How Does Height Influence Torque in Static Equilibrium?

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So with this problem, I have been thinking about it. Since it is in equilibrium, I know the sum of forces and torques will be equal to zero. However, I'm not sure how the height, h, fits into this. I know the weights have a force and are a distance away from that fulcrum, d and 2d, so I could see those as torques, and the 800 N one at 2d should counteract the 200 N one at d, but I just don't see how h fits into this, which is what we aim to find.
 

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Woopydalan said:
So with this problem, I have been thinking about it. Since it is in equilibrium, I know the sum of forces and torques will be equal to zero. However, I'm not sure how the height, h, fits into this. I know the weights have a force and are a distance away from that fulcrum, d and 2d, so I could see those as torques, and the 800 N one at 2d should counteract the 200 N one at d, but I just don't see how h fits into this, which is what we aim to find.

I am not sure but try considering h as the height of rightmost pulley.
 
This is not a rigid-body problem. The masses are connected to a string, and the tension of the string counteracts with the weights. Draw the free body diagram for both weights.
ehild
 
can you explain to me the difference between a ''rigid-body problem'' and a non-rigid body problem? Also, with my free body diagram, I see the tensions are to counteract the weights, but I have no other equations. How does d come into play with the problem in order to get closer to solving for h?
 
Woopydalan said:
can you explain to me the difference between a ''rigid-body problem'' and a non-rigid body problem? Also, with my free body diagram, I see the tensions are to counteract the weights, but I have no other equations. How does d come into play with the problem in order to get closer to solving for h?

The string and the two weights do not make a rigid body, that keeps shape during motion.
The forces of the string have both horizontal and vertical components. For equilibrium, both components of the net force have to cancel at each weight. The components depend on the angle the string makes with the horizontal. How does that angle depend on h and d?
You can take the tension equal everywhere along the string. That gives two equations for the unknown tension and h.

ehild
 
Your solution is not correct. The tension acts along the string at the point where it is attached to the object. The tension is the same along the whole string, at both sides of the pulley. If there are two strings, both exert force. ehild
 
When you say two strings, are you referring to the string attached to the 200 N crate? it looks like there is only 1 string throughout the whole drawing unless you consider that one. If it is a separate string, is its tension separate from the one connecting the two pulleys?
 
Woopydalan said:
When you say two strings, are you referring to the string attached to the 200 N crate? it looks like there is only 1 string throughout the whole drawing unless you consider that one. If it is a separate string, is its tension separate from the one connecting the two pulleys?

You are right, it is the same string at both sides of the middle pulley, but two pieces, and the same tension in both. Both pieces act on the pulley and balance the force of the vertical string attached to the 200 N crate.
The tension in the vertical string balances the weight of 200 N. ehild
 
Ok, now I think I understand what you wrote. I think this one is a good one, I finally considered the string on both sides of the middle pulley
 

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