Moment of force and static equilibrium

  • Thread starter lalm22
  • Start date
  • #1
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1) Problem:
An individual holds a ball of mass 3 kg in his hand in static equilibrium. The biceps brachii attaches a distance of 2 cm from the centre of rotation of the elbow joint, and the muscle's line of action is directed at an angle of 115° counter clockwise from the forearm. The forearm has a mass of 1.5 kg. The forearm's centre of mass is located a distance of 18 cm from the elbow joint. The distance from the elbow joint to the point at which the ball is held in the individual's hand is 45 cm. Assuming that the biceps brachii is the only muscle contributing to elbow flexor moment, what is the magnitude of the tension running through the biceps brachii?

http://imgur.com/1GQ3nNv

2) Relevant equations:

f=ma, Static Equilibrium: Em = 0

3) My solution
I know that i have to convert the mass of the forearm and the weight to Newtons, but what do I do with the angle. Fb = force biceps:

-29.43(0.45) - 14.715(0.18) + Fb(0.02) = 0
fb = 794.61

However, I can't figure out when to incorporate the angle (115) at which the biceps is flexed :/
 

Answers and Replies

  • #2
291
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There is an equation that relates the angle at which the force is exerted and the torque that will result from this force. This equation can be easily found. I don't see it in the introductory physics formulary... I'm sure that is for a reason, though.
 
  • #3
291
33
F=ma is always relevant, but for this problem there are other equations you will need to know. You seem to be ill-prepared to solve this type of problem, but you can remedy that by checking out the Wikipedia articles on torque and moment of inertia. Once you are able to list the relevant equations I can do more to help you.
 

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