Solve Two-Pulley Problem: Tensions, Forces & Diagrams

[scipioaffric]

Homework Statement

A pulley hangs from the ceiling. A string is run over the pulley and one end is attached to the ceiling, while a 10 lb weight hangs from the other end. In between the first pulley and the end of the string that is attacked to the ceiling rest another pulley on the string with a 5 lb weight hanging from it, as in the diagram below. Find the tensions of the various segments of string, in terms of applied force.

http://www.energeia.us/pulley1.png

Homework Equations

T1= ?
T2= ?
T3= ?
T4= ?
W1= 10lbs.
W2= 5lbs.

This is the section I'm having the most trouble with. Which tensions are part of which system? Does T1+T2+T3=15pounds or 5 pounds, does T4 = the weight of the whole system, or only part of it?

The Attempt at a Solution

I can't begin to solve the problem without being able to identify all the systems. Also, I'm not sure how to draw the force diagrams without the systems. Here are some things I think I know about the problem:

There is no normal force counteracting the weights from underneath. All normal forces will counteract the forces operating against them in the system that is suspended from them at the points attached to the ceiling.

so, we have F_n1 at the leftmost ceiling-mounted point, and F_n2 at the rightmost ceiling-mounted point. So, we have a force diagram for each of these with an N axis. The magnitude of the vector for normal force is given as the equal and opposite force of the system hanging from the string.

Other than this, the only thing I'm fairly sure of is that the 10lb weight exerts 10lbs of force downward on the right side of the ceiling mounted pulley, and 10lbs upward on the left side of the pulley.

Someone please help me out, this is the first problem I've seen like this. Thanks.

 

[Doc Al]

Other than this, the only thing I'm fairly sure of is that the 10lb weight exerts 10lbs of force downward on the right side of the ceiling mounted pulley, and 10lbs upward on the left side of the pulley.

Well... not exactly. The 10lb weight would exert 10lbs of force on the rope if there were no acceleration, but that's not the case here. You have to solve for the tension in the rope.

Identify the forces acting on each mass and apply Newton's 2nd law. Combining the two equations (one for each mass) will allow you to solve for the acceleration and rope tension. Hint: The tension is the same throughout the rope.

 

[thomate1]

I would recommend breaking down the problem into smaller systems and then applying Newton's laws of motion to solve for the tensions in each segment of string.

First, let's identify the different systems in this setup. We have the entire system, which includes both weights and the pulleys. Then we have two smaller systems: the first pulley with the 10 lb weight hanging from it, and the second pulley with the 5 lb weight hanging from it.

Next, let's draw free body diagrams for each of these systems. For the entire system, we have two forces acting on it: the weight of the entire system (15 lbs) and the tension in the string connecting the two pulleys (T4). The direction of T4 will be towards the right, since the weight of the system is pulling down on the right side of the pulley.

For the first pulley system, we have three forces acting on it: the weight of the 10 lb weight (10 lbs), the normal force from the ceiling (Fn1), and the tension in the string connecting to the second pulley (T1). The direction of T1 will be towards the left, since the weight of the 10 lb weight is pulling down on the left side of the pulley.

For the second pulley system, we have three forces acting on it: the weight of the 5 lb weight (5 lbs), the normal force from the ceiling (Fn2), and the tension in the string connecting to the first pulley (T2). The direction of T2 will be towards the right, since the weight of the 5 lb weight is pulling down on the right side of the pulley.

Now, we can apply Newton's second law (F=ma) to each of these systems. Since the entire system is not accelerating, the sum of all the forces acting on it must be zero. This means that T4 must equal 15 lbs.

For the first pulley system, the sum of the forces in the y-direction must be zero since it is not accelerating in that direction. This means that Fn1 must equal 10 lbs, since it is the only force acting in the y-direction. In the x-direction, we have T1 acting towards the left and T4 acting towards the right. Since the system is not accelerating in the x-direction, these two forces must be equal and opposite, meaning that