Torque question in Deltoid muscle

[dcmf]

Homework Statement

A woman lifts a 3.6-kg barbell in each hand with her arm in a horizontal position at the side of her body and holds it there for 3 s (see the figure below). What force does the deltoid muscle in her shoulder exert on the humerus bone while holding the barbell? The deltoid attaches 13 cm from the shoulder joint and makes a 13 degree angle with the humerus. The barbell in her hand is 0.55 m from the shoulder joint, and the center of mass of her 4.0-kg arm is 0.24 m from the joint.

Relevant Equations

Tnet, T=Flsinθ, Fg=mg

Here's a picture the question provided.

I tried solving this question two ways (assuming the axis of rotation is at the shoulder joint) and am getting wildly different answers.

Some potential reasons there's a discrepancy:
- I'm not super confident about my use of the torque equation (T=FlSinθ), especially the angle part, which I think needs radians as an input
- Question mentioned something about 3 seconds but I never used that in any of my calculations and this is a one-part question so its not as if it could be used later on

Thanks in advance!

 

[Orodruin]

$\sin(13^\circ) < 1$ so you cannot possibly have done the math right in the first attempt as dividing something by a positive number < 1 cannot possibly give a smaller number.

 

[Lnewqban]

If you draw a free body diagram, it would be evident how far from reality that first result of 17 N is, considering that the barbell weights 35 N.

 

[dcmf]

Thanks for the input.

$\sin(13^\circ) < 1$ so you cannot possibly have done the math right in the first attempt as dividing something by a positive number < 1 cannot possibly give a smaller number.

You're absolutely correct, there was a calculation error there but that results in about 985.2N, but I think the answer needs about 2 significant figures so there's still (at least what seems to me) a discrepancy between the two values.

Also is 3s really not being used at all in the question? Just a red herring?

 

[dcmf]

If you draw a free body diagram, it would be evident how far from reality that first result of 17 N is, considering that the barbell weights 35 N.

I had a really poorly done sketch so I omitted it The numbers didn't look right to me either though. But I just have terrible intuition with math and it didn't click that dividing by a decimal should result in a bigger number :')
Thank you

 

[Lnewqban]

View attachment 341914
I had a really poorly done sketch so I omitted it The numbers didn't look right to me either though. But I just have terrible intuition with math and it didn't click that dividing by a decimal should result in a bigger number :')
Thank you

Your FBD for summation of moments is correct.
The greater the deviation of the line of action of the muscle from the vertical, the greater the actual force respect to the vertical component (which you have properly calculated).
My calculation of the actual muscle effort gives 986.25 N.

 

[Orodruin]

You're absolutely correct, there was a calculation error there but that results in about 985.2N, but I think the answer needs about 2 significant figures so there's still (at least what seems to me) a discrepancy between the two values.

Your 2.9 radians is far too imprecise to represent 167°. It is closer to 166°, which makes a big difference in the sine (0.225 vs 0.239).

Also is 3s really not being used at all in the question? Just a red herring?

For computing the static force in equilibrium, yes, it is a red herring.

 

[erobz]

View attachment 341914
I had a really poorly done sketch so I omitted it The numbers didn't look right to me either though. But I just have terrible intuition with math and it didn't click that dividing by a decimal should result in a bigger number :')

I know it’s not necessary for the computation of $T$, but a FBD should include the reaction forces at the shoulder joint too.