While exercising in a fitness center, a man lies face down on a bench and lifts

Question

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the magnitude of the angular acceleration produced given the mass lifted is 9.5 kg at a distance of 26.5 cm from the knee joint, the moment of inertia of the lower leg is 0.89 kg m^2, the muscle force is 1250 N, and its effective perpendicular lever arm is 2.85 cm. alpha = 12.3 (b) How much work is done if the leg rotates through an angle of 19.5 degree with a constant force exerted by the muscle?

Explanation / Answer

Torque is given by:

T = I*alpha = Sum of (F*r)

alpha = angular acceleration = ?

I = moment of inertia

I = 0.89 kg-m^2 + 9.5 kg*0.265^2 m^2

I = 1.557 kg-m^2

Torque = 1250*0.0285 - 9.5*9.81*0.265 = 10.93 N-m

So,

alpha = 10.93/1.557 = 7.02 rad/sec^2

B.

Workdone = torque*thera

W = 10.93*19.5*2*pi/360

W = 3.72 J

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