Given: m = 55.1 kg F_g =(55.1kg)(9.8N/kg) = 540 N What is the net torque about t

Question

Given: m = 55.1 kg F_g =(55.1kg)(9.8N/kg) = 540 N What is the net torque about the climber's center of mass? Is it positive, negative or zero? Note that it was assumed that the vertical distances between the climbers feet and center-of-mass and between the climber's back and center-of-mass were both 037 m. Given these distances, is the minimum 300 N horizontal force the climber exerts on the walls of the chimney to achieve linear stability also sufficient to achieve rotational stability? Explain.

Explanation / Answer

a) assume counterclockwise torque to be positive.

then total torque about climber's center of mass=Fna*ya+Fnb*yb+Ffb*d-Ffa*(w-d)

=300*0.37+300*0.37+210*0.2-330*(1-0.2)=0

hence net torque about center of mass is zero.

b)as we saw in part a, using Fna=Fnb=300 N, total torque is 0

hence rotational stability is achived.

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