- #1
Emmanems08
- 1
- 0
This problem is trying to find out how much weight can be placed in the middle of a system before the weights on the outside go over the pulleys.
A diagram of the problem can be found here http://www.physics.umn.edu/classes/...ds/208321-1201Lab3_P4_Equilibrium_Walkway.pdf
We are trying to find d (the distance of the sag) in terms of known quantities. We will know m (the mass of each counterweight), L (the length), and M the mass of the center weight.
So far I have...
∑F=Fc+Fa-Fb=0
ƩFx=Fc cosθ+Fa cosθ=0
∑Fy=Fc sinθ+Fa sinθ-Fb=0
All from the middle point P, where Fb is the force down, Fa is the force up to the left, and Fc is the force up to the right.
sinθ=d/L1 =d/√(d2+(L/2)2)
That check mark is supposed to be a squareroot if that wasn't clear.
I am not sure where to go from here to solve for d.
A diagram of the problem can be found here http://www.physics.umn.edu/classes/...ds/208321-1201Lab3_P4_Equilibrium_Walkway.pdf
We are trying to find d (the distance of the sag) in terms of known quantities. We will know m (the mass of each counterweight), L (the length), and M the mass of the center weight.
So far I have...
∑F=Fc+Fa-Fb=0
ƩFx=Fc cosθ+Fa cosθ=0
∑Fy=Fc sinθ+Fa sinθ-Fb=0
All from the middle point P, where Fb is the force down, Fa is the force up to the left, and Fc is the force up to the right.
sinθ=d/L1 =d/√(d2+(L/2)2)
That check mark is supposed to be a squareroot if that wasn't clear.
I am not sure where to go from here to solve for d.