## College Physics (4th Edition)

(a) The tension in the rope is $729~N$ (b) The magnitude of the force exerted by the wall on the climber's feet is $327~N$ and this force is directed at an angle of $19.5^{\circ}$ above the horizontal.
(a) We can consider the torques about the rotation axis where the climber's feet meet the wall. The weight of the climber exerts a clockwise torque about this axis. The tension exerts a counterclockwise torque about this axis. Since the system is in equilibrium, the counterclockwise torque is equal in magnitude to the clockwise torque. We can find the tension $F_T$: $\tau_{ccw} = \tau_{cw}$ $(1.06~m)(F_T)~sin~65^{\circ} = (0.91~m)(770~N)$ $F_T = \frac{(0.91~m)(770~N)}{(1.06~m)~sin~65^{\circ}}$ $F_T = 729~N$ The tension in the rope is $729~N$ (b) Since the system is in equilibrium, the sum of the horizontal forces equals zero. We can find $F_x$, the horizontal component of the force exerted by the wall on the climber: $F_x = F_T~cos~65^{\circ}$ $F_x = (729~N)~cos~65^{\circ}$ $F_x = 308~N$ Since the system is in equilibrium, the sum of the vertical forces equals zero. We can find $F_y$, the vertical component of the force exerted by the wall on the climber: $F_y + F_T~sin~65^{\circ} - 770~N = 0$ $F_y = 770~N - F_T~sin~65^{\circ}$ $F_y = 770~N - (729~N)~sin~65^{\circ}$ $F_y = 109~N$ We can find the magnitude of the force exerted by the wall on the climber's feet: $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(308~N)^2+(109~N)^2}$ $F = 327~N$ We can find the angle $\theta$ above the horizontal that the force is directed: $tan~\theta = \frac{109}{308}$ $\theta = tan^{-1}(\frac{109}{308})$ $\theta = 19.5^{\circ}$ The magnitude of the force exerted by the wall on the climber's feet is $327~N$ and this force is directed at an angle of $19.5^{\circ}$ above the horizontal.