In the figure, the horizontal lower arm has a mass of 2.9kg and its center of gr

Question

In the figure, the horizontal lower arm has a mass of 2.9kg and its center of gravity is 14cm from the elbow joint pivot. How much force F_M must the vertical extensor muscle in the upper arm exert on the lower arm to hold a 9.5 kg shot put in static equilibrium? As you open a door, it pushes away from you, and the hangers are on your right. In what direction is the angular velocity vector of the door as it opens? Down up left right If two forces of equal magnitude act on an object that is hinged at a pivot, the force acting farther from the pivot must produce the larger torque about the pivot. True False If the torque on a ball adds up to zero, the ball could be both rotating and accelerating linearly. True Flase A wheel is rolling without slipping, with a linear velocity v. If we attach a piece of tape to the wheel, when the piece of tape is touching the road (at the bottom of the wheel), its velocity with respect to the road is: 2v v 1.5v zero The angular momentum of a system remains constant when a constant torque acts upon it. True False

Explanation / Answer

Fm = ?

rm = distance of force Fm from elbow joint = 2.5 cm = 0.025 m

m = mass of arm = 2.9 kg

r = distance of center of arm from elbow joint = 14 cm = 0.14 m

M = mass of shotput = 9.5 kg

R = distance of shotput from elbowjoint = 30 cm = 0.30 m

using equilibrium of torque about elbow joint

Fm rm = mg r + Mg R

Fm (0.025) = (2.9) (9.8) (0.14) + (9.5) (9.8) (0.30)

Fm = 1276.4 N

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