Vectors and Motion
From Lm
Newtonian Physics by Benjamin Crowell
This is a version of the book that has been converted to wiki format from the original at lightandmatter.com, where it is available in PDF, HTML, and LaTeX formats. The version at lightandmatter.com is the one that is actively maintained by the author. Please see this page for information about the purpose of this wiki version.
Chapter 8 - Vectors and Motion
-
Vectors and Motion
In 1872, capitalist and former California governor Leland Stanford asked photographer Eadweard Muybridge if he would work for him on a project to settle a $25,000 bet (a princely sum at that time). Stanford's friends were convinced that a galloping horse always had at least one foot on the ground, but Stanford claimed that there was a moment during each cycle of the motion when all four feet were in the air. The human eye was simply not fast enough to settle the question. In 1878, Muybridge finally succeeded in producing what amounted to a motion picture of the horse, showing conclusively that all four feet did leave the ground at one point. (Muybridge was a colorful figure in San Francisco history, and his acquittal for the murder of his wife's lover was considered the trial of the century in California.)
The losers of the bet had probably been influenced by Aristotelian reasoning, for instance the expectation that a leaping horse would lose horizontal velocity while in the air with no force to push it forward, so that it would be more efficient for the horse to run without leaping. But even for students who have converted wholeheartedly to Newtonianism, the relationship between force and acceleration leads to some conceptual difficulties, the main one being a problem with the true but seemingly absurd statement that an object can have an acceleration vector whose direction is not the same as the direction of motion. The horse, for instance, has nearly constant horizontal velocity, so its ax is zero. But as anyone can tell you who has ridden a galloping horse, the horse accelerates up and down. The horse's acceleration vector therefore changes back and forth between the up and down directions, but is never in the same direction as the horse's motion. In this chapter, we will examine more carefully the properties of the velocity, acceleration, and force vectors. No new principles are introduced, but an attempt is made to tie things together and show examples of the power of the vector formulation of Newton's laws.
a / Example 1.
Contents |
The Velocity Vector
For motion with constant velocity, the velocity vector is
Failed to parse (unknown function\vc): \vc{v} = \Delta \vc{r}/\Delta t \qquad . \qquad \hfill \shoveright{\text{[only for constant velocity]}}
The Δr vector points in the direction of the motion, and dividing it by the scalar Δ t only changes its length, not its direction, so the velocity vector points in the same direction as the motion. When the velocity is not constant, i.e., when the x-t,y-t, and z-t graphs are not all linear, we use the slope-of-the-tangent-line approach to define the components vx, vy, and vz, from which we assemble the velocity vector. Even when the velocity vector is not constant, it still points along the direction of motion.
Vector addition is the correct way to generalize the one-dimensional concept of adding velocities in relative motion, as shown in the following example:
Example 1: Velocity vectors in relative motion
◊ You wish to cross a river and arrive at a dock that is directly across from you, but the river's current will tend to carry you downstream. To compensate, you must steer the boat at an angle. Find the angle θ, given the magnitude, |vWL|, of the water's velocity relative to the land, and the maximum speed, |vBW|, of which the boat is capable relative to the water.
◊ The boat's velocity relative to the land equals the vector sum of its velocity with respect to the water and the water's velocity with respect to the land, vBL = vBW+ vWL . If the boat is to travel straight across the river, i.e., along the y axis, then we need to have vBL,x=0. This x component equals the sum of the x components of the other two vectors, vBL,x = vBW,x + vWL,x , or 0 = -|vBW| sin θ + |vWL| . Solving for θ, we find
Failed to parse (unknown function\vc): \begin{align} \sin \theta &= |\vc{v}_{WL}|/|\vc{v}_{BW}| \qquad ,\\ \intertext{so} \theta &= \sin^{-1}\frac{|\vc{v}_{WL}|}{\vc{v}_{BW}}\qquad . \end{align}
◊ Solved problem: Annie Oakley — problem 8
”Discussion Questions”
◊ Is it possible for an airplane to maintain a constant velocity vector but not a constant Failed to parse (unknown function\vc): |\vc{v}|?
How about the
opposite -- a constant |v| but not a constant velocity vector? Explain.
◊ New York and Rome are at about the same latitude, so the earth's rotation carries them both around nearly the same circle. Do the two cities have the same velocity vector (relative to the center of the earth)? If not, is there any way for two cities to have the same velocity vector?
b / A change in the magnitude of the velocity vector implies an acceleration.
c / A change in the direction of the velocity vector also produces a nonzero Δv vector, and thus a nonzero acceleration vector, Δv/Δ t.
The Acceleration Vector
When all three acceleration components are constant, i.e., when the vx-t, vy-t, and vz-t graphs are all linear, we can define the acceleration vector as
Failed to parse (unknown function\vc): \vc{a} = \Delta\vc{v}/\Delta t \qquad , \qquad \shoveright{\text{[only for constant acceleration]}}
which can be written in terms of initial and final velocities as
Failed to parse (unknown function\vc): \vc{a} = (\vc{v}_f-\vc{v}_i)/\Delta t \qquad . \qquad \shoveright{\text{[only for constant acceleration]}}
If the acceleration is not constant, we define it as the vector made out of the ax, ay, and az components found by applying the slope-of-the-tangent-line technique to the vx-t, vy-t, and vz-t graphs.
Now there are two ways in which we could have a nonzero acceleration. Either the magnitude or the direction of the velocity vector could change. This can be visualized with arrow diagrams as shown in figures b and c. Both the magnitude and direction can change simultaneously, as when a car accelerates while turning. Only when the magnitude of the velocity changes while its direction stays constant do we have a Δ v vector and an acceleration vector along the same line as the motion.
self-check: (1) In figure b, is the object speeding up, or slowing down? (2) What would the diagram look like if vi was the same as Failed to parse (unknown function\vc): \vc{v}_f?
(3) Describe how the Δ v vector is
different depending on whether an object is speeding up or slowing down. (answer in the back of the PDF version of the book)
If this all seems a little strange and abstract to you, you're not alone. It doesn't mean much to most physics students the first time someone tells them that acceleration is a vector, and that the acceleration vector does not have to be in the same direction as the velocity vector. One way to understand those statements better is to imagine an object such as an air freshener or a pair of fuzzy dice hanging from the rear-view mirror of a car. Such a hanging object, called a bob, constitutes an accelerometer. If you watch the bob as you accelerate from a stop light, you'll see it swing backward. The horizontal direction in which the bob tilts is opposite to the direction of the acceleration. If you apply the brakes and the car's acceleration vector points backward, the bob tilts forward.
After accelerating and slowing down a few times, you think you've put your accelerometer through its paces, but then you make a right turn. Surprise! Acceleration is a vector, and needn't point in the same direction as the velocity vector. As you make a right turn, the bob swings outward, to your left. That means the car's acceleration vector is to your right, perpendicular to your velocity vector. A useful definition of an acceleration vector should relate in a systematic way to the actual physical effects produced by the acceleration, so a physically reasonable definition of the acceleration vector must allow for cases where it is not in the same direction as the motion.
self-check: In projectile motion, what direction does the acceleration vector have? (answer in the back of the PDF version of the book)
d / Example 2.
Example 2: Rappelling
In figure d, the rappeller's velocity has long periods of gradual change interspersed with short periods of rapid change. These correspond to periods of small acceleration and force, and periods of large acceleration and force.
e / Example 3.
Example 3: The galloping horse
Figure e on page 206 shows outlines traced from the first, third, fifth, seventh, and ninth frames in Muybridge's series of photographs of the galloping horse. The estimated location of the horse's center of mass is shown with a circle, which bobs above and below the horizontal dashed line.
If we don't care about calculating velocities and accelerations in any particular system of units, then we can pretend that the time between frames is one unit. The horse's velocity vector as it moves from one point to the next can then be found simply by drawing an arrow to connect one position of the center of mass to the next. This produces a series of velocity vectors which alternate between pointing above and below horizontal.
The Δv vector is the vector which we would have to add onto one velocity vector in order to get the next velocity vector in the series. The Δv vector alternates between pointing down (around the time when the horse is in the air, B) and up (around the time when the horse has two feet on the ground, D).
”Discussion Questions”
◊ When a car accelerates, why does a bob hanging from the rearview mirror swing toward the back of the car? Is it because a force throws it backward? If so, what force? Similarly, describe what happens in the other cases described above.
◊ Superman is guiding a crippled spaceship into port. The ship's engines are not working. If Superman suddenly changes the direction of his force on the ship, does the ship's velocity vector change suddenly? Its acceleration vector? Its direction of motion?
f / Example 4.
g / The applied force FA pushes the block up the frictionless ramp.
h / Three forces act on the block. Their vector sum is zero.
i / If the block is to move at constant velocity, Newton's first law says that the three force vectors acting on it must add up to zero. To perform vector addition, we put the vectors tip to tail, and in this case we are adding three vectors, so each one's tail goes against the tip of the previous one. Since they are supposed to add up to zero, the third vector's tip must come back to touch the tail of the first vector. They form a triangle, and since the applied force is perpendicular to the normal force, it is a right triangle.
- Discussion question A.
j / Discussion question B.
The Force Vector and Simple Machines
Force is relatively easy to intuit as a vector. The force vector points in the direction in which it is trying to accelerate the object it is acting on.
Since force vectors are so much easier to visualize than acceleration vectors, it is often helpful to first find the direction of the (total) force vector acting on an object, and then use that information to determine the direction of the acceleration vector. Newton's second law, Ftotal=ma, tells us that the two must be in the same direction.
Example 4: A component of a force vector
Figure f, redrawn from a classic 1920 textbook, shows a boy pulling another child on a sled. His force has both a horizontal component and a vertical one, but only the horizontal one accelerates the sled. (The vertical component just partially cancels the force of gravity, causing a decrease in the normal force between the runners and the snow.) There are two triangles in the figure. One triangle's hypotenuse is the rope, and the other's is the magnitude of the force. These triangles are similar, so their internal angles are all the same, but they are not the same triangle. One is a distance triangle, with sides measured in meters, the other a force triangle, with sides in newtons. In both cases, the horizontal leg is 93% as long as the hypotenuse. It does not make sense, however, to compare the sizes of the triangles --- the force triangle is not smaller in any meaningful sense.
Example 5: Pushing a block up a ramp
◊ Figure (a) shows a block being pushed up a frictionless ramp at constant speed by an applied force FA. How much force is required, in terms of the block's mass, m, and the angle of the ramp, θ?
◊ Figure (b) shows the other two forces acting on the block: a normal force, FN, created by the ramp, and the weight force, FW, created by the earth's gravity. Because the block is being pushed up at constant speed, it has zero acceleration, and the total force on it must be zero. From figure (c), we find
Failed to parse (unknown function\vc): \begin{align} |\vc{F}_A| &= |\vc{F}_W| \sin \theta \\ &= mg \sin \theta \qquad . \end{align}
Since the sine is always less than one, the applied force is
always less than mg, i.e., pushing the block up the
ramp is easier than lifting it straight up. This is
presumably the principle on which the pyramids were
constructed: the ancient Egyptians would have had a hard
time applying the forces of enough slaves to equal the full
weight of the huge blocks of stone.
Essentially the same analysis applies to several other simple machines, such as the wedge and the screw.
◊ Solved problem: A cargo plane — problem 9 ◊ Solved problem: The angle of repose — problem 11 ◊ Solved problem: A wagon — problem 10
”Discussion Questions”
◊ The figure shows a block being pressed diagonally upward against a wall, causing it to slide up the wall. Analyze the forces involved, including their directions.
◊ The figure shows a roller coaster car rolling down and then up under the influence of gravity. Sketch the car's velocity vectors and acceleration vectors. Pick an interesting point in the motion and sketch a set of force vectors acting on the car whose vector sum could have resulted in the right acceleration vector.
Calculus With Vectors (optional calculus-based section)
Using the unit vector notation introduced in section 7.4, the definitions of the velocity and acceleration components given in chapter 6 can be translated into calculus notation as
Failed to parse (unknown function\vc): \begin{align} \vc{v} &= \frac{d x}{d t} \hat{\vc{x}} + \frac{d y}{d t} \hat{\vc{y}} + \frac{d z}{d t} \hat{\vc{z}}\\ \intertext{and} \vc{a} &= \frac{d v_x}{d t} \hat{\vc{x}} + \frac{d v_y}{d t} \hat{\vc{y}} + \frac{d v_z}{d t} \hat{\vc{z}} \qquad . \end{align}
To make the notation less cumbersome, we generalize the concept of the derivative to include derivatives of vectors, so that we can abbreviate the above equations as
Failed to parse (unknown function\vc): \begin{align} \vc{v} &= \frac{d \vc{r}}{d t} \\ \intertext{and} \vc{a} &= \frac{d \vc{v}}{d t} \qquad . \end{align}
In words, to take the derivative of a vector, you take the derivatives of its components and make a new vector out of those. This definition means that the derivative of a vector function has the familiar properties
Failed to parse (unknown function\vc): \begin{align} \frac{d(c\vc{f})}{d t} &= c \frac{d(\vc{f})}{d t} \qquad \shoveright{\text{[$c$ is a constant]}} \\ \intertext{and} \frac{d(\vc{f}+\vc{g})}{d t} &= \frac{d(\vc{f})}{d t} + \frac{d(\vc{g})}{d t} \qquad . \end{align}
The integral of a vector is likewise defined as integrating component by component.
Example 6: The second derivative of a vector
◊ Two objects have positions as functions of time given by the equations
Failed to parse (unknown function\vc): \begin{align} \vc{r}_1 &= 3t^2\hat{\vc{x}} + t\hat{\vc{y}} \intertext{and} \vc{r}_2 &= 3t^4\hat{\vc{x}} + t\hat{\vc{y}} \qquad . \end{align}
Find both objects' accelerations using calculus. Could either answer have been found without calculus?
◊ Taking the first derivative of each component, we find
Failed to parse (unknown function\vc): \begin{align} \vc{v}_1 &= 6t\hat{\vc{x}} + \hat{\vc{y}} \\ \vc{v}_2 &= 12t^3\hat{\vc{x}} + \hat{\vc{y}} \qquad , \end{align}
and taking the derivatives again gives acceleration,
Failed to parse (unknown function\vc): \begin{align} \vc{a}_1 &= 6\hat{\vc{x}}\\ \vc{a}_2 &= 36t^2\hat{\vc{x}} \qquad . \end{align}
The first object's acceleration could have been found without calculus, simply by comparing the x and y coordinates with the constant-acceleration equation
. The
second equation, however, isn't just a second-order
polynomial in t, so the acceleration isn't constant, and
we really did need calculus to find the corresponding acceleration.
Example 7: The integral of a vector
◊ Starting from rest, a flying saucer of mass m is observed to vary its propulsion with mathematical precision according to the equation
Failed to parse (unknown function\vc): \vc{F} = bt^{42}\hat{\vc{x}} + ct^{137}\hat{\vc{y}} \qquad . \\
(The aliens inform us that the numbers 42 and 137 have a special religious significance for them.) Find the saucer's velocity as a function of time.
◊ From the given force, we can easily find the acceleration
Failed to parse (unknown function\vc): \begin{align} \vc{a} &= \frac{\vc{F}}{m} \\ &= \frac{b}{m}t^{42}\hat{\vc{x}} + \frac{c}{m}t^{137}\hat{\vc{y}} \qquad . \end{align}
The velocity vector v is the integral with respect to time of the acceleration,
Failed to parse (unknown function\vc): \begin{align} \vc{v} &= \int \vc{a} \:d t \\ &= \int \left( \frac{b}{m}t^{42}\hat{\vc{x}} + \frac{c}{m}t^{137}\hat{\vc{y}} \right) d t \qquad , \\ \intertext{and integrating component by component gives} &= \left( \int \frac{b}{m}t^{42} \: d t \right) \hat{\vc{x}} + \left( \int \frac{c}{m}t^{137} \: d t\right) \hat{\vc{y}}\\ &= \frac{b}{43m}t^{43} \hat{\vc{x}} + \frac{c}{138m}t^{138}\hat{\vc{y}} \qquad , \end{align}
where we have omitted the constants of integration, since the saucer was starting from rest.
Example 8: A fire-extinguisher stunt on ice
◊ Prof. Puerile smuggles a fire extinguisher into a skating rink. Climbing out onto the ice without any skates on, he sits down and pushes off from the wall with his feet, acquiring an initial velocity Failed to parse (unknown function\vc): v_text{o}\hat{\vc{y}} . At t=0, he then discharges the fire extinguisher at a 45-degree angle so that it applies a force to him that is backward and to the left, i.e., along the negative y axis and the positive x axis. The fire extinguisher's force is strong at first, but then dies down according to the equation |F|=b-ct, where b and c are constants. Find the professor's velocity as a function of time.
◊ Measured counterclockwise from the x axis, the angle of the force vector becomes 315°. Breaking the force down into x and y components, we have
Failed to parse (unknown function\vc): \begin{align} F_x &= |\vc{F}| \cos 315\degunit \\ &= (b-ct) \\ F_y &= |\vc{F}| \sin 315\degunit \\ &= (-b+ct) \qquad . \end{align}
In unit vector notation, this is
Failed to parse (unknown function\vc): F = (b-ct)\hat{\vc{x}} + (-b+ct)\hat{\vc{y}} \qquad .
Newton's second law gives
Failed to parse (unknown function\vc): \begin{align} \vc{a} &= \vc{F}/m \\ &= \frac{b-ct}{\sqrt{2}m}\hat{\vc{x}} + \frac{-b+ct}{\sqrt{2}m}\hat{\vc{y}} \qquad . \end{align}
To find the velocity vector as a function of time, we need to integrate the acceleration vector with respect to time,
Failed to parse (unknown function\vc): \begin{align} \vc{v} &= \int \vc{a} \: d t \\ &= \int \left( \frac{b-ct}{\sqrt{2}m} \: \hat{\vc{x}} + \frac{-b+ct}{\sqrt{2}m} \: \hat{\vc{y}} \right) d t \\ &= \frac{1}{\sqrt{2}m} \int \bigl[ (b-ct) \: \hat{\vc{x}} \: + \: (-b+ct) \: \hat{\vc{y}} \bigr] d t \\ \end{align}
A vector function can be integrated component by component, so this can be broken down into two integrals,
Failed to parse (unknown function\vc): \begin{align} \vc{v} &= \frac{\hat{\vc{x}}}{\sqrt{2}m} \int (b-ct) \: d t \: + \: \frac{\hat{\vc{y}}}{\sqrt{2}m}\int (-b+ct) \: d t \\ &= \left(\frac{bt-\frac{1}{2}ct^2}{\sqrt{2}m}+\text{constant \#1}\right)\hat{\vc{x}} + \left(\frac{-bt+\frac{1}{2}ct^2}{\sqrt{2}m}+\text{constant \#2}\right)\hat{\vc{y}} \end{align}
Here the physical significance of the two constants of integration is that they give the initial velocity. Constant
- 1 is therefore zero, and constant #2 must equal
vo. The final result is
Failed to parse (unknown function\vc): \vc{v} = \left(\frac{bt-\frac{1}{2}ct^2}{\sqrt{2}m}\right)\hat{\vc{x}} + \left(\frac{-bt+\frac{1}{2}ct^2}{\sqrt{2}m}+v_text{o}\right)\hat{\vc{y}} \qquad .
Summary
Summary
[] The velocity vector points in the direction of the object's motion. Relative motion can be described by vector addition of velocities.
The acceleration vector need not point in the same direction as the object's motion. We use the word “acceleration” to describe any change in an object's velocity vector, which can be either a change in its magnitude or a change in its direction.
An important application of the vector addition of forces is the use of Newton's first law to analyze mechanical systems.
m / Problem 5.
n / Problem 9.
o / Problem 10.
p / Problem 12.
q / Problem 13 (Millikan and Gale, 1920).
r / Problem 15.
Homework Problems
k / Problem 1.
1. A dinosaur fossil is slowly moving down the slope of a
glacier under the influence of wind, rain and gravity. At the same time, the glacier is moving relative to the continent underneath. The dashed lines represent the directions but not the magnitudes of the velocities. Pick a scale, and use graphical addition of vectors to find the magnitude and the direction of the fossil's velocity relative to the continent. You will need a ruler and protractor. (answer check available at lightandmatter.com)
2. Is it possible for a helicopter to have an acceleration
due east and a velocity due west? If so, what would be going on? If not, why not?
3. A bird is initially flying horizontally east at 21.1
m/s, but one second later it has changed direction so that it is flying horizontally and 7° north of east, at the same speed. What are the magnitude and direction of its acceleration vector during that one second time interval? (Assume its acceleration was roughly constant.) (answer check available at lightandmatter.com)
l / Problem 4.
4. A person of mass M stands in the middle of a tightrope,
which is fixed at the ends to two buildings separated by a
horizontal distance L. The rope sags in the middle,
stretching and lengthening the rope slightly.
(a) If the
tightrope walker wants the rope to sag vertically by no more
than a height h, find the minimum tension, T, that the
rope must be able to withstand without breaking, in terms of
h, g, M, and L.(answer check available at lightandmatter.com)
(b) Based on your equation, explain why it
is not possible to get h=0, and give a physical interpretation.
5. Your hand presses a block of mass m against a wall
with a force FH acting at an angle θ, as shown in the figure. Find the
minimum and maximum possible values of |FH| that can keep
the block stationary, in terms of m, g, θ, and
μs, the coefficient of static friction between
the block and the wall.(answer check available at lightandmatter.com)
6. A skier of mass m is coasting down a slope inclined at
an angle θ compared to horizontal. Assume for
simplicity that the treatment of kinetic friction given in
chapter 5 is appropriate here, although a soft and wet
surface actually behaves a little differently. The
coefficient of kinetic friction acting between the skis and
the snow is μk, and in addition the skier experiences
an air friction force of magnitude bv2, where b is a
constant.
(a) Find the maximum speed that the skier will
attain, in terms of the variables m, g, θ, μk, and
b.(answer check available at lightandmatter.com)
(b) For angles below a certain minimum angle θmin,
the equation gives a result that is not mathematically
meaningful. Find an equation for θmin, and give a
physical explanation of what is happening for θ <θmin.
7. A gun is aimed horizontally to the west, and fired at
t=0. The bullet's position vector as a function of time is
Failed to parse (unknown function\vc): \vc{r}=b\hat{\vc{x}}+ct\hat{\vc{y}}+dt^2\hat{\vc{z}}
, where b, c, and d are
positive constants.
(a) What units would
b, c, and d need to have for the equation to make sense?
(b) Find the bullet's velocity and acceleration as functions
of time.
(c) Give physical interpretations of b, c, d,
Failed to parse (unknown function\vc): \hat{\vc{x}}
,
Failed to parse (unknown function\vc): \hat{\vc{y}}
, and
Failed to parse (unknown function\vc): \hat{\vc{z}}
.
∫
8. (solution in the pdf version of the book) Annie Oakley, riding north on horseback at 30 mi/hr,
shoots her rifle, aiming horizontally and to the northeast.
The muzzle speed of the rifle is 140 mi/hr. When the bullet
hits a defenseless fuzzy animal, what is its speed of
impact? Neglect air resistance, and ignore the vertical
motion of the bullet.
9. (solution in the pdf version of the book) A cargo plane has taken off from a tiny airstrip in
the Andes, and is climbing at constant speed, at an angle of
θ=17° with respect to horizontal. Its engines
supply a thrust of Fthrust=200 kN, and the lift from
its wings is Flift=654 kN. Assume that air resistance
(drag) is negligible, so the only forces acting are thrust,
lift, and weight. What is its mass, in kg?
10. A wagon is being pulled at constant speed up a slope
θ by a rope that makes an angle φ with the
vertical.
(a) Assuming negligible friction, show that the
tension in the rope is given by the equation
where FW is the weight force acting on the wagon.
(b)
Interpret this equation in the special cases of φ=0
and φ =180°-θ.(solution in the pdf version of the book)
11. The angle of repose is the maximum slope on which an
object will not slide. On airless, geologically inert bodies
like the moon or an asteroid, the only thing that determines
whether dust or rubble will stay on a slope is whether the
slope is less steep than the angle of repose.
(a) Find an
equation for the angle of repose, deciding for yourself what
are the relevant variables.
(b) On an asteroid, where g
can be thousands of times lower than on Earth, would rubble
be able to lie at a steeper angle of repose?(solution in the pdf version of the book)
12. The figure shows an experiment in which a cart is
released from rest at A, and accelerates down the slope
through a distance x until it passes through a sensor's
light beam. The point of the experiment is to determine the
cart's acceleration. At B, a cardboard vane mounted on the
cart enters the light beam, blocking the light beam, and
starts an electronic timer running. At C, the vane emerges
from the beam, and the timer stops.
(a) Find the final
velocity of the cart in terms of the width w of the vane
and the time tb for which the sensor's light beam was
blocked.(answer check available at lightandmatter.com)
(b) Find the magnitude of the cart's acceleration
in terms of the measurable quantities x, tb, and w.(answer check available at lightandmatter.com)
(c) Analyze the forces in which the cart participates, using
a table in the format introduced in section 5.3. Assume
friction is negligible.
(d) Find a theoretical value for the
acceleration of the cart, which could be compared with the
experimentally observed value extracted in part b. Express
the theoretical value in terms of the angle θ of the
slope, and the strength g of the gravitational field.(answer check available at lightandmatter.com)
13. The figure shows a boy hanging in three positions: (1) with his arms straight up,
(2) with his arms at 45 degrees, and (3) with his arms at 60 degrees with
respect to the vertical. Compare the tension in his arms in the three cases.
14. Driving down a hill inclined at an angle θ with
respect to horizontal, you slam on the brakes to keep from
hitting a deer.
(a) Analyze the forces. (Ignore rolling
resistance and air friction.)
(b) Find the car's maximum
possible deceleration, a (expressed as a positive number),
in terms of g, θ, and the relevant coefficient of
friction.(answer check available at lightandmatter.com)
(c) Explain physically why the car's mass has no
effect on your answer.
(d) Discuss the mathematical behavior
and physical interpretation of your result for negative
values of θ.
(e) Do the same for very large positive
values of θ.
15. The figure shows the path followed by Hurricane Irene in 2005 as it moved north. The dots show the
location of the center of the storm at six-hour intervals, with lighter dots at
the time when the storm reached its greatest intensity. Find the time when the
storm's center had a velocity vector to the northeast and an acceleration vector
to the southeast.
