Question

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effective in pulling the sled horizontally along the ground?

Answer 1

When a force, F, is applied to an object at an angle,θ, to compute for the horizontal and vertical components, we have the following

$F_{vertical} = Fsin(\theta)$
$F_{horizontal} = Fcos(\theta)$

So, for this particular problem, given that the force applied to the sled is 4.00 N at 62⁰. Thus, the horizontal component of the force is equal to

F = 4.00(cos62⁰) = 1.88 N

Therefore, the horizontal force is 1.88 N.
Answer: 1.88 N

Answer 2

You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.

If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.

The angle made with the x-axis is cosine of the angle theta.

Please see picture attached.