Question

What is the normal force exerted by a road inclined 8 degrees from the horizontal on a `1500kg` car?

Answer 1

`1.46xx10^4N`, rounded to two decimal places.

Explanation:

We know from the figure given below that
![dev.physicslab.org]()

When an object rests on an incline plane of angle `theta` with the horizontal, the normal force supplied by the surface of the incline is equal to the `costheta` component of its weight, `mg`, and is calculated from the expression

`F_n = mg cosθ`
the mnemonic "`n`" represents "normal" which is perpendicular to the incline.

Given `theta=8^@`,
`:.F_n = 1500xx9.81xx cos8^@`
`=>F_n = 1.46xx10^4N`, rounded to two decimal places.