PSC 151 - Lab 4
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Virtual Dynamics Track
PENCIL
VPL Lab – Dynamics – Freefall – mod Rev 12/19/18
Mod: 2/9/19
1
Physical Science Laboratory: PSC 151 Lab 4
Name Dynamics – Net Force, Equilibrium, and Acceleration
Date P
URPOSE
To learn to apply Newton’s Second Law to systems of masses.
To understand the concept of equilibrium.
To understand how an x-vs-t graph looks like for an accelerating object.
E
QUIPMENT
E
XPLORE THE A
PPARATUS
Open the Virtual Dynamics Track. You’ll 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.
Drag the cart near the left end of the track.
Drag the spool of massless red string somewhere just to the right of the cart.
(Fig. 1a.) Release it by releasing your mouse button. A segment of string will
attach to the cart and find its way over the pulley. A small loop (Fig. 1b.) will form at its lower end.
Drag the mass hanger until its curved handle is a bit above the loop (Fig. 1c). Release your mouse button and the hanger will attach. (Fig. 1d) The cart will take off. It’s alive! Drag the cart back and forth. Everything should work just as you’d expect.
Give it a toss or use Go
→ to launch it with a known initial negative velocity.
Check out the Brakes On/Off toggle button.
Try Reset All.
Reattach the string and hanger.
Turn on the brake and move the cart near the middle of the track. (Just so you can see the hanger.) You can add to, and remove masses from, the cart or hanger as needed. Drag the largest mass, 200 grams, and drop it when it’s somewhat above the base of the spindle of the cart. (Fig. 2) Drag a 50-g mass onto the hanger.
We refer to this group – cart, string, and hangers – as a system
. You can actually call any group of objects a system – the solar system for example. We’re going to investigate how our cart, string, and hanger move together as a system. The massless string transmits forces between the cart and hanger.
Figure 1a
1b
1c
1d
Figure 2
Toss the cart around and notice how it moves. Experiment with a variety of masses on both the cart and hanger. Investigate what determines the system’s acceleration. You know how to add masses. You can remove them just by clicking. Try this. Turn on the brakes to hold the cart still and then add two masses to the cart and to each hanger. Click once on each of the three. The top mass will disappear in each case. Another three clicks will empty them all. Clicking on an empty hanger will remove it. The Reset All button will then remove the massless string.
VPL Lab – Dynamics – Freefall – mod 2
Rev 12/19/18
Mod: 2/9/19
Physical Science Laboratory: PSC 151 Lab 4
We’ll now use this and other arrangements of masses to look at various systems, how they move, and why. We’ll quantify what we find using Newton’s second law, F= ma
, where F
is the net, external force on the system
, m is the total mass of the system, and
a
is the acceleration of the system
.
Restore your system to its simplest configuration – empty cart, string, and empty hanger. Turn on the brake and move the cart
near
the left bumper.
P
ROCEDURE
IA. Effect of mass on motion
NOTE: For many parts of this lab, you may want to put the brakes on the cart when you move it to make sure that the cart starts at rest and that you are not providing it with an initial velocity.
1.
Move the cart with strings and hangers attached to the center of the track. Turn off the brakes.
a.
What happens to the cart? The cart does not move.
b.
Why does this happen? The strings and hangers on each end have equal weight so the force acting on the cart is equal to 0.
2.
Add 10g mass to the hanger on the right side. Move the cart back to the center and then turn off the brakes.
a.
What happens to the cart? The cart moves to the right.
b.
Why does this happen? More weight is on the right side so the cart moves to the right.
3.
Add 50g mass to the hanger on the left side. Move the cart back to the center and then turn off the brakes.
a.
What happens to the cart? The cart moves to the left.
b.
Why does this happen? There is more weight on the left side so the cart moves left.
c.
How does this compare to what you observed in IA2? The cart moves to the side that has more weight acting on the cart.
4.
Add 40g (2x 20g) mass to the hanger on the right side. Move the cart back to the center and then turn off the
brakes.
a.
What happens to the cart? The cart does not move.
b.
Why does this happen? Both sides have equal weight acting on the cart.
c.
How does this compare to what you observed in IA2 and IA3
? For IA4, the masses were equal after we added the 10g to the right (for IA2), 50g to the left (for IA3), and 40g to the 10g on the right (for IA4). After we added the weight to each side, the cart moved into the direction with the most weight. This happened until the weights are equal.
5.
Experiment with different masses on either side of the cart.
a.
In general, what do you notice about the relationship between the mass on either side of the cart and the motion of the cart? The cart will move in the direction of the side with the most mass. The more mass on one side the faster the cart will move in that direction.
b.
What force is responsible? The weight in the form of the weights and the hanger are responsible
for the motion of the cart.
IB. Equilibrium
1.
Remove all masses from the hangers (remember that you click on the mass to remove it) and return the cart to the center of the track with the brakes on.
2.
Calculate the net force on the cart.
a.
The force pulling the cart to the left is the force on the left side of the cart. At this moment is it just hanger (with a mass of 50g) pulling down on the string which pulls the cart to the left. [Note that we
have to convert to kg and 50 g = 0.05 kg]
VPL Lab – Dynamics – Freefall – mod 3
Rev 12/19/18
Mod: 2/9/19
Physical Science Laboratory: PSC 151 Lab 4
F
left
= m
left x g = (0.050kg) x (9.81m/s
2
) = 0.49 N
b.
Similarly, the hanger on the right pulls the cart down with the same force. F
right
= m
right x g = (0.050kg) x (9.81m/s
2
) = 0.49 N
c.
The net force is the difference between these two forces. It does not matter which way we do the subtraction, so we’ll do F
right
-F
left
. If we get a positive number, the net force will pull to the right. If we get a negative number, the net force pulls to the left. If we get 0, then the forces are balanced and
the system is in equilibrium.
F
net = F
right
-F
left
= 0.49 N – 0.49 N = 0 N
d.
Given there is no net force on the cart, what should happen when you release the brakes? The cart will not move.
e.
Release the brakes. Were you right? Yes
3.
Return the cart to the center of the track and turn the brakes on. Then click the track and drag it such that it is at an angle of 3.9
º
. a.
What do you think will happen to the cart when you release the brakes? The cart will move to the left because the track is tilted down on the left. The mass and gravity will act together to move
the cart down the track. _______
b.
Release the brakes. Were you right? Yes
c.
What has changed between now and IB2? The track is tilted down to the left. This allows gravity
to act as well.
4.
Gravity wants to pull objects straight down. Each of the hangers is connected by a string over a pulley, so they always pull left or right. They are still balanced as before the track was placed at an angle. We ignored this in the previous part, but gravity also pulls down the cart. When the track is level, there is no effect and so we ignored it. Now, however, because the track is slanted, when gravity pulls straight down on the cart, the cart slides a little down and to the left. This means (a part of) the weight of the cart now pulls to the left as an additional force. 5.
Turn the brakes on and move the cart back to the center of the track. Add 20g mass to the hanger on the right and release the brakes.
a.
What happens to the cart? The cart moves up the track to the right.
b.
Between this and part 3, what can you say about the relative strength of the additional force from the weight of the cart when it is at an angle? The additional force on the right side was enough to
offset the angle of the track and the weight of the hanger. If there is more force on the side of the track that is higher, then the cart will move up the track.
6.
Turn the brakes on and move the cart back to the center of the track. Add 60g mass to the cart itself and release the brakes.
a.
What happens to the cart? The cart moves down the track to the left.
b.
Why does this happen? The mass on the cart acts with gravity and moves to the left because the
track is at an angle.
c.
Calculate the net force on the cart:
F
left
= F
hanger
+ F
cart
F
left = (0.05 kg)(9.81 m/s
2
) + (0.310kg)x(9.81m/s
2
)x sin(3.9
º
) [Note: sin(3.7
º
) = 0.0645]
F
left
= 0.491 N + 0.207 N = 1.188 N
F
right
= F
hanger
+ F
added mass
F
right
= (0.050 kg)(9.81 m/s
2
) + (0.020 kg)x(9.81m/s
2
) = 0.491 N + 0.196 N
F
right
= 1.177N
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VPL Lab – Dynamics – Freefall – mod 4
Rev 12/19/18
Mod: 2/9/19
Physical Science Laboratory: PSC 151 Lab 4
F
net
= F
right
– F
left
= 0.011 N
d.
How does the value you found for F
net
compare to 0? The net value is less than 0. The force shows
movement to the left.
________________________________________________________________________________
e.
Does this explain your answer in part a? Why or why not? It does because there is force still acting on the cart. The cart is moving to the left because the net force is negative.
7.
Try different combinations of masses and angles and observe the motion of the cart. In general, what can you say about the effect each variable (mass on left, mass on cart, mass on right, angle) has on the motion of
the cart and the forces involved? With the track tilted, gravity will act with the side that is tilted down as well as the mass on the cart. The more mass added to the tilted side and the cart, the more the cart will move to that tilted side. If the mass on the higher side is greater than the mass on the cart, gravity, and the mass on the lower side then the cart will move higher
.
IC. Acceleration
1.
Reset the system to the following:
a.
Brakes on.
b.
Track at an angle of 3.7
º
c.
Cart at the bottom left
d.
String and hangers on both left and right
e.
Mass on cart of 50g (total of 300g)
f.
Mass on right hanger of 20g (total of 70g)
g.
T
max
to at least 25 s
2.
Given that the settings are identical to the equilibrium position in part IB, with a slightly smaller force pulling to the left, what do you think will happen when you release the brakes? The cart will move to the right.
3.
Release the brakes. What do you observe? Focus on how the behavior of the cart is at the beginning and how it is at the end of its motion. The cart moves slowly at first towards the right before it speeds up a little until reaching the end. 4.
Repeat the set up but turn on the sensor before you release the breaks. Include the x-vs-t graph here:
PLACE GRAPH HERE
VPL Lab – Dynamics – Freefall – mod 5
Rev 12/19/18
Mod: 2/9/19
Physical Science Laboratory: PSC 151 Lab 4
5.
We saw in the last few labs that for x-vs-t graphs, a horizontal line meant the cart was not moving and a straight line means the cart is moving with a constant velocity. When the line changes as above, the cart is accelerating
. A net force causes an acceleration.
6. Calculate F
net
that caused the acceleration with ramp angle at 3.7
o
shown on Fig. 1 graph:
F
left
= 0.49 N + (0.30 kg)(9.81 m/s
2
)(sin3.7
o
) = 0.49 N + 0.190 N = 0.680 N
F
right
= (0.070 kg)(91.81 m/s
2
) = 0.687 N
F
net
= F
right
– F
left
= 0.687 N - 0.680 N = 0.007 N
7. What does the net force you calculated say about the shape of the acceleration graph?
8. Try different masses and angles (use values below). Include the x-vs-t graphs under the given data:
m
L
= 50 g; M
C
= 300g; m
R
= 70 g;
ramp
= 3.6
o
m
L
= 50 g; M
C t
= 300 g; m
R
= 70 g;
ramp
= 3.8
o
PLACE GRAPH HERE PLACE GRAPH HERE
Fig. 2 Acceleration Graph: Ramp angle = 3.6
o. Fig. 3 Acceleration Graph: Ramp angle = 3.8
o
m
L
= 50 g; M
C
= 300 g; m
R
= 70 g;
ramp
= 3.9
o
m
L
= 0.050 kg; M
C
= 0.300 kg; m
R
= 0.090 kg
PLACE GRAPH HERE PLACE GRAPH HERE
VPL Lab – Dynamics – Freefall – mod 6
Rev 12/19/18
Mod: 2/9/19
Physical Science Laboratory: PSC 151 Lab 4
Fig. 4 Acceleration graph: Ramp angle = 3.9
o
Fig. 5 Acceleration graph: Ramp angle = 3.9
o
9. Calculate the F
net
that caused the shape of each acceleration graph
9a. Calculate F
net
that shaped Fig. 2 with m
L
= 0.05 kg; M
C
= 0.300 kg; m
R
= 0.070 kg; ramp angle
ramp
=3.6
o
F
left
= 0.49 N + (0.30 kg)(9.81 m/s
2
)(sin3.6
o
) = 0.49 N + 0.185 N = 0.675 N
F
right
= (0.070 kg)(9.81 m/s
2
) = 0.687 N
F
net
= F
right
– F
left
= 0.687 N - 0.675 N = 0.012 N
9b. Calculate F
net
that shaped Fig. 3 with m
L
= 0.05 kg; M
C
= 0.300 kg; m
R
= 0.070 kg; ramp angle
ramp
= 3.8
o
F
left
= 0.49 N + (0.30 kg)(9.81 m/s
2
)(sin3.8
o
) = 0.49 N + 0.195 N = 0.685 N
F
right
= (0.070 kg)(9.81 m/s
2
) = 0.687 F
net
= F
right
– F
left
= 0.687 N - 0.685 N = 0.001 N
9c. Calculate F
net
that shaped Fig. 4 with m
L
= 0.05 kg; M
C
= 0.300 kg; m
R
= 0.070 kg; ramp angle
ramp
= 3.9
o
F
left = 0.49 N + (0.300 kg)x(9.81 m/s
2
)x sin(3.9º) = 0.49 N + 0.2001 N = 0.690
F
right
= (0.070 kg)(9.81 m/s
2
) = 0.687 N
F
net
= -F
left
+ F
right
= -0.690 N + 0.687 N = - 0.002 N
9d. Calculate F
net
that shaped Fig. 5 with m
L
= 0.05 kg; M
C
= 0.300 kg; m
R
= 0.090 kg; ramp angle
ramp
= 3.9
o
, F
left
= 0.49 N + (0.30 kg)(9.81 m/s
2
)(sin3.9
o
) = 0.49 N + 0.207 N = 0.697 N
F
right
= (0.090 kg)(9.81 m/s
2
) = 0.883 F
net
= F
right
– F
left
= 0.883 N - 0.697 N = 0.186 N
10. What effects contribute to the greatest acceleration as see on the graphs and on F
net
calculations? ________________________________________________________________________________
________________________________________________________________________________
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