Question

a 230-newton box rests on a plank 5 meters long. One end is 1.2 meters higher than the other end. Find the components of the force the box exerts

Answer 1

Answer:

55.3 N, 223.3 N

Explanation:

First of all, we can find the angle of the inclined plane.

We have:

L = 5 m the length of the incline

h = 1.2 m is the height

We also have the relationship

$h = L sin \theta$

where $\theta$ is the angle of the incline. Solving for the angle,

$\theta= sin^{-1} (\frac{h}{L})=sin^{-1} (\frac{1.2 m}{5 m})=13.9^{\circ}$

Now we can find the components of the weight of the box, which is the force that the box exerts on the plank. Calling W = 230 N the weight of the box, we have:

- Component parallel to the incline:

$W_{par} = W sin \theta = (230 N)(sin 13.9^{\circ}) =55.3 N$

- Component perpendicular to the incline:

$W_{per} = W cos \theta = (230 N)(cos 13.9^{\circ}) =223.3 N$