Question

A box is pulled across a frictionless floor by a 50 N force directed 50° above the horizontal. How much is the net force along the horizontal direction?

Answer 1

Answer:

$F_x=32.14N$

Explanation:

In this case the only force on the horizontal direction would be the horizontal component of the force that it's being applied. Since the angle given is above the horizontal, this horizontal component can be calculated as $F_x=Fcos\theta$, where F is the magnitude of the force and $\theta$ the angle given.

For our values we have:

$F_x=Fcos\theta=(50N)cos(50^{\circ})=32.14N$