A man holds a 185-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force , which is applied perpendicular to the forearm. The forearm weighs 24.3 N and has a center of gravity as indicated. Find (a) the magnitude of and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint.

Question

May I please get help with this question. I'm very confused. I appreciate all the assistance.

Answer 1

To solve this problem, we can apply the principle of equilibrium for the forearm to find the force exerted by the upper arm bone at the elbow joint.

First, let's set up the problem:

- Let Fm be the flexor muscle force
- Let Fe be the force exerted by the upper arm bone at the elbow joint
- Let Fa be the force applied by the upper arm bone to the forearm at the elbow joint
- Let d1 be the distance between the elbow joint and the center of gravity of the forearm (assume d1 = 0.17 m, as this is roughly the midpoint of an average human forearm)
- Let d2 be the distance between the elbow joint and the center of gravity of the ball (assume d2 = 0.4 m for this problem)
- Let d3 be the distance between the elbow joint and the point where the flexor muscle force is applied (assume d3 = 0.05 m)

Now, we have the following equations from equilibrium:

ΣFy = 0 => Fa - Fm - 185 = 0 (1)
ΣFx = 0 => Fe = 0 (2)
ΣM (elbow) = 0 => -Fm*d3 + 185*d2 - 24.3*d1 = 0 (3)

From equation (2) we can say that there is no horizontal force at the elbow joint exerted by the upper arm bone, therefore the force is purely vertical.

Now substituting the assumed values of d1, d2, and d3 in equation (3), we can find the value of Fm:

- Fm * 0.05 + 185 * 0.4 - 24.3 * 0.17 = 0 => Fm = (185 * 0.4 - 24.3 * 0.17) / 0.05 = 1348 N

Now, using equation (1) we can find the value of Fa:

Fa - 1348 - 185 = 0 => Fa = 1533 N

So the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint is 1533 N. Since there is no horizontal component to this force, the angle is simply 0° counterclockwise from the horizontal.

To summarize:

a) Magnitude of force exerted by the upper arm bone at the elbow joint: Fe = 0 N
b) Magnitude of force applied by the upper arm bone to the forearm at the elbow joint: Fa = 1533 N
c) Direction of force exerted by the upper arm bone at the elbow joint: 0° counterclockwise from the horizontal