To treat fractures of the humerus shaft, a cast and traction device is used; such that the cast of the upper arm is bent by 100 degrees to 110 degrees at the elbow joint. The traction device is an object of mass M that is hung at the elbow, it keeps the upper arm under tension. What is the range of masses for the traction device used if the muscle force it must compensate ranges from 5N to 10N?

My ans:

Using fbd: Fnet: T- Mg = 0
5- Mg =0
-M =-5/g

Of course I realized this is wrong since I never used the angles given, but I have no idea...

To determine the range of masses for the traction device used, we need to consider the forces acting on the upper arm in both cases where the muscle force it must compensate ranges from 5N to 10N.

First, let's understand the forces involved. The primary force acting on the upper arm is the tension force (T) provided by the traction device. The weight of the upper arm, represented by the force of gravity (Mg), also acts downward.

Now, let's analyze the situation for both cases:

Case 1: Muscle force = 5N
In this case, the tension force provided by the traction device (T) must balance the muscle force (5N) and the weight of the upper arm (Mg). Therefore, the equation becomes: T - Mg = 5.

Case 2: Muscle force = 10N
Similarly, in this case, the equation becomes: T - Mg = 10.

Now, let's introduce the angles into the equations. The information provided states that the cast of the upper arm is bent by 100 degrees to 110 degrees at the elbow joint. However, since the range of angles is not specified, we will assume the maximum angle of 110 degrees. This angle will affect the vertical component of the muscle force and the tension force.

To incorporate the angles, we need to consider the vertical and horizontal components of the muscle force as well as the tension force. The vertical component of the muscle force can be calculated using the formula: F_vertical = F * cos(angle), where angle is the angle of bend at the elbow joint.

Let's consider case 2 (muscle force = 10N) to demonstrate the calculations:

Vertical component of the muscle force (Fm_vertical) = 10 * cos(110 degrees)

Now, our equation becomes:

T - Mg = 10 + Fm_vertical

Rearranging the equation, we have:

T = Mg + 10 + Fm_vertical

Substituting Fm_vertical with the calculated value, we have:

T = Mg + 10 + (10 * cos(110 degrees))

Since the cosine of 110 degrees is negative, we have:

T = Mg + 10 - (10 * abs(cos(110 degrees)))

Performing similar calculations for case 1 (muscle force = 5N), we can find the lower limit of the range of masses for the traction device.

By evaluating these equations, you can determine the range of masses for the traction device used based on the given conditions.