circular motion and gravitation An arm is holding a 50 N hammer in equilibrium a
ID: 1338141 • Letter: C
Question
circular motion and gravitation An arm is holding a 50 N hammer in equilibrium as shown in the picture. The bicep attaches to tin- forearm 3 cm from the elbow joint, and the hammer is held 35 cm from the elbow joint. Ignore the weight of bones and muscles in this problem. Find t he upward force F the bicep exerts on the forearm bone. Find the downward force N the elbow joint exerts on the forearm bone. How do these forces compare to the weight of the hammer? Using your answer to (b), calculate the torque around the bicep attachment point.Explanation / Answer
a)
for the force is F
balancing the moment of forces about the elbow joint
F * 3 - 50 * 35 = 0
F = 583.3 N
the upwards force F is 583.3 N
b)
Now , balancing the forces in vertical direction
583.3 - N - 50 = 0
N = 533.3 N
the normal force N is 533.3 N
c)
F/w = 583.3/50
F/w = 11.7 N
force F is 11.7 times of weight of hammer
N/w = 533.3/50
N/w = 10.7
the force N is 10.7 times of weight of hammer
d)
Torque = 533.3 * 0.03 - 50 * (0.35 - 0.03)
Torque = 0 N.m
Torque about bicep muscle is 0 N.m
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