Question

How much force is needed from the bicep which is 5 cm from the joint to overcome a 59 kg weight in the hand if the .7m long forearm has a mass of 5kg and the elbow is flexed to 90 degrees?

Answer 1

A diagram would have been really helpful.

This is basically a torque problem. Here we have two force acting down (weight in the hand and weight of forearm)

One force is acting up (force of bicep, we need to find this, Let's call this Fb)

The point of action of force of bicep is 5 cm away from joint.

so, torque about elbow joint will be

Fb * 0.05 ( counter clockwise)

Now,

weight in hand = 59*9.8 = 578.2 N

now this is placed on the hand which is at a distance of 0.7 m from elbow joint ( That's why i said that a diagram would have been helpful. I am assuming that 0.7 m is from the elbow joint to the weight in hand)

so, torque created will be 578.2 * 0.7 (clockwise)

weight of forearm = 5*9.8 = 49 N

Now, treating the forearm as a beam , the weight force will act exactly at the middle which is 0.7 / 2 = 0.35 m from elbow joint

so,

torque = 49*0.35 (clockwise)

so,

we can now equate torques ( counter clockwise = clockwise)

Fb * 0.05 = 49*0.35 + 578.2 * 0.7

Fb = 49*0.35 + 578.2 * 0.7 / 0.05

Now, just solve for Fb

Note - This is the method to solve for force exerted by bicep. Now, if you have a diagram available and then make sure the moment arm of torques are proper ( I mean to say that confirm the distance of each force from elbow joint).

Let me know if you have any doubt.