Question

 

PROCEDURE
Part 1
1. Make sure the spring scales are calibrated using the standard masses.
2. Hook the handle of the 5 N spring scale to the hook of the 10 N spring scale.
3. Holding the 10 N spring scale stationary, pull the hook of the 5 N spring scale until the force reads 5 N on it. Record the
force on the 10 N spring scale in Table 3.
4. Repeat Steps 2 and 3 with the 10 N spring scale hanging from the 5 N spring scale. Record the force on the 5 N spring
scale in Table 3.you believe the rocket will be propelled from its original position if you squeeze the bulb. Record your
prediction in Table 5.

Part 2
5. Suspend the 0.5 kg mass in the air using the 10 N spring scale. Record the force on the 10 N spring scale in Table 4.
6. Tie one end of one of the pieces of string to the 0.5 kg mass and the other end to the hook of the 10 N spring scale.
7. Suspend the mass in the air by lifting the 10 N spring scale. Record the force on the 10 N spring scale in Table 4.
8. Untie the end of the string attached to the 0.5 kg mass and tie it to the
hook of the 5 N spring scale.
9. Hook the 0.5 kg mass to the handle of the 5 N spring scale. Suspend the
mass, scales, and string by holding the handle of the 10 N spring scale.
Record the values of the spring scales in Table 4.
10. Secure the pulley on a table top by tying string to one of the hooks. Then,
use masking tape to secure the string to a table top so that the hook on
the top of the pulley lays flat on the side of the table top (Figure 4).
11.Using the mass setup from Step 5, place the string over the pulley by
unhooking one of the spring scales, feeding the string through the pulley
and reattaching the string to the hook of the spring scale (Figure 5).
Figure 4: Pulley setup

12. Hold the 10 N spring scale in place so that the scales and mass are stationary.
Record the values for both spring scales in Table 4.

Experiment 2: Newton’s Third Law and Force Pairs

Data Tables and Post-Lab Assessment

Table 3: Forces on Stationary Spring

Force on Stationary 10 N Spring Scale (N)

5.4 N

Force on Stationary 5N Spring Scale (N)

5.0 N

Table 4: Spring Scale Force Data

Suspension Set Up	Force (N) on 10 N Spring Scale	Force (N) on 5 N Spring Scale
0.5 kg Mass on 10 N Spring Scale	5 N
0.5 kg Mass with String on 10 N Spring Scale	5.2 N
0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale	5.6 N	5 N
0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale on Pulley	6 N	5 N

Post-Lab Questions

1. How did the magnitude of the forces on both spring scales compare after you moved the 10 N spring scale?

 

 

 

2. How did the magnitude of the forces on both spring scales compare after you move the 5 N spring scale?

 

 

 

3. Use Newton’s 3rd Law to explain your observations in Questions 1 and 2.

 

 

 

4. Compare the force on the 10 N spring scale when it was directly attached to the 0.5 kg mass and when there was a string between them.

 

 

5. Compare the force on the two spring scales in Steps 5 and 6. What can you conclude about the tension in a string?

Answer 1

1. When pulling the 10 N scale, the magnitude of the forces in both the scales, 5 N as well as 10 N scale, would be same. Ideally, both scales should read 5 N.
2. When pulling 5 N scale in second part, the magnitude of the forces in both scales, would be same. Ideally, both scales should read 5 N.
3. Newton's third law states that for every action, there would be equal and opposite reaction. This is applicable to the above both situation. That is why in both cases, we should get same, 5 N, readings. No matter what way we arrange the scales or our systems, Newton's law would always hold. In this case, both the scales become one system and force would be same as 5 N. When we pull one scale by 5 N, the other scale would be oppose it by same force but in the opposite direction. Therefore, both the scales would read same reading of 5 N.
4. The reading in scale for both the cases would be same. Here, we assume that the string has negligible mass. If string has some mass, the scale reading would be more than the weight of 0.5 kg mass.
5. Again, the reading in both the scales would be equal. The scale to which the weight is attached would read the weight. Since the weight is stationary, it is balanced by the tension in the string. This tension is transmitted to the other end of the string where second scale is attached. Here, we assume that the string is ideal. So there is no loss of tension in the string. Hence, the second scale would read the same reading as the first scale. However, If string has some mass, the scale reading would be more than the weight of 0.5 kg mass.