Below is a diagram of a baseball/forearm system at rest. It is acted on by four

Question

Below is a diagram of a baseball/forearm system at rest. It is acted on by four different forces: the weight of the forearm, the weight of the baseball, the bicep force, and a force from the upper arm bone (attached at the elbow).

"A Physics 210-HW#9-Sp. x ( e Two Masses (mA= 3 Kg" -> Dwww.webassign.net/web/Student/Assignment-Responses/submit?dep-13701 713 Below is a diagram of a baseball/forearm system at rest. It is acted on by four different forces: the weight of the forearm, the weight of the basebal, the bicep force, and a force from the upper arm bone (attached at the elbow) bicep Fois not shown) ebow is pivot Mg L/2 ie length of forearm L = 20 cm elbow to bicep distance d = 2 cm mass of forearm = M = 4 kg mass of baseball = m = 1 kg a.) Based on torque considerations, the force of the bicep must be greater thanthe combined weight of the baseball/forearm (Make sure you understand the reasoning.) b.) How large is the force exerted by the bicep? 196 c.) The upper arm bone exerts a force (not shown above) on the forearm. This force must be: size: 222.4 × N, dir: downwards

Explanation / Answer

b)
Balance torque at elbow:
Fbicep*d = Mg*L/2 + mg*L
Fbicep*0.02 = 4*9.8*0.1 + 1*9.8*0.2
Fbicep = 294 N
Answer: 294 N

c)
Balance forces
F + Fbicep -Mg -mg = 0
F + 294 -4*9.8 - 1*9.8
F = -245 N
negative sign shows its downward
Answer: 245 N

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