# Broken leg & pulleys

1. Sep 20, 2015

### stunner5000pt

1. The problem statement, all variables and given/known data
assume that the total downward force exerted at A & B by the leg is 38N. find the weight W, attached on the lower puller

2. Relevant equations
please see attached diagram

3. The attempt at a solution

i have written roman numerals to help determine which equation refers to which point

at point I
$$W = T$$

at point II
in the X direction
$$T \cos 25 + F_{1} = F_{A} cos \theta$$
in the y direction

$$T \sin 25 = F_{A} \sin \theta$$

at point III
in the X direction
$$0=T \cos 20 +T \cos 25$$

in the Y direction

$$T \sin 20 + T \sin 25$$

at point IV

in the X direction

$$T \cos 20 + T \cos 60 = 0$$

in the Y direction

$$-T \sin 20 - T\sin 60 = 0$$

And that the end of this, can we say that T = 38N?

Please let me know if all the equations are correct?
Thank you for your input and assistance!

#### Attached Files:

• ###### leg question.jpg
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2. Sep 23, 2015

### PhanthomJay

Well I'm not sure of this problem but if the lower leg is horizontal and takes only an axial pulling force F1 , then draw the FBD of the pulley at the foot and sum vertical forces to solve for T and thence mg. F-a is given I think as a vert force of 38 N at A so I don't know why you are breaking it into components. You are also missing a tension force in your equation at that pulley.