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Honors Lab 4.5–Freefall, Apparent Weight, and Friction

Purpose

To investigate the vector nature of forces
To practice the use free-body diagrams (FBDs)
To learn to apply Newton’s Second Law to systems of masses
To develop a clear concept of the idea of apparent weight
To explore the effect of kinetic friction on motion

Equipment

Virtual Dynamics Track

PENCIL

Explore the Apparatus

Before starting this activity, watch the introductory video in the eBook at 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 50g. 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 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.) /
Figure 2

We’ll now use this and other arrangements of masses to look at various systems, how they move, and why. The why part will always come from Newton’s second law, , where is the net, external force on the system, m is the mass of the system, and is the acceleration of the system.

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

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.
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.
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 inoursimplified, linear representation the net force is just to the right. / / (Actual)

(Simplified) / (Simplified)
Figure (6a) / (6b) / (6c)

From Newton’s second law, , we have

ΣF = (Wr – T) + T = Wr

Wr = (mhanger + mcart) a

mhanger g = (mhanger + mcart) a

or

Substituting in .050kg for mh, .250 kg for mc, and 9.80m/s2 for g, we have a system acceleration of

=1.63 m/s2

If we now add the extra 200 g to the cart we get

=.98 m/s2

which as we observed is smaller than before. That’s because there’s the same external force, Wh accelerating more mass.

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 allposition 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 isaZombie 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 nearthe left end of the track. Now click on the Z-cart icon below the track. The cart will become translucentindicating that it has become massless. Zoom! Clickingthe Z-carticon again willrestore 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 colorand 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 freelyfallinghanger. Find itsaccelerationas follows.

  1. Set Tmax to 5 seconds.
  2. Set Recoil to zero to eliminate bouncing.
  3. [Normal mode](Not Z-mode.)Empty cart and hanger. Cartnear the left end of the track with the brake on.
  4. [Motion Sensor on]
  5. [Z mode]Click on the Z-icon. The cart should quickly accelerate to the right end of the track.
  6. [Motion Sensor Off]

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 Δt2Equation1

Wecan 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 firstdata pairafter 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.

  1. Record this first data pair as x1, and t1 in the table.
  2. 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.
  3. Calculate and record Δx, and Δt.
  4. Calculate and record the acceleration in Table 1using Equation 1.

Table 1Free-fall acceleration
Mass, m (kg) / Position, x (m) / Time, t (s) / Δx (m) / Δt (s) / a (m/s2)
.050 / x1= / t1=
x2= / t2=
.100 / x3= / t3=
x4= / t4=

4.For your system (mass hanger), m = kg,ΣF = mg = N,predicted a = F/m = m/s2

Your experimental and predicted values should correspond fairly well. If not, be sure to try again. Actually the experimental value in the table should be a bit larger than the theoretical value since t1 was recorded a bit after the cart started so Δx is a bit small.

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 sincethe 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.)

5.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.)

  1. 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.
  2. 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 forceacting 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.
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.
Add 50 grams to each hanger.Your mass (mr) is 100 g! /
Figure 7

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.