Finding Torque within the Human Arm

[KakashiT]

Homework Statement

The arm in the figure below weighs 37.1N. The force of gravity acting on the arm acts through point A. Assume that L1 = 0.0720m, L2 = 0.300m and α = 11.8deg. Point A is 0.300m away from the pivot point. Fs is unknown angle degree from the horizontal.


a) Determine the magnitude of the tension force Ft in the deltoid muscle.
b) Determine the magnitude of the tension force Fs of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.
c) Determine the angle of tension force Fs relative to the x-axis.

Homework Equations

t = r T \perp
Σt =0
F⃗ net=ΣF⃗ =0
(F⃗ net)x=ΣFx=0
(F⃗ net)y=ΣFy=0
sin^2 \theta + cos^2 \theta = 1
\frac{sin \theta}{cos \theta} = tan \theta

The Attempt at a Solution

a) This was straightforward. I found the torque at Point A. Because this system is in rigid body equillibrium, the net torque is zero. As a result, I found the tension force in Ft by subbing the r value and the torque into the torque equation. Which came out to be 756 N.

b) To find the tension in Fs, I understood that because the system was in rigid body equilibrium, the net force is zero. I divided Fs into its horizontal and vertical components. The horizontal component was equal to 740 N, while the vertical component was equal to 117.49 N. Once I got the components, I added them together to find the tension force of Fs. Here I used the trigonometry identity: sin^2 \theta + cos^2 \theta = 1 so that I could eliminate the unknown angle to solve for Fs. Which came out to be 349 N.

c)This is where I am stuck currently. I used the formula: arctan\frac{Fs sin \theta}{Fs cos \theta} = \theta. I plugged in my values that I found in part B, which give me the answer of 9.02 deg. However my homework site says that this answer is wrong. Even though the previous two parts are considered correct. I do not know what I am doing wrong here.

arctan (117.9/740) = 9.02 deg

 

[haruspex]

I divided Fs into its horizontal and vertical components. The horizontal component was equal to 740 N, while the vertical component was equal to 117.49 N. Once I got the components, I added them together to find the tension force of Fs. Here I used the trigonometry identity: sin^2 \theta + cos^2 \theta = 1 so that I could eliminate the unknown angle to solve for Fs. Which came out to be 349 N.

If the components are 740 and 117.5 the magnitude of the force must be more than either. Maybe you meant 749? If so, I agree with your answer for the angle.

 

[Simon Bridge]

To find the tension in Fs, I understood that because the system was in rigid body equilibrium, the net force is zero. I [resolved] Fs into its horizontal and vertical components. The horizontal component was equal to 740 N, while the vertical component was equal to 117.49 N. Once I got the components, I added them together to find the tension force of Fs. Here I used the trigonometry identity:
$\sin^2\theta+\cos^2\theta=1$
so that I could eliminate the unknown angle to solve for Fs. Which came out to be 349 N.

Was this confirmed as correct?
If 117.49N and 740N are the components of a vector, then shouldn't the magnitude of the vector be bigger than each of these? Typo?

arctan (117.9/740) = 9.02 deg

... it isn't, arctan(117.49/740)=9.02deg - is this another typo?

[edit]There seems to be a number of typos in your writeup above - make sure you are not misstyping your final answer.

Also check the format of the answer - number of decimal places etc... computer mediated marking can get pedantic.