The slide is inclined at an angle of 52.0°. Danny weighs 46.0 pounds. He is sitting in a cardboard box with a piece of wax paper on the bottom. Stacey is at the top of the slide holding on to the cardboard
box. Find the magnitude of the force Stacey must pull with, in order to keep Danny from sliding down
the slide. (We are assuming that the wax paper makes the slide into a frictionless surface, so that the
only force keeping Danny from sliding is the force with which Stacey pulls.)

Step-by-step explanation:

Since the slide is inclined at an angle of 52.0° to the horizontal, Danny's weight (mass, m) of 46.0 pounds has a component W = mgcos52.0° perpendicular to the incline and W' = mgsin52.0° parallel to the incline where g = acceleration due to gravity = 32 ft/s²

The perpendicular component of Danny's weight is the normal force to the incline. This normal force would determine the magnitude of the friction along the incline.

Since the wax paper makes the incline surface frictionless, so, there is no friction along the surface and thus the only horizontal force acting along the surface is the component of Danny's weight along the surface.

For Stacey to keep Danny from sliding down along the incline, the force she applies along the incline, F must be equal to the component of Danny's weight along the incline.

So, F = W'

F = mgsin52.0°

F = 46lb × 32 ft/s² × sin52°

F = 1472 lb-ft/s² 0.7880

F = 1159.95 lb-f

F ≅ 1160 lb-f