Calculating force required of the deltoid muscle - statics problem

  • Thread starter omega5
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Homework Statement


Hello and thank you in advance for your help!
I am taking an introductory algebra-based physics class and am trying to solve this problem:
Suppose an arm holds an 5.8-kg mass. The total mass of the arm is 3.3 kg. Gravity acts from the center of the arm, which is 24 cm away from the joint. To support the mass, the deltoid muscle exerts a leftward force 15° above the horizontal, and 12 cm away from the joint. The joint itself exerts a rightward force subtending an unspecified angle from the horizontal.
What force, FM, is required of the deltoid muscle, assuming the mass is 52 cm from the shoulder joint?

Homework Equations



Since there is no motion,
Tau = 0
ƩFy= 0
ƩFx= 0
Torques about the shoulder joint:
Tau = perpendicular force * length from joint

Gravitational forces:
FG = mg

For forces at an angle:
Fnet = √(Fx)2+(Fy)2

The Attempt at a Solution



There seem to be too many variables to calculate Fm.
The best equation system I could come up with was:
(y+ is up, x+ is rightward)

Taunet = (FM sin 15°)(0.12 m) - (FJsin θ)(x) - (3.3g)(0.24 m) - (5.8g)(0.52 m)
FXnet = (FM sin 15°) - (FJsin θ) - (9.1g)
FYnet = (FJcos θ) - (FM cos 15°)

I know there must be some way to cut the variables down to three and then solve the system, but neither FJ, the angle it subtends, or the lever arm of its vertical force is given. I can't figure out how to work around this.

I hope the image upload works:
giancoli_cp7.ch9.p35.jpg
 
Last edited:

Answers and Replies

  • #2
TSny
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Hello omega5 and welcome to PF.
Taunet = (FM sin 15°)(0.12 m) - (FJsin θ)(x) - (3.3g)(0.24 m) - (5.8g)(0.52 m)
You should be able to see what value to use for x in your torque expression.
 
  • #3
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Thanks for responding!
Do you mean FJ cos θ°? I didn't think it would give me the lever arm distance, just the horizontal force.
 
  • #4
TSny
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What location are you choosing for the axis of rotation for setting up your torques?
 
  • #5
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The joint itself - I tried setting the axis as the point where FJ acts, but thought the lever arms would still have to be include x to define them from that frame of reference.
 
Last edited:
  • #6
TSny
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If the axis is chosen at the joint and if FJ is located at the joint, then what is the lever arm distance, x, for this force?
 
  • #7
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Oh! It just hit me. For some reason I had realized the lever arm was at the point the force acted at for FM but didn't realize that for FJ. I am very grateful!
 

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