Download III. Apparent Weight - KET Virtual Physics Labs

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Honors Lab 4.5 – Freefall, Apparent Weight, and Friction
Purpose
To investigate the vector nature of forces
To develop a clear concept of the idea of apparent weight
To practice the use free-body diagrams (FBDs)
To explore the effect of kinetic friction on motion
To learn to apply Newton’s Second Law to systems of masses
Equipment
Virtual Dynamics Track
PENCIL
Explore the Apparatus
Before starting this activity, watch the introductory video in the eBook at http://www.dl.ket.org/ePhysics/ap-hon.html.
There’s some information that doesn’t apply to this Honors lab, but you’ll need to watch it all anyway. Sorry.
Open the Virtual Dynamics Lab on the website. You should see the low-friction track and cart at the top of the screen. At the
bottom you’ll see a roll of massless string, several masses and a mass hanger. Roll your pointer over each of these to view the
behavior of each. Note the values of the masses. Also note that the empty hanger’s mass is 50 g. If you move your pointer
over the cart you’ll see that its mass is 250 g.
Let’s take a trial run with the apparatus. Remember, everyone needs to take a turn at this.

Turn on the cart brake (stop sign) and then move the cart to the middle of the track.

Drag the roll of massless red string up to the cart and release it with your pointer (mouse button up) somewhere just
to the right of the cart. (Fig. 1a.) You should see a short segment of string connecting the cart to the pointer.
(Fig. 1b.) Without pressing any mouse buttons, move your pointer to the right. The string will follow.
Figure 1a
1b
1c
1d
1e
1f
1f
1g
Continue until your pointer passes the pulley by a bit. (Fig. 1c) Now move downward. When the scissors appear,
click with your mouse. (Fig. 1d) The string will extend downward and a loop will appear at the end. (Fig. 1e)

Drag the mass hanger until its curved handle is a bit above the loop and release (Fig. 1f). The hanger will attach.
(Fig. 1g) Drag the cart back and forth. It’s alive! Everything should work just as you’d expect.
Now turn off the brake and explore the cart’s behavior. It even bounces a little off the ends of the track.
We refer to this group – cart, string, and hangers – as a system. You can actually call any group of objects a system
– solar system, digestive system, etc. In this case, since the objects move together as a group, we’ll focus on their
common motion and look at them as a system.

Add some mass to the cart as follows.
Turn on the brake. Drag the largest mass, 200 grams, and drop it when it’s
somewhat above the base of the spindle of the cart. If you miss, just try again.
We now have a cart with a total mass of 450 g, a 50-g mass hanger, and a
massless string connecting them.
Now observe the more massive system’s motion. You should notice that it
moves with less acceleration than before when its mass was 200 g less. We now
have the (weight of the) same 50-g mass, moving a larger total mass (500 g.)
Honors Lab 4.5 - Dynamics
1
Figure 2
May 12, 2017
We’ll now use this and other arrangements of masses to look at various systems, how they move, and why. The why part will
⃑ = 𝒎𝒂
⃑ is the net, external force on the system, m is the mass of
⃑ , where 𝚺𝑭
always come from Newton’s second law, 𝚺𝑭
⃑ is the acceleration of the system.
the system, and 𝒂
I. Drawing force vectors – “A picture’s worth a thousand points.”
We’ve found that writing down what we know and want to know is essential to successful problem-solving. But with forces,
there’s the added directional information that’s best described with a figure. Figure 2 is a nice representation of what our
apparatus looks like, but it doesn’t show us the forces. Drawing force vectors can make it easier to identify the forces acting
on systems. They also guide us in setting up the equation ΣF = ma.
Click on “Remove Masses” to remove the 200-g mass from the cart. Turn the brake on and move the cart near the left
bumper.
Click
to turn on Dynamic Vectors. The vectors in the middle of the track represent the forces on the cart as well
as its velocity and acceleration. (You’ll generally want to leave Dynamic Vectors on most of the time in this lab.)
In the vertical you should see the normal force, N, and the weight, W, which are equal and
opposite. Thus there is no vertical acceleration by the cart, and there won’t ever be with a level
track. (N and W aren’t to the same scale as the horizontal forces because they’re so much larger.)
In the horizontal you should see the friction force, F(f), and the tension in the (right) string, Tr.
They are also equal but opposite. Thus there is currently no acceleration in the horizontal.
In the horizontal there are also vectors for velocity, Fnet, a, W(x), and Tl. Since these are all zero,
you just see their names.
The forces on the mass hanger are also shown. You just see the opposite and equal forces Tr, and
Wr. Wr is the weight of the right mass hanger.
For simplicity, let’s turn some of them off. Click the check boxes on the left side of the screen to
make it look like Figure 3.
Figure 3
You should now have something like this.
Figure 4
Forces
The friction force, F(f), is only present because of the brake. When you turn it off for an actual run the vectors should like
Figure 5. This means the brake’s force will not be involved in any of what follows.
Figure 5
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The hanger’s weight, Wr, is gravity’s pull downward on the hanger, so it’s an external force. That is, it’s a force exerted on
the system from outside. (Gravity is a very mysterious force that’s still not clearly understood, but we speak of it as a force
acting on the hanger by the earth below it.)
The two tension forces, Tr, are internal to our cart-string-hanger system, so they don’t affect its motion. Got that? The
system includes the string. The pulley changes the direction of the tension force but otherwise doesn’t affect it.
So the only external force acting is Wr. So the net force, Fnet is equal to Wr. Note that their lengths are equal.
Masses
The mass being accelerated is the sum of the cart’s mass and the hanger’s mass. After all, it’s all accelerating.
Free-body diagram
We can clarify this a bit with a free-body diagram (FBD) which represents the mass being accelerated as a dot and each force
as a vector arrow. We’ll draw an FBD for each object and then one for the whole system. The vector arrows on our apparatus
actually create the diagrams for us on each individual part of the system. Since the pulley just changes the direction of the
force we can simplify our drawings by treating the system is if it were entirely horizontal with Wr acting to the right on the
hanger. (Both representations are shown in figure 6b.)
Note: We will ignore forces not acting along the direction of motion. The direction of motion is in the horizontal for the cart,
and in the vertical for the hanger.
a. The Cart. Once the brake is released, the only force acting on the
cart is Tr, the tension in the right string. Hence its acceleration will be
to the right.
(Actual)
b. The Hanger. The two forces on the hanger, Wr, and Tr, exert
opposite, but not equal forces on the hanger. As a result its
acceleration is in the direction of the larger force, Wr.
(Simplified)
c. The System. The net force acts on the total system mass. The
pulley just changes the direction of the motion of the different parts
of the system. So the net force doesn’t have a specific direction, but
in our simplified, linear representation the net force is just to the
right.
Figure (6a)
(6b)
(Simplified)
(6c)
⃑ = 𝒎𝒂
⃑ , we have
From Newton’s second law, 𝚺𝑭
ΣF = (Wr – T) + T = Wr
Wr = (mhanger + mcart) a
mhanger g = (mhanger + mcart) a
or
𝑎=
𝑚ℎ 𝑔
𝑚ℎ + 𝑚 𝑐
Substituting in .050kg for mh, .250 kg for mc, and 9.80m/s2 for g, we have a system acceleration of
𝑎=
.050 𝑘𝑔𝑥9.8𝑚/𝑠 2
.050 𝑘𝑔+ .250 𝑘𝑔
= 1.63 m/s2
If we now add the extra 200 g to the cart we get
𝑎=
.050 𝑘𝑔𝑥9.8𝑚/𝑠 2
.050 𝑘𝑔+ .450 𝑘𝑔
= .98 m/s2
which as we observed is smaller than before. That’s because there’s the same external force, W h accelerating more mass.
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Don’t panic! That’s all just a = F/m where F is the weight of the hanger and m is the total mass, including the hanger. And
does the .98 m/s2 ring any bells? That’s g/10. The system is accelerating at 1/10 the acceleration it would have if you just
dropped it. That’s because the weight of only 1/10 of the system’s mass (.050 kg g) is “pulling” the entire system (500 g).
What if you just snipped the string and left the cart behind? That’s called freefall.
IIa. Freefall Acceleration
We’ve found that all bodies in free fall have the same downward acceleration, g. Let’s use our apparatus to find the value of g
for a falling mass hanger. We need a system of one object – the mass hanger. Hmm. You can’t have a mass hanger without
a string to attach it to. And all position measurements are associated with the cart, not the hanger. So we need a cart, but then
again we don’t. We need a cart that won’t interfere with the falling hanger; a sort of vacant, empty, soulless, massless cart.
What we need is a Zombie CART! Or as Dr. Seuss would call it, little cart Z.
Set up the dynamics track with an empty cart and hanger. Drag the cart to the left and let it go.
That looks OK, but as noted, the hanger is not really free-falling? It’s dragging the cart behind it.
Put the brake on and move the cart near the left end of the track. Now click on the Z-cart icon below the track. The cart will
become translucent indicating that it has become massless. Zoom! Clicking the Z-cart icon again will restore its mass. Repeat
in and out of Z-mode to convince yourself that the whole system seems to move with a greater acceleration, possibly equal to
g, when the cart’s mass is eliminated. That’s because the system here is really just the mass hanger. So the Z-car is doing just
what we want it to do.
Try this. Turn on the motion sensor and record the motion in normal mode (250-g cart) in one color. Then select another line
color and try a Z-mode run. Do the graphs show different accelerations? They should.
1. You may have noticed that the brake doesn’t work in Z-mode. Check it out. Why doesn’t it work in Z-mode?
2. Since we can eliminate the mass of the cart with z-mode, we can now look at the motion of the freely falling hanger. Find
its acceleration as follows.
a.
Set Tmax to 5 seconds.
b.
Set Recoil to zero to eliminate bouncing.
c.
[Normal mode]
d.
[Motion Sensor on]
e.
[Z mode]
f.
[Motion Sensor Off]
(Not Z-mode.) Empty cart and hanger. Cart near the left end of the track with the brake on.
Click on the Z-icon. The cart should quickly accelerate to the right end of the track.
3. Let’s find the approximate acceleration using one of our kinematics equations. The cart and hanger travel with identical
motions. The pulley just puts a 90° turn in the middle. We’ll find the acceleration of the cart which is the same as that of the
hanger. For the cart we can say
Δx = vot + ½ a Δt2
But if we choose our initial time, t1, as just about when we turn off the brake, vo ≈ 0. So
Δx = ½ a Δt2
Equation 1
We can get Δx and Δt from the on-screen data table. Look at the position values. They’re all the same until you turned off the
brake. Find the first data pair after the first change in position. Ex. 0.14, 0.14, … ,0.14, 0.15  that’s the one we want.
This will be our initial value, when vo ≈ 0. The cart’s motion actually started between this instant and the one before it, but
we’ll get a fairly accurate number if we ignore this error.
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a.
Record this first data pair as x1, and t1 in the table.
b.
Scroll through the data to find a point somewhat before the cart hit the right bumper. The first position reading
greater than 1.8 m is a good spot. Record the data for this point as x2, and t2 in the table.
c.
Calculate and record Δx, and Δt.
d.
Calculate and record the acceleration in Table 1 using Equation 1.
Table 1 Free-fall acceleration
Mass, m (kg)
.050
.100
Position, x (m)
Time, t (s)
x1 =
t1 =
x2 =
t2 =
x3 =
t3 =
x4 =
t4 =
4. For your system (mass hanger), m =
kg,
ΣF = mg =
Δx (m)
N,
Δt (s)
a (m/s2)
predicted a = F/m =
m/s2
Your experimental and predicted values should correspond fairly well. Actually the experimental value in the table may be as
much as 15% larger than the theoretical value since t1 was recorded a bit after the cart started, so Δx is a bit small. But the
important thing to notice is how well the two trials compare. You should find that the acceleration is independent of mass.
IIb. Freefall – the effect of the weight of the falling object on its acceleration
Throughout most of recorded history it was believed that the heavier a body is the faster it will fall. (Those of that opinion
also didn’t quibble over the definition of the terms “fast” and “slow.”) F = ma would seem to bear their theory out since the
acceleration is directly proportional to F. Let’s test this old theory.
In part IIa) we observed just the fall of the mass hanger. Let’s make it heavier to see if it goes “faster.” (Greater acceleration.)
1. Double the weight of the hanger by adding a 50-g mass to it. (You’ll have to turn off Z-mode temporarily to allow you to
put mass on the hanger.)
a.
Change graph colors so that you can see the new graph along with the previous one. Take your data and record it as
before but in the .100-kg part of Table 1. Calculate and record the free-fall acceleration of the 100-g mass.
b.
Hmm. Doubling the force did not seem to double the acceleration. If a ∝ F, we must be missing something.
“The
of a body is directly proportional to the net force acting on it.
So if you double the force on it, its acceleration should
.
When you doubled the mass in this case, the acceleration
This is because doubling the weight also doubled the
.
.”
This is a very special relationship that is still not clearly understood. Why should the force of gravity on a body be directly
proportional to its mass (inertia)? Or vice versa? Perhaps this will be solved in your lifetime. Stay tuned!
III. Apparent Weight
In Figure 6b we had two opposing forces. Are they an action-reaction pair? No! And not just because they aren’t equal.
Newton’s 2nd Law: ΣF = ma where ΣF is the sum of the forces on a body with mass m.
Newton’s 3rd Law: Body 1 exerts a force on body 2. Body 2 exerts an equal but opposite on body 1.
FBD’s are about the 2nd Law: It shows the external force on one body or a system.
In figure 6b, the dot represents the weight hanger. There are two external forces on it.
Fig 6b
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The weight, Wr is the force that the earth exerts on the weight hanger. The reaction force is the weight hanger’s
force on the earth which is not shown since it doesn’t act on the hanger.
F(Earth on hanger), Wr = -F(hanger on Earth)
A-R pair #1
Likewise the string exerts an upward tension force on the weight hanger and the weight hanger acts equally
downward on the string.
F(string on hanger), Tr = -F(hanger on string)
A-R pair #2
(6b)
Newton’s second law relates the two forces acting on the hanger, the force on the left sides of the equations.
Tr – Wr = mra
Now suppose that instead of the mass hanger, you were clinging to the end of string. You are the dot in the figure. Gravity
pulls you downward with a force Wr. The string pulls up with a force Tr.
That force Tr is what we call your apparent weight.
If you were standing on a scale in an elevator it would work the same way. It’s the force upward on you needed to hold you
at rest, move you up or down at constant speed, or accelerate you up or down. And it changes depending on the acceleration.
Let’s explore this with our apparatus. To make it more vivid, imagine yourself as the right mass hanger and attached
masses. You’re holding the string to keep from falling.
To make you accelerate up and down we need a hanger on the left side.
3.
Set up the system shown in Figure 7. You’ll want a string and empty
hanger on each side.
You’ll want Z-mode for this activity.
With Dynamic Vectors on you’ll notice a set of check boxes on the left side
of the apparatus that allow you to show and hide vectors. The column of
arrows identifies them by color. This column is also a toggle to show/hide
the check boxes. You might want to hide them.
Figure 7
Add 50 grams to each hanger. Your mass (mr) is 100 g!
4. Let’s look at the initial situation - hanging from the rope at rest. That is, vo = 0.
a = 0 So Tr = Wr
Note the length of these vectors on the right side. They’re equal since a = 0.
5. Now give the cart a good shove in either direction. Go at Vo might be better. (It’s a bit jerky. Vectors are slow to draw.)
5a. How does your apparent weight (Tr) at rest compare to Tr when moving at a constant, non-zero velocity?
Tr(rest) < Tr(constant velocity)
Tr(rest) > Tr(constant velocity)
Tr(rest) = Tr(constant velocity)
6. Now try an upward acceleration. Add 100 grams to the left hanger. You should now have a total of 100 g on the right and
200 g on the left. Drag the cart to near the right end of the track. Note that Tr = Wr when the cart is being held.
6a. When you release the cart, what happens to Tr and Wr? For each, state increases, decreases, or remains the same.
6b. When you release the cart, what happens to your apparent weight? (Same choices.)
6c. When you release the cart, what happens to your actual weight? (Same choices.)
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7. Now try a downward acceleration. Adjust the masses so that the right side is still 100 g and the left side is an empty 50-g
hanger. Drag the cart to near the left end of the track. Note that Tr = Wr when the cart is being held.
7a. When you release the cart, what happens to Tr and Wr? For each, state increases, decreases, or remains the same.
7b. When you release the cart, what happens to your apparent weight?
8. When you accelerate upward your apparent weight
.
9. When you accelerate downward your apparent weight
.
10. When you are at rest or moving at a constant speed your apparent weight equals
IV. Kinetic Friction
And finally, let’s have a quick look at kinetic friction. For this study we’ll just be working with the cart. The only force
affecting its motion will be kinetic friction. So click [Remove All] to remove all the masses and hangers.
So far the only friction force we’ve encountered is the one supplied by the cart brake. It’s much too strong a force for our
purposes. Instead we’ll use a friction pad like the brake. But this one is adjustable. In Figure 8 you see a snapshot of the cart
traveling to the right at velocity vo. A friction force acts to the left, thus slowing the cart down.
In Figure 9 you see the friction pad control. When “wheels” is selected this control show the familiar Vo controls. By
clicking on “friction pad” the controls change to allow you to adjust μk, and μs. We’ll use only kinetic friction which is
controlled by the μk stepper. We’ll start with it set just as it is in the figure. After one trial you’ll try to determine an unknown
kinetic friction coefficient.
1. Set the friction control as shown in figure 9.
2. Turn on the ruler.
3. Move the cart to x = 10.0 cm. (This is the position of the cart mast.)
4. Turn on the dynamic vectors:
5. Click “wheels” and set Vo to 200 cm/s. Then switch back to “friction pad.”
Figure 8
Figure 9
6. [Go at known Vo]
The cart should launch to the right and slow to a stop before reaching the right bumper. If not, retry the preceding
instructions.
Repeat the preceding step as needed to answer the following.
7. Clearly the cart slows down. What about the friction vector, F(f), tells you that why it slows down.
8. What about the velocity vector tells you that it does slow down.
The cart slows down due to the friction force acting on it. From Newton’s second law we know
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ΣF = ma
Ff = ma
Ff = -μk FN = -μk mg
-μk mg = ma
μk = -a/g
Equation 2
So if we knew a, we could divide by g to determine μ k.
How can we find the acceleration, a? If we launch the cart down the track, letting it come to a halt we know
vo = 2.00 m/s
vf = 0.00 m/s
Δx can be found with our ruler
a can be found from a single kinematics equation. You figure that part out.
9. Got it! Take the data you need and record it in Table 2. If your value for μ k is not close to .12, then you’ve done something
wrong with your technique or your calculations. You should get a negative value for the acceleration.
Table 2 Determination of known μk = .12
Show calculations below for a and μk
g = 9.80 m/s2
vo = 2.00
m/s
x1 = .100
m
x2 =
m
Δx =
m
a=
m/s2
μk(experimental) =
(no units)
10. Let’s try an unknown friction coefficient. Change the μ ks? value to 1 using the numeric stepper. You’ll quickly find that
your 2.00 m/s is not a good choice this time. You can change it to whatever you like. But letting the cart go most of the way
along the track will give better results than a short run. Note also that the new μk should be much smaller than the previous
value.
Table 3 Determination of known μks? (1)
Show calculations below for a and μk
g = 9.80 m/s2
vo = 1.30
m/s
x1 = .100
m
x2 =
m
Δx =
m
a=
m/s2
μk(experimental) =
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(no units)
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