Assignment 2: Newton's Laws of Motion

Due: 2:00am on Friday, February 15, 2013

Note: To understand how points are awarded, read your instructor's Grading Policy.

Pulling Three Blocks

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . The magnitude of the

tension in the string between blocks B and C is = 3.00 . Assume that each block has

mass = 0.400 .

Part A

What is the magnitude of the force?

Express your answer numerically in newtons.

Hint 1. Find the acceleration of block B

What is the magnitude of the acceleration of block B?

Express your answer numerically in meter per second squared.

Hint 1. Consider blocks A and B as a unit

Since blocks A and B are not moving with respect to each other, you can treat them as one larger "object." This larger object hasthe same acceleration as either block A or block B alone. The advantage of such an approach is that the larger object has only oneforce acting on it (the tension 3.00 in the rope).

ANSWER:

Hint 2. Find the acceleration of all three blocks

Which of the following expressions gives the magnitude of the acceleration of the three blocks?

Hint 1. Consider all three blocks as a unit

Since the three blocks are not moving with respect to one another, you can treat them as one larger "object" of mass equal to the

sum of the masses of all three blocks. The only horizontal force acting on this larger object is , so you can use Newton's 2nd law

to determine an expression for its acceleration.

ANSWER:

PC1431AY1213SEM2

Assignment 2: Newton's Laws of Mo... Resources

=

Signed in as Mikael Lemanza Help Close

ANSWER:

Correct

Part B

What is the tension in the string between block A and block B?

Express your answer numerically in newtons

Hint 1. How to approach the question

The tension is the only horizontal force acting on block A. Thus, you can find the acceleration of block A and then apply Newton's 2nd

law. Note that all three blocks have the same acceleration.

ANSWER:

Correct

± A Ride on the Ferris Wheel

A woman rides on a Ferris wheel of radius 16 that maintains the same speed throughout its motion. To better understand physics, she takes along a

digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph ofscale reading versus time. Note that the graph has a minimum value of 510 and a

maximum value of 666 .

Part A

What is the woman's mass?

Express your answer in kilograms.

Hint 1. How to approach the problem

The woman is moving in a circle with constant speed. To maintain this motion she must experience a net acceleration (called centripetalacceleration) directed toward the center of the Ferris wheel.

Draw and analyze the woman's free-body diagram at a wisely chosen point on the circular path and use Newton's 2nd law to determine her

= 4.50

= 1.50

mass.

Hint 2. Find the extreme points on the circular path

The bathroom scale does not record the gravitational force acting on the woman. If it did, the reading would not vary as she rides the Ferriswheel. Instead, the scale records the normal force acting on the woman, which can vary as she moves along the circular path andexperiences different accelerations. This normal force is sometimes referred to as an apparent weight, because it mimics the feelings ofbeing heavier or lighter.

Note that the normal force is equal in magnitude to the gravitational force on the flat surface of the earth, so the apparent weight is justcalled the weight in this static situation.

As the woman travels along the circular path, her apparent weight fluctuates between a maximum value and a minimum value. At whatlocation (A - D) will the apparent weight be a maximum? Where will it be a minimum?

Enter the letters that correspond to the correct positions separated by a comma.

Hint 1. Analyze the free-body diagram

Draw a free-body diagram for the woman. Assume that the y direction points vertically upward. Which of the following statements

are true for every point on the circle traveled by the woman?

Check all that apply.

ANSWER:

Hint 2. Analyze the acceleration

At all times during the woman's motion, she experiences a net acceleration (called centripetal acceleration) directed toward thecenter of the Ferris wheel. At what location (A - D) will the acceleration be in only the vertical direction?

Check all that apply.

ANSWER:

The gravitational force acting on the woman points in the y direction.

The gravitational force acting on the woman points in the y direction.

The magnitude of the gravitational force acting on the woman is constant.

The normal force points in the y direction.

The normal force points in the y direction.

The magnitude of the normal force is constant.

Hint 3. Applying Newton's 2nd law

Since the normal force and graviational force are the only forces acting on the woman in the vertical (y) direction,

,

where and are the magnitudes of the normal force and the gravitational force, respectively.

Apply what you know about the acceleration of the woman in the y direction at certain points along the circular path to determinewhen will be a maximum or minimum.

ANSWER:

Hint 3. Find the acceleration of the woman

What is the acceleration of the woman?

Express your answer in meters per second squared.

Hint 1. How to approach the problem

You can use the information from the problem statement and the graph to determine the woman's speed. This can be used to findher centripetal acceleration:

.

Hint 2. Find the woman's speed

What is the speed of the woman?

Express your answer in meters per second.

Hint 1. Determining the speed

Since the Ferris wheel turns at constant speed, the distance the woman travels during some time interval is given by

, where is the speed and is the time. For a complete cycle, the distance traveled is the circumference of the Ferris

wheel and the time required is one period . Thus for a Ferris wheel of radius ,

.

Hint 2. Find the period

The easiest way to determine the speed of the woman is to calculate the distance she travels during one complete cycle ofmotion and divide this by the time that it takes to complete this cycle. The time for a complete cycle is called the period .

What is the period of the Ferris wheel?

Express your answer in seconds.

ANSWER:

ANSWER:

ANSWER:

A

B

C

D

=

=

=

ANSWER:

Correct

The Window Washer

A window washer of mass is sitting on a platform suspended by a system of cables and pulleys as shown . He is pulling on the cable with a force of

magnitude . The cables and pulleys are ideal (massless and frictionless), and the platform

has negligible mass.

Part A

Find the magnitude of the minimum force that allows the window washer to move upward.

Express your answer in terms of the mass and the magnitude of the acceleration due to gravity .

Hint 1. Find a simple expression for the tension

Find an expression for the tension in the cable on which the man is pulling.

Express your answer in terms of some or all of the variables , , and .

ANSWER:

Hint 2. Upward force on the platform

The tension along the cable is equal to the force . The 3 sections of the cable (strands) are seperated by the pulleys. What is the force

exerted on the platform , by the pulley to the left of the diagram in the upward direction? Remember that the pulley to the left is supported

by two strands of the cable.

Express your answer in terms of , the tension in the cable.

ANSWER:

Hint 3. Upward forces on window washer

What forces pull the window washer upward?

ANSWER:

= 60

=

=

A force equal to and a force equal to

A force equal to and the normal force acting on the window washer from the platform equal to .

The normal force acting on the window washer from the platform equal to and a force equal to

A force equal to and the normal force acting on the window washer from the platform equal to and a force equal to

Hint 4. All forces on window washer

What objects exert forces on the window washer?

ANSWER:

ANSWER:

Correct

Skydiving

A sky diver of mass 80.0 (including parachute) jumps off a plane and begins her descent.

Throughout this problem use 9.80 for the magnitude of the acceleration due to gravity.

Part A

At the beginning of her fall, does the sky diver have an acceleration?

Hint 1. Free fall

The speed of an object in free fall through a medium (liquid or gas) increases as the object initially begins to descend. If the resistance ofthe medium is neglected, the speed of the object increases at a constant rate. If one takes into account the drag force due to resistance ofthe medium, the acceleration of the object is no longer constant. The speed of the object increases as the object falls down, but only up toa certain value, called the terminal speed.

ANSWER:

Correct

This applet shows the sky diver (not to scale) with her position, speed, and acceleration graphed as functions of time. You can see how heracceleration drops to zero over time, giving constant speed after a long time.

Part B

At some point during her free fall, the sky diver reaches her terminal speed. What is the magnitude of the drag force due to air resistance that

acts on the sky diver when she has reached terminal speed?

Express your answer in newtons.

Hint 1. Dynamic equilibrium

A body moving at constant velocity is in dynamic equilibrium: The net force acting on the body is zero. Since the only forces acting on thesky diver are her weight and the drag force, when she falls at constant speed these two forces must balance.

ANSWER:

Correct

Only the platform and the string being pulled

Only the platform and the earth

Only the earth and the cable supporting the platform

Only the cable being pulled and the cable supporting the platform

Only the platform; the earth; and the cable.

=

No; the sky diver falls at constant speed.

Yes and her acceleration is directed upward.

Yes and her acceleration is directed downward.

= 784

Part C

For an object falling through air at a high speed , the drag force acting on it due to air resistance can be expressed as

,

where the coefficient depends on the shape and size of the falling object and on the density of air. For a human body, the numerical value for

is about 0.250 .

Using this value for , what is the terminal speed of the sky diver?

Express you answer in meters per second.

Hint 1. Terminal speed

When an object in free fall has reached terminal speed, it has no acceleration and the net force acting on it is zero. Since the only forcesacting on the skydiver are the drag force and her weight, they must balance each other. To find the terminal speed of the skydiver, expressthe condition of dynamic equilibrium mathematically and solve for the speed.

ANSWER:

Correct

Recreational sky divers can control their terminal speed to some extent by changing their body posture. When oriented in a headfirst dive, asky diver can reach speeds of about 54 meters per second (120 miles per hour). For maximum drag and stability, sky divers often will orientthemselves "belly-first." In this position, their terminal speed is typically around 45 meters per second (100 miles per hour).

Part D

When the sky diver descends to a certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute isdeployed when the sky diver reaches an altitude of about 900 --3000 .) Immediately after deploying the parachute, does the skydiver have a

nonzero acceleration?

Hint 1. How to approach the problem

The parachute acts to decrease the speed of the sky diver. A decrease in speed corresponds to an acceleration opposite to the direction ofmotion.

ANSWER:

Correct

Part E

When the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 . What is the drag force

acting on the sky diver immediately after she has opened the parachute?

Express your answer in newtons.

Hint 1. How to approach the problem

The drag force depends of the speed of the falling object and the coefficient . You simply need to determine the speed of the sky diver

when she opens the parachute.

Hint 2. Find the speed of the sky diver when the parachute is deployed

When the parachute is deployed, what is the speed of the sky diver?

Express your answer in meters per second.

Hint 1. How to approach the question

= 56.0

No; the sky diver keeps falling at constant speed.

Yes and her acceleration is directed downward.

Yes and her acceleration is directed upward.

When she opens the parachute, she is still falling at a constant speed equal to the terminal speed calculated in Part C.

ANSWER:

ANSWER:

Correct

Part F

What is the terminal speed of the sky diver when the parachute is opened?

Express your answer in meters per second.

Hint 1. How to approach the problem

Calculate the terminal speed of the sky diver as you did in Part C, but take into account the new value of the coefficient .

ANSWER:

Correct

A typical "student" parachute for recreational skydiving has a drag coefficient that gives a terminal speed for landing of about 2 meters persecond (5 miles per hour). If this seems slow based on video or real-life sky divers you have seen, that may be because the sky divers yousaw were using high-performance parachutes; these offer the sky divers more maneuverability in the air but increase the terminal speed up to4 meters per second (10 miles per hour).

Exercise 4.30

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. The light turns green and you stomp onthe gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. (Assume that the truck is accelerating to theright.)

Part A

Draw clearly labeled free-body diagram for the box. (The bed of the truck is not frictionless.)

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors willnot be graded but the relative length of one to the other will be graded.

ANSWER:

=

= 1.88×105

= 3.61

Correct

Part B

Draw clearly labeled free-body diagram for the truck. (The bed of the truck is not frictionless.)

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors willnot be graded but the relative length of one to the other will be graded.

ANSWER:

Correct

Block on an Incline Adjacent to a Wall

A wedge with an inclination of angle rests next to a wall. A block of mass is sliding down the plane, as shown. There is no friction between the

wedge and the block or between the wedge and the horizontal surface.

Part A

Find the magnitude, , of the sum of all forces acting on the block.

Express in terms of and , along with any necessary constants.

Hint 1. Direction of the net force on the block

The net force on the block must be the force in the direction of motion, which is down the incline.

Hint 2. Determine the forces acting on the block

What forces act on the block? Keep in mind that there is no friction between the block and the wedge.

ANSWER:

Correct

Hint 3. Find the magnitude of the force acting along the direction of motion

Consider a coordinate system with the x direction pointing down the incline and the y direction perpendicular to the incline. In thesecoordinates, what is , the component of the block's weight in the x direction?

Express in terms of , , and .

ANSWER:

Correct

"Normal," in this context, is a synonym for "perpendicular." The normal force has no component in the direction of the block's motion(down the incline).

ANSWER:

Correct

Part B

Find the magnitude, , of the force that the wall exerts on the wedge.

Express in terms of and , along with any necessary constants.

The weight of the block and friction

The weight of the block and the normal (contact) force

The weight of the block and the weight of the wedge

The weight of the block and the force that the wall exerts on the wedge

=

=

Hint 1. The force between the wall and the wedge

There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the net horizontal force on thewedge must be zero. If the block exerts a force with a horizontal component on the wedge, some other horizontal force must act on thewedge so that the net force is zero.

Hint 2. Find the normal force between the block and the wedge

What is the magnitude, , of the normal (contact) force between the block and the wedge? (You might have computed this already while

answering Part A.)

Express in terms of , , and .

ANSWER:

Correct

Hint 3. Find the horizontal component of the normal force

In the previous hint you found the magnitude of the normal force between the block and the wedge. What is the magnitude, , of the

horizontal component of this normal force?

Express in terms of and .

ANSWER:

Correct

ANSWER:

Correct

Your answer to Part B could be expressed as either or . In either form, we see that as gets very small or as

approaches 90 degrees ( radians), the contact force between the wall and the wedge goes to zero. This is what we should expect; in

the first limit ( small), the block is accelerating very slowly, and all horizontal forces are small. In the second limit ( about 90 degrees), the

block simply falls vertically and exerts no horizontal force on the wedge.

Velocity from Force Diagram Ranking Task

Below are birds-eye views of six identical toy cars moving to the right at 2 . Various forces act on the cars with magnitudes and directions indicated

below. All forces act in the horizontal plane and are either parallel or at 45 or 90 degrees to the car's motion.

Part A

Rank these cars on the basis of their speed a short time (ie. before any car's speed can reach zero) after the forces are applied.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. How to approach the problem

First, added up the force vectors to find the net force acting on the car. Since you are asked about the speed a short time after the forcesare applied, the speeds of all cars will be close to 2 . For each car, the small difference from 2 will be due to the acceleration

caused by the net force acting on the car. A net force acting to the right will result in a speed greater than 2 , while a net force acting

to the left will result in a speed less than 2 . Recall that the acceleration may be found from the net force by using Newton's 2nd law,

.

Hint 2. Summing force vectors

Forces are vectors and sum in the same way that velocity or acceleration vectors sum. Be careful when resolving vectors into components,

if you need to. Once the force vectors are summed into a single total, or net force , the acceleration of the car of mass can be

=

=

=

determined by Newton’s 2nd law:

.

ANSWER:

Correct

± Mass on Turntable

A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram .

The small metal cylinder has a mass of 0.20 , the coefficient of static friction between the

cylinder and the turntable is 0.080, and the cylinder is located 0.15 from the center of the

turntable.

Take the magnitude of the acceleration due to gravity to be 9.81 .

Part A

What is the maximum speed that the cylinder can move along its circular path without slipping off the turntable?

Express your answer numerically in meters per second to two significant figures.

Hint 1. Centripetal acceleration

If you know a body is in uniform circular motion, you know what its acceleration must be. If a body of mass is traveling with speed in a

circle of radius , what is the magnitude of its centripetal acceleration?

ANSWER:

Correct

Hint 2. Determine the force causing acceleration

Whenever you see uniform circular motion, there is a real force that causes the associated centripetal acceleration. In this problem, whatforce causes the centripetal acceleration?

ANSWER:

Correct

Hint 3. Find the maximum possible friction force

The magnitude of the force due to static friction satisfies . What is in this problem?

Express your answer numerically in newtons to three significant figures.

ANSWER:

Correct

Hint 4. Newton's 2nd law

To solve this problem, relate the answers to the previous two hints using Newton's 2nd law:

.

ANSWER:

Correct

Two Blocks and Two Pulleys

A block of mass is attached to a massless, ideal string. This string wraps around a massless pulley and then wraps around a second pulley that is

attached to a block of mass that is free to slide on a frictionless table. The string is firmly anchored to a wall and the whole system is frictionless.

Use the coordinate system indicated in the figure when solving this problem.

normal force

static friction

weight of cylinder

a force other than those above

= 0.157

= 0.34

Part A

Assuming that is the magnitude of the horizontal acceleration of the block of mass , what is , the tension in the string?

Express the tension in terms of and .

Hint 1. Which physical principle to use

You should use Newton's 2nd law:

,

where , , ... are forces acting on the block of mass . Keep in mind that the whole system is frictionless.

Hint 2. Force diagram for the block of mass

Which figure correctly illustrates the forces acting on the block of mass ?

The vectors , , , and denote the normal force, the gravitational force, the

tension in the string, and the friction force, respectively.

ANSWER:

Correct

ANSWER:

Correct

Part B

Figure 1

Figure 2

Figure 3

None of the above

=

Given , the tension in the string, calculate , the magnitude of the vertical acceleration of the block of mass .

Express the acceleration magnitude in terms of , , and .

Hint 1. Which physical principle to use

Apply Newton's 2nd law:

,

where , , ... are forces acting on the block of mass .

Hint 2. Force diagram for the block of mass

Which figure correctly illustrates the forces acting on the block of mass ?

The vectors , , and denote the gravitational force, the tension in the string,

and the inertial force, respectively.

ANSWER:

Correct

ANSWER:

Correct

Part C

Given the magnitude of the acceleration of the block of mass , find , the magnitude of the horizontal acceleration of the block of mass .

Express in terms of .

Hint 1. Method 1: String constraint (uses calculus)

Define and as the vertical coordinate of the block of mass and the horizontal coordinate of the block of mass ,

respectively. It is clear that , the length of the string, is

,

where is a constant that accounts for the wound portions of the string and the length of string between the y axis and the wall. Do not

worry about the value of , as it will vanish in the next step. By differentiating this equation twice with respect to time, you should obtain a

relation between and . The variables and will vanish upon differentiation.

Figure 1

Figure 2

Figure 3

None of the above

=

Hint 2. Method 2: Intuition (does not involve calculus)

You should notice that, while the block of mass descends a height , the other will move only half of . Hence, at each instant,

, where and are the speeds of the blocks of masses and , respectively. The formula for versus should be

obvious.

ANSWER:

Correct

Part D

Using the result of Part C in the formula for that you previously obtained in Part A, express as a function of .

Express your answer in terms of and .

ANSWER:

Correct

Part E

Having solved the previous parts, you have all the pieces needed to calculate , the magnitude of the acceleration of the block of mass . Write

an expression for .

Express the acceleration magnitude in terms of , , and .

Hint 1. How to approach this problem

In Part B, using Newton's 2nd law, you derived a relation between and the tension in the string, . In Part D you found as a function of

. Now eliminate from this system of two linear equations and solve for .

ANSWER:

Correct

Hanging Chandelier

A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural

decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceilingdirectly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension and makes an angle of with the

ceiling. Cable 2 has tension and makes an angle of with the ceiling.

=

=

=

Part A

Find an expression for , the tension in cable 1, that does not depend on .

Express your answer in terms of some or all of the variables , , and , as well as the magnitude of the acceleration due to gravity

.

Hint 1. Find the sum of forces in the x direction

The chandelier is static; hence the vector forces on it sum to zero. Type in the sum of the x components of the forces acting on thechandelier, using the coordinate system shown.

Express your answer in terms of some or all of the variables , , , ,

and .

ANSWER:

Correct

Hint 2. Find the sum of forces in the y direction

Now type the corresponding equation relating the y components of the forces acting on the chandelier, again using the coordinate systemshown.

Express your answer in terms of some or all of the variables , , , , and , as well as the magnitude of the acceleration

due to gravity .

ANSWER:

Correct

Hint 3. Putting it all together

There are two unknowns in this problem, and . Each of the previous two hints leads you to an equation involving these two unknowns.

Eliminate from this pair of equations and solve for .

ANSWER:

=

=

Correct

At the Test Track

You want to test the grip of the tires on your new race car. You decide to take the race car to a small test track to experimentally determine thecoefficient of friction. The racetrack consists of a flat, circular road with a radius of 45 . The applet shows the result of driving the car around the track

at various speeds.

Part A

What is , the coefficient of static friction between the tires and the track?

Express your answer to two significant figures.

Hint 1. How to approach the problem

You need to find the point at which the force of friction is just strong enough to keep the car on the circular track. Then, you can set theexpression for the frictional force equal to the centripetal force needed to keep the car going in a circle. Solving this equation for gives the

answer. Use for the mass of the car in your calculations.

Hint 2. Use the applet to find the speed

When the car successfully goes around the track, without leaving the track at all, then the needed centripetal force must be less than orequal to the maximum possible static friction. When the car leaves the track during a lap, then the needed centripetal force must be greaterthan the maximum possible static friction. What is the lowest speed at which the car leaves the track in the applet?

Express your answer in meters per second as an integer.

ANSWER:

Correct

If the lowest speed at which the car leaves the track is , then the speed at which the needed centripetal force equals the maximum

frictional force must be between and the closest speed below that you can measure. In this applet, you can only measure

in integer multiples of meters per second, so you would have . You can use any speed greater than and less

than to calculate the centripetal force that equals the maximum static friction.

Hint 3. Find an expression for

Find an expression for , the coefficient of static friction between the car's tires and the road. Use for the mass of the car, for the

magnitude of the acceleration due to gravity, for the speed at which the car is just about to leave the track, and for the radius of the

track.

Express your answer in terms of , , , and .

Hint 1. Expression for centripetal acceleration

The centripetal acceleration required to move an object in a circular path of radius at speed is .

Hint 2. Expression for the force of static friction

Recall that static friction will equal the force attempting to move an object unless the magnitude of that force exceeds , where

is the magnitude of the normal force and is the coefficient of static friction. Since the track is flat, the normal force is equal in

magnitude to the weight of the car, giving . Thus, the magnitude of the maximum force of static friction is .

ANSWER:

Correct

=

= 21

=

ANSWER:

Correct

Score Summary:

Your score on this assignment is 100%.You received 50 out of a possible total of 50 points.

= 0.91