Calculating the Force of the Deltoid Muscle to Raise an Arm

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SUMMARY

The discussion focuses on calculating the force exerted by the deltoid muscle (Fm) to raise an arm, considering the forces of gravity (Fg) and the shoulder socket (Fs). The deltoid muscle acts at an angle of approximately 13.2 degrees with respect to the negative x-axis. The mass of the arm is given as 1.57 kg, leading to a gravitational force of Fg = 1.57 kg * 9.8 m/s² = 15.39 N. The participant initially calculated Fm incorrectly as 14.98 N using cosine, while the correct magnitude of Fm is determined to be 67.38 N.

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Homework Statement


The main muscle responsible for raising an arm is the deltoid. The deltoid muscle connects at the upper end of the shoulder, extends over the upper arm bone (humerus), and attaches near the elbow. Effectively there are three forces involved in raising the arm: (i) the force of the deltoid muscle, Fm, acting at an angle of approximately 13.2o with respect to the negative x axis, (ii) the force of gravity, Fg, acting effectively at the centre of mass of the arm located close to the elbow, and (iii) the force of the shoulder socket, Fs, acting on the humerus effectively along the x axis. Assuming the mass of the arm is 1.57 kg, calculate the magnitude of Fm in Newton.




Homework Equations


F=ma



The Attempt at a Solution


I'm not quite sure where to start. I tried using F_x=ma and F_y=ma to work out the net force but not getting the correct answer. Any help would be greatly appreciated. Thanks.
 

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Sorry I made a mistake you can do it with just Newtons law no torques needed . Can you show your actual work force decomposition should work if you didn't do any calculation mistakes?
 
Last edited:
bp_psy said:
Sorry I made a mistake you can do it with just Newtons law no torques needed . Can you show your actual work force decomposition should work if you didn't do any calculation mistakes?
After going over my work abit more I see I have no clue what I am doing:frown:
Im not sure how to implement the angle into the equation or what a is in the x direction(Isnt it 0?)
Would Fy just be Fy=1.56*9.8?
Any help on where to start would be great. Thanks:smile:
 
Eggphys said:
After going over my work abit more I see I have no clue what I am doing:frown:
Im not sure how to implement the angle into the equation or what a is in the x direction(Isnt it 0?)
Would Fy just be Fy=1.56*9.8?
Any help on where to start would be great. Thanks:smile:


So i decided to come at this from another angle. First i solved for Fg,
Fg=1.57*9.8 = 15.39
Then to solve for Fm,
Fm=15.39cos 13.2 = 14.98

But the correct answer is 67.38. I am not sure where to go from here to get the correct answer. Any help would be appreciated. Thanks:smile:
 
Just bumping back to the top.
 

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