What Is the Required Muscle Force to Hold an Outstretched Arm?

[Angie913]

I can get pretty far in this question, but I can't seem to get the final answer. The problem is:
Calculate the force required by the "deltoid" muscle, Fm, to hold up an outstretched arm. The total mass of the arm is 3.33kg. On the diagram, the pivot point I chose is the shoulder socket, and the diagram shows the center of mass to be the elbow (with an arrow pointing down labeled 'mg'). By the shoulder socket, there is a drawing of the force Fm we need to find and it's d=.119 m away from the shoulder socket and the force points from .119 meters away upwards at an angle of alpha=14.8deg towards the socket. There is another force Fj (I assume force of the joint?) pointing downwards from the socket at an angle and it appears that the angle it points downwards at would be alpha as well (interior angles) but it is not stated or shown. The center of mass at the elbow is D distance from the socket, D=.238m. I know that we need to find the sum of the torques, but I can't figure out is Fj is supposed to be equal to Fm but opposite sign or what? I have that Sigma Torque= -mgD(torque at elbow)-(Fj)d sin alpha + (Fm) d sin alpha=0 I can't have two variables though, so what am I missing? Any hints are appreciated! Thanks.



Please please help!

 

[Angie913]

I got it! woo hoo...nevermindnow

 

[wais]

It seems like you are on the right track with your approach to finding the torque of the human arm. Let's break down the problem step by step to see if we can find a solution.

First, we need to define some variables:

- Fm: force required by the deltoid muscle
- Fj: force of the joint (possibly due to the weight of the arm)
- m: mass of the arm (3.33 kg)
- g: acceleration due to gravity (9.8 m/s^2)
- D: distance from the shoulder socket to the center of mass at the elbow (0.238 m)
- d: distance from the shoulder socket to the point where Fm is applied (0.119 m)
- alpha: angle between Fm and the arm (14.8 degrees)

Now, let's consider the forces acting on the arm. We have the weight of the arm, mg, acting downwards at the center of mass at the elbow. We also have Fm acting upwards at an angle of 14.8 degrees from the shoulder socket. And finally, we have Fj acting downwards from the shoulder socket, but we don't know the magnitude or direction of this force.

To find the torque, we need to sum up the torques of these forces about the shoulder socket. We can do this by using the equation:

Sigma Torque = F x d (where F is the force and d is the perpendicular distance from the force to the pivot point)

So, we have:

Sigma Torque = -(mg)(D) + (Fj)(d sin alpha) + (Fm)(d sin alpha)

We can see that the torques due to Fj and Fm are in opposite directions, so we can rewrite the equation as:

Sigma Torque = -(mg)(D) + (Fm-Fj)(d sin alpha)

Now, we also know that the arm is in equilibrium, which means that the sum of the forces in the x-direction must be equal to 0. In other words, the force Fj must be equal in magnitude to Fm, but in the opposite direction. So we can write:

Fj = -Fm

Plugging this into our torque equation, we get:

Sigma Torque = -(mg)(D) + (Fm-(-Fm))(d sin alpha)

Simplifying this, we get:

Sigma Torque = -(mg