Rotational dynamics/Rigid objects in equilibrium

2 replies · 4K views · last reply
giant016
Messages
20
Reaction score
0

Homework Statement



Textbook diagram of a forearm horizontal on a scale: triceps muscle force M acts upward near the elbow joint, the forearm center of gravity cg is marked, with distances 0.0250 m, 0.150 m and 0.300 m labeled.


Homework Equations


Net force of the Y vector: -F1 + F2 - weight = 0


The Attempt at a Solution



I tried to solve the equation thinking that m could = torque (torque=force X length of lever arm) but I couldn't figure out how to fit in that .025m spot in between the joint and the muscle. Not sure if the equation provided will really help. Any help is appreciated!
 
Last edited:
The question is tricky because the M force is drawn in the wrong direction. The M force is the blue arrow up, not a black arrow going right. Put an equation equal to 0 where clockwise moments are positive and counterclockwise are negative
0 = (0.025)(M) + (0.15)(22) - (0.3)(111)
M = 1200N

That doesn't seem right. That's equivalent to 122kg, resting on 1 muscle.
 
ShawnD said:
The question is tricky because the M force is drawn in the wrong direction. The M force is the blue arrow up, not a black arrow going right. Put an equation equal to 0 where clockwise moments are positive and counterclockwise are negative
0 = (0.025)(M) + (0.15)(22) - (0.3)(111)
M = 1200N

That doesn't seem right. That's equivalent to 122kg, resting on 1 muscle.

We have an answer sheet (it's a multiple choice question) and one of the answers is 1.2X10^3, and your equation makes sense, so...Thanks! I'm on some other forums (mostly car junk) and I've never gotten an answer that quick!