Thread: Mechanics problem: finding forces acting on a point

I'm having trouble with this problem:
http://i.imgur.com/0gPKd.jpg
So I presume the force by the arm is acting horizontally, towards the left, but are these the only forces acting on the elbow flexors? Do I need to take into account the resistance of the wrist strap first?

I asked if you needed to know the distance to the elbow flexors, and was giving a response that I originally took to mean no, but now, as I struggle with it, I think it could've meant that you do need to make some assumption there.
Here's an example question that was shown:
http://i.imgur.com/nghwI.jpg
any help would be greatly appreciated

Originally Posted by lupy

I'm having trouble with this problem:
http://i.imgur.com/0gPKd.jpg
So I presume the force by the arm is acting horizontally, towards the left, but are these the only forces acting on the elbow flexors? Do I need to take into account the resistance of the wrist strap first?

I asked if you needed to know the distance to the elbow flexors, and was giving a response that I originally took to mean no, but now, as I struggle with it, I think it could've meant that you do need to make some assumption there.
Here's an example question that was shown:
http://i.imgur.com/nghwI.jpg
any help would be greatly appreciated

1. For further informations see here: Lever - Wikipedia, the free encyclopedia

2. to a): You are dealing with a lever whose pivot is the elbow joint. Use the law of levers to determine the force of the muscle.

3. to b): You are dealing with a lever whose pivot is the arm wrist (I've marked the pivot by dark red dot). Use the law of levers to determine the force which is acting horizontally on the elbow joint. In addition the weight of the arm is acting vertically on the elbow joint (drawn in green). That means you have to use a parallelogram of forces to determine the resultant force on the elbow joint.

oh man,
big thanks; I was way off.

I'm still confused for b: if the pivot is at the wrist, why is the force at the elbow in the same direction as that at the muscle, not the opposite?

EDIT: Sorry, think I get it now; it's the force acting on it, that's why it's in the same direction, right? So it's like the force at the muscle over the distance to the pivot, is equal to the force at the joint over the distance to the pivot, or 175/20 = x/25, to get the 218.75?

And then the parallelogram of forces is just pythagoras with the 2 also?

thanks again!

Originally Posted by lupy

oh man,
big thanks; I was way off.

I'm still confused for b: if the pivot is at the wrist, why is the force at the elbow in the same direction as that at the muscle, not the opposite?

EDIT: Sorry, think I get it now; it's the force acting on it, that's why it's in the same direction, right? So it's like the force at the muscle over the distance to the pivot, is equal to the force at the joint over the distance to the pivot, or 175/20 = x/25, to get the 218.75?

And then the parallelogram of forces is just pythagoras with the 2 also?

thanks again!

Correct!