Question

A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grips the pole with each hand a distance d=0.595 m from the center of the pole. A bird of mass mb=560 g lands on the very end of the left‑hand side of the pole. Assuming the daredevil applies upward forces with the left and right hands in a direction perpendicular to the pole, what magnitude of force Fleft and Fright must the left and right hand exert to counteract the torque of the bird?

Answer 1

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

          W_bird  L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

          F_left = F_right = F

          W_bird L / 2 - 2 F 0.595 = 0

          F = $\frac{m_{bird} \ g L} { 4 \ 0.595}$

   

we calculate

         F = 0.560 9.8 14.0 /2.38

         F = 32.28 N