Question

If a crate weighing 350. N is at rest on an inclined plane whose angle provides the crate with 150. N of normal force, at what angle to the horizontal would the ramp be?

Answer 1

Answer:
`25.37\°`

Step-by-step explanation:

If we draw a free body diagram of the crate we will have the following related to the net force in the y-axis:

`N- W sin \theta=0` (1)

Hence:

`N=W sin \theta` (2)

Where:

`N=150 N` is the normal force

`W=350 N` is the weight of the crate

`\theta` is the angle of the inclined plane where the crate is

Findind
`\theta`:

`\theta=sin^(-1) ((N)/(W))` (3)

`\theta=sin^(-1) ((150 N)/(350 N))` (4)

Finally:

`\theta=25.37\°`