Question

A 62kg rock climber is attached to a rope that is allowing him to hang horizontally with his feet against the wall. The tension in the rope is 710N and the rope makes an angle of 32 degrees with the horizontal. Determine the force exerted by the wall on the climber's feet.
Answer is 710N, Why?

Answer 1

It is only the horizontal component of the force that is 
Fcos\theta=602N

Answer 2

Answer:

F = 645 N

Explanation:

For equilibrium in vertical direction we can make force balance

so we will have

$F_y + Tsin32 = mg$

$F_y + 710 sin32 = (62\times 9.8)$

$F_y = 607.6 - 376.2$

$F_y = 231.3 N$

Now for horizontal equilibrium we know that

$F_x = Tcos32$

$F_x = 710cos32$

$F_x = 602.11 N$

now we have net force due to wall given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{231.3^2 + 602.11^2}$

$F = 645 N$