How Do Newton's Laws Apply to a Pulley System with Unequal Masses?

[crhscoog]

Homework Statement

http://img213.imageshack.us/img213/6969/physicsproblemhq3.png

A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on the floor. The opposite end of the rope is held by Student B of mass 60 kg, who is suspended at rest above the floor.

1. Calculate the magnitude of the force exerted by the floor on Student A.

2. Student B now climbs up the rope at a constant acceleration of .25 m/s^2 with respect to the floor. Calculate the tension in the rope while Student B is accelerating.

3. As Student B is accelerating, is Student A pulled upward off the floor? Justify your answer.

4. With what minimum acceleration must Student B climb up the rope to lift Student A upward off the floor?

The Attempt at a Solution

1. Because it is at rest:
T + N = Mg, or N = Mg - T = (70 - 60)g ~ 98 N

I am currently doing 2-4 right now... I'll post what I get after I finish them.

 

[lippyka]

2. T = Mg - N = (70 + 60)g - T = 128g - T = 128(9.8) - T = 1253.4 - T = T = 1253.4 N 3. No, because the tension in the rope is not enough to lift both masses off the ground. 4. T = Mg - N = (70 + 60)g - N = 128g - N = 128(9.8) - N = 1253.4 - N = N = 1253.4 N Acceleration must be greater than or equal to 1253.4/130 = 9.64 m/s^2

 

[baba89]

Your attempt at solving the problem is correct so far. Now, let's apply Newton's second law, F=ma, to calculate the tension in the rope as Student B is accelerating. We know that the net force on Student B is equal to the tension in the rope minus their weight, so we can write the equation as:

T - 60g = 60a

Substituting in the acceleration of 0.25 m/s^2, we get:

T - 60g = 60(0.25)

T = 60(0.25) + 60g

T = 75 + 60g ~ 735.8 N

This is the tension in the rope while Student B is accelerating.

Now, let's consider whether Student A is pulled upward off the floor as Student B accelerates. According to Newton's third law, for every action, there is an equal and opposite reaction. In this case, the action is the force that Student B exerts on the rope as they climb, and the reaction is the force that the rope exerts on Student B. Since the rope is connected to both Student A and Student B, the rope will also exert a force on Student A in the opposite direction. However, since Student A is heavier than Student B, the force exerted on Student A by the rope will not be enough to lift them off the floor. Therefore, Student A will not be pulled upward off the floor as Student B accelerates.

Finally, let's determine the minimum acceleration that Student B must have in order to lift Student A off the floor. We can use the same equation as before, but this time we set T equal to the force exerted by the floor on Student A, which is 98 N. So we have:

98 - 60g = 70a

Substituting in the acceleration of 0.25 m/s^2, we get:

98 - 60g = 70(0.25)

98 - 60g = 17.5

60g = 98 - 17.5

g = 80.5/60 ~ 1.34 m/s^2

Therefore, Student B must have a minimum acceleration of 1.34 m/s^2 in order to lift Student A off the floor. This is because at this acceleration, the tension in the rope will be equal to the force exerted by