EK physics/ torque pulley

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The question says that the mass sits in the exact middle of the board, so regardless of how long the board is, the pulley system which is twice as far from the pivot as the mass is will only have to use have the force. Here's a more math based explanation, with x=length of board

Torque from mass=(10kg)(10m/s^2)(x/2)

Torque from Pulley System=(F)(2)(x) (The 2 comes from the mechanical advantage of the pulley)

When you set them equal, the variables cancel and you get F=25N.
 
The question says that the mass sits in the exact middle of the board, so regardless of how long the board is, the pulley system which is twice as far from the pivot as the mass is will only have to use have the force. Here's a more math based explanation, with x=length of board

Torque from mass=(10kg)(10m/s^2)(x/2)

Torque from Pulley System=(F)(2)(x) (The 2 comes from the mechanical advantage of the pulley)

When you set them equal, the variables cancel and you get F=25N.

Can you elaborate further?

I get where the x/2 comes from. It's the lever arm. However, I'm wondering about the 2x in the "torque from pulley system". Why is it not only x? Its lever arm length is x.
 
The extra 2 comes from the mechanical advantage of our pulley. Draw a free body diagram for the pulley system. In this system, if a tension is applied to the end of the string, there will be two upward pointing tension forces on each side in the pulley system, thus when you pull outward with force F, the pulley generates an upward force of 2F.
 
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