Answer
(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.
Work Step by Step
(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.
