, then

∑Fx = −mg sin θ = max.

Thus, ax = −g sin θ. So, d =v20

2g sin θ. Plugging in our expression for h gives

h = d sin θ =v2

0

2g sin θsin θ =

v20

2g,

which is independent of the mass and angle, as advertised.

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7. A block of mass m is being lifted vertically by a uniform rope of mass M and lengthL. The rope is being pulled upward by a force applied to its top end, and the ropeand block are accelerating upward with an acceleration of magnitude a. Show thatthe tension in the rope at a distance x (where x < L) above the block is given by(a+ g) [m+ (x/L)M ].

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Solution

The system is seen in the figure to theright. The total amount of mass below thepoint x is the mass of the block, m, plusthe mass of the segment of rope of lengthx. The trick is figuring out how much thatmass is. Since the mass is uniform, ev-ery meter of the rope has the same mass.So, M/L is a constant number, and sayshow much mass a unit length of the ropecarries. So, if we take this number, andmultiply it by the length, x, this this tellsus how much mass in in the segment. So,msegment = M

Lx. Thus, the total mass be-

low the point x is

Mtotal = m+M

Lx.

m

ML

x(m+(M/L)x)

Fg

T

Now that we know what the mass is, we can determine the tension from a free-bodydiagram, as seen above. The sum of the forces along y is∑

Fy = T −Mtotalg = T −(m+

M

Lx

)g.

But, the force is the mass times the acceleration, and the mass is the total mass of theblock and segment, Mtotal. So, setting

∑Fy =

(m+ M

Lx)a and solving for T gives

T =

(m+

M

Lx

)(g + a) .

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8. The figure shows a 20-kg block sliding ona 10-kg block. All surfaces are frictionlessand the pulley is massless and frictionless.Find the acceleration of each block andthe tension in the string that connects theblocks.

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Solution

We can draw free-body diagramsfor each block, as seen to the left.The diagram for the 20 kg blockis first, while the 10 kg is on theright. The top block has threeforces acting on it: the tensionpulling it along x, the gravitationalforce pulling it down, and thenormal force pushing it up alongy. The bottom block is a bit morecomplicated, having four forces.There is still the tension, thegravitational force, and the normalforce, but the top block is sittingon the bottom block, weighing itdown a bit. So, there’s an extraforce, ~F20.

20

Fg

x

y

m

Fn

T

20

20

20F

g

x

y

m

Fn

T

10

10

F20

Let’s start with the top block. Since the acceleration is in the x direction,∑Fx =

m20ax. So, working out the forces in the x direction gives∑Fx = T −m20g sin 20◦ = −m20ax,

where we let the acceleration be negative, since the top block will likely slide backwards,since it’s heavier. Working out the forces for the second block gives∑

Fx = T −m10g sin 20◦ = +m10ax,

where the acceleration is positive because it’s opposite to that of the top block (if the topblock slides down, then the bottom one is sliding up, etc.). So, solving the top for the

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tension and substituting in to the second gives (m20 −m10) g sin 20◦ = (m20 +m10) ax.Solving for ax gives

ax =

(m20 −m10

m20 +m10

)g sin 20◦.

This is the acceleration. We can find the tension by plugging this back into our forceequation for, say, the bottom block.

T = m10ax +m10g sin 20◦ = m10

(m20 −m10

m20 +m10

)g sin 20◦ +m10g sin 20◦,

or,

T = 2m10m20

m10 +m20

g sin 20◦.

So, plugging in the numbers gives

ax =

(m20 −m10

m20 +m10

)g sin 20◦ =

(20− 10

20 + 10

)(9.8) sin 20◦ = 1.12 m/s2.

So, the bottom block slides up at 1.12 m/s2. The top block slides down at the sameacceleration. Furthermore,

T = 2m10m20

m10 +m20

g sin 20◦ = 2

(20× 10

20 + 10

)(9.8) sin 20◦ = 45 N.

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9. A 2.0 kg block rests on a frictionless wedgethat has a 60◦ incline and an acceleration~a to the right such that the mass remainsstationary to the wedge. (a) Draw thefree-body diagram of the block and useit to determine the magnitude of theacceleration. (b) What would happenif the wedge were given an accelerationlarger than this value? Smaller than thisvalue?

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Solution

(a) The ramp is accelerating to the right. Thispushes against the block leading to a nor-mal force. So, there are two forces act-ing on the block, the normal force andthe gravitational force. Since the acceler-ation is in the x direction, let’s work outthe sum of the forces in the x direction,∑Fx = ma.∑

Fx = Fn cos 30◦ = ma.

So, in order to determine the acceleration,we need the normal force, which we candetermine from the forces along the y di-rection. Since there is no acceleration inthe y direction,

∑Fy = 0. Now,∑

Fy = Fn sin 30◦−mg = 0⇒ Fn =mg

sin 30◦.

Fg

Fn

y

x

30

Plugging this in for our acceleration gives

a =Fnm

cos 30◦ = g cot 30◦ = (9.8)× cot 30 = 17 m/s2.

(b) The normal force comes from the acceleration. An acceleration greater than 17m/s2 would provide a bigger normal force than needed to balance the gravitationalforce. If this is the case, then, because there is a vertical component to the normalforce, which is bigger than the gravitational force, there is an acceleration upwards,due to the forces not balancing. So, the block would slide up the ramp. Conversely,if the acceleration was less than 17 m/s2, then the normal force wouldn’t quitebalance the gravitational force, and the block would slide down the ramp.

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10. Elvis Presley has supposedly been sightednumerous times since his death on August16, 1977. The following is a chart ofwhat Elvis’s weight would be if here weresighted on the surfaces of other objectsin our solar system. Use the chart todetermine: (a) Elvis’s mass on Earth, (b)Elvis’s mass on Pluto, and (c) the free-fallacceleration on Mars. (d) Compare thefree-fall acceleration on Pluto to thefree-fall acceleration on the moon.

Planet Elvis’s Weight (N)Mercury 431Venus 1031Earth 1133Mars 431

Jupiter 2880Saturn 1222Pluto 58Moon 191

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Solution

(a) The weight, W , of an object on a planet with gravitational acceleration g is justW = mg. So, on Earth, Elvis has a mass m = W/g = 1133/9.8 = 116 kg.

(b) Because the mass is an intrinsic property of an object, it doesn’t change fromplace to place. So, Elvis’s mass on Pluto is exactly the same as his mass onEarth, m = 116 kg.

(c) The free-fall acceleration on Mars, gMars can be found from Elvis’s weight on Mars,divided by his mass, gMars = WMars/m = 431/116 = 3.72 m/s2.

(d) The free-fall acceleration on Pluto, gPluto, is again, Elvis’s weight on Pluto, dividedby his mass, gPluto = WPluto/m. The free-fall acceleration on the Moon is, by thesame reasoning, gMoon = WMoon/m. Solving the second for the mass, and pluggingit into the first gives

gPluto =

(WPluto

WMoon

)gMoon.

Plugging in the weights tells us that gPluto = (58/191)gMoon = 0.3gMoon. So, thefree-fall acceleration on Pluto is less than 1/3 that of the Moon!

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