Given: m = 55.1 kg F_g = (55.1kg)(9.8N/kg) = 540 Recall that according to the de

Question

Given: m = 55.1 kg F_g = (55.1kg)(9.8N/kg) = 540 Recall that according to the definition of torque (for both r_A and F_A in the x-y plane), the torque on the climber's feet about her center-of-mass is given by tau_A = (r_A, x F_A, y - r_A, y F_A, x)z Apply the right hand rule to r_A and F_A to determine whether the torque vector points along the +z (out of the paper) or -z (into the paper) axis. Is the rotation that would result from this torque, if it were unbalanced, clockwise or counterclockwise from your perspective? Use the information in the preceding diagram to explain why, using vector notation r_A = -0.80mx - 0.37 m y and F_A = 300x + 330N y Use the definition of torque given in part (a) to show that r_A = - 153m Nz

Explanation / Answer

torque = r*F

where * indicate cross product of r and F

so for two dimensional case it turn out torque = rxFy-ryFx

if we express r and F in vector notation than it will be easy for us to identify component of r (that is rx and ry ) and F(that is Fx and Fy)

so from given expression of r and F

rx = 0.2 and ry =0.37; Fx = -300 and Fy = 210

torque = 0.2*210 - 0.37*(-300)=0.2*210+0.37*300= 153 Nm towards positive z direction

and it produces counter clockwise rotation

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