Solve Forearm Equilibrium: Calculate Force Held by Biceps

[Tina20]

Homework Statement

A forearm can be modeled as a 1.36 kg, 32.8 cm long "beam" (denoted by l) that pivots at the elbow which has a width of 2.26 cm (denoted by w) and is supported by the biceps. How much force must the biceps exert to hold a 540.0 g ball with the forearm parallel to the floor?

The diagram is similar to the one below except that the values on the diagram are not the values pertaining to this question. Use the values stated in the above question. The link below is just a reference as to what the diagram should look like
http://session.masteringphysics.com/problemAsset/1013607/12/jfk.Figure.P08.37.jpg

Homework Equations

Torque = I*alpha
where I is moment of inertia and alpha is angular acceleration

Net torque = 0
Fnet = 0 because the arm is in static equilibrium

The Attempt at a Solution

I don't know how to go about this.
Torque = Ia = 0
Torque = r*F
= 0.0226m *F

Tgravity = Mgx
where M = total mass
g = 9.8
x = distance of centre of mass to axis

Any ideas as to how to solve this? I need to complete it by tonight, so any help would be greatly appreciated!

 

[willem2]

You must first find the torques of the known force on the ball and on the arm around the elbow. The torque from the biceps must make the net torque 0.

 

[Tina20]

Ok,

so Torque of arm = Torque of ball?
r*F = r*F
(0.0226)*F = (0.3054)(9.8)(0.540kg)
F = 71.5N ?

That force is wrong according to the computer. I may have equated it wrong? Any hints you could please give, because I am completely confused.

Tina

 

[willem2]

Ok,

so Torque of arm = Torque of ball?
r*F = r*F
(0.0226)*F = (0.3054)(9.8)(0.540kg)
F = 71.5N ?

That force is wrong according to the computer. I may have equated it wrong? Any hints you could please give, because I am completely confused.

Tina

There's also the torque from the weight of the arm itself