# Vector Diagram of Forces Acting on a Forearm

1. Oct 1, 2009

### Emethyst

1. The problem statement, all variables and given/known data
A patient in therapy has a forearm that weighs 20.5N and lifts a 112.0N weight. The only other significant forces acting on the forearm come from the biceps muscle (which acts perpendiculary to the forearm) and the force at the elbow. If the biceps produce a pull of 232.0N when the forearm is raised 43 degrees above the horizontal, find the magnitude and direction of the force that the elbow exerts on the forearm.

2. Relevant equations

3. The attempt at a solution
My only problem with this question is drawing the correct vector diagram. I know that the elbow and biceps together will produce a resultant force of 132.5N upwards to keep the forearm balanced (equilibrium), but i'm not sure where to draw these vectors :tongue:. If someone could help me out here it would be greatly appreciated, thanks in advance.

PS: I included a diagram of what I drew, so hopefully it will pop up.

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2. Oct 1, 2009

### kuruman

Your diagram is incomplete. You have not drawn the elbow force that acts at the end of the forearm and the weight of the forearm that acts at the forearm's center of mass, presumably the middle. However, drawing all these in will not result in a solution. You need the point of application of the biceps force which apparently is not given in the statement of the problem. Without it you cannot find an answer because you need to balance torques in addition to balancing forces. You cannot calculate the torque generated by the biceps without knowing where the biceps force is applied.

3. Oct 1, 2009

### Emethyst

No, it's not an equilibrium situation, even though it seems like it, this question was given in my vector section of the course, I haven't gotten to the equilibrium and torque section yet. I need to find the vector diagram to solve for this question because my prof has stated that we should only have to use vectors to solve for this question.

4. Oct 1, 2009

### kuruman

Yes, I see now. You have two equations and two unknowns.

It doesn't matter where you draw the forces, just make sure that you draw all of them and in the right direction as I indicated earlier. Then say that that the sum of all the x-components is zero and the sum of all the y-components is zero.

5. Oct 1, 2009

### Emethyst

Would the elbow force be acting on the horizontal axis, or would this be incorrect?

6. Oct 2, 2009

### kuruman

The elbow force has two components, one horizontal and one vertical. You can find what these are by saying that the sum of all the forces in the horizontal direction is zero and the sum of all the forces in the vertical direction is zero. You have two equations and two unknowns.

7. Oct 3, 2009

### Emethyst

Ok i'm still having trouble. I did the sums for the vertical and horziontal, but came out with 171.0N roughly instead of the correct answer of 160.0N. What I did was this:

sum of Fy = 0
Fbicepsy + Felbowy + Fweight = 0
232sin43 + Felbowy - 132.5 = 0
Felbowy = -25.72N

sum of Fx = 0
Fbicepsx + Felbowx = 0
-232cos43 + Felbowx = 0
Felbowx = 169.67N

I then used Pythagorean's Theorem to try and find the resultant vector of Felbow but came up with the wrong answer of 171N. I think I might be missing some horizontal force(s), but i'm not sure.

8. Oct 3, 2009

### kuruman

I can't find anything. I redid your numbers and I agree with the 171 N. It's time to take your work to your instructor and ask where the 160 N comes from.