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

 

[rwisz]

I would approach this problem by first identifying the key variables and principles involved. In this case, we are dealing with a static equilibrium situation, where the forces acting on the forearm must balance out in order for it to remain still. The key variables are the mass and length of the forearm, the width of the elbow, and the mass of the ball being held. The key principles involved are torque and moment of inertia.

To solve for the force exerted by the biceps, we can use the equation for torque, which is equal to the product of the force and the distance from the pivot point. In this case, the pivot point is the elbow and the distance from the pivot point to the center of mass of the forearm is half of its length, or 16.4 cm.

Using the net torque equation, we know that the torque due to the force exerted by the biceps must be equal and opposite to the torque due to the weight of the ball. This gives us the equation:

Fb * 16.4 cm = (0.54 kg * 9.8 m/s^2) * 16.4 cm

Solving for Fb, we get:

Fb = (0.54 kg * 9.8 m/s^2) * 16.4 cm / 16.4 cm

Fb = 5.29 N

Therefore, the biceps must exert a force of 5.29 N to hold the 540 g ball with the forearm parallel to the floor. It is important to note that this is a simplified model and does not take into account the actual biomechanics and muscle activation involved in holding an object. However, it provides a basic understanding of the forces involved in this scenario.