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While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contracting the muscles in the back of the upper leg. (a) Find the angular acceleration (in rad/s2) produced given the mass lifted is 12.5 kg at a distance of 28.0 cm from the knee joint. The moment of inertia of the lower leg is 0.900 kg · m2, the muscle force is 1475 N, and its effective perpendicular lever arm is 3.00 cm.

Respuesta :

Answer:

[tex]\alpha = 5.27 rad/s^2[/tex]

Explanation:

Given data:

m = 12.5 kg

distance[tex] r_w = 28.0 cm = 0.28 m[/tex]

I = 0.900 kg m^2

force  = 1475 N

lever arm [tex]= 3.00 cm = 0.03 m[/tex]

[tex]tau_{net} = I \alpha [/tex]

[tex]\alpha = \frac{\tau_{net}}{I}[/tex]

[tex]\alpha = \frac{F r_m - F_w r_w}{I + m r_w^2}[/tex]

[tex]\alpha = \frac{1475 \times 0.03 - 12.5b\times 9.81 \times 0.28}{0.900  + (12.5 \times 0.28^2)}[/tex]

[tex]\alpha = 5.27 rad/s^2[/tex]

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