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

Sep 10, 2016

$1.46 \times {10}^{4} N$, rounded to two decimal places.

#### Explanation:

We know from the figure given below that

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 $\cos \theta$ component of its weight, $m g$, and is calculated from the expression

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

Given $\theta = {8}^{\circ}$,
$\therefore {F}_{n} = 1500 \times 9.81 \times \cos {8}^{\circ}$
$\implies {F}_{n} = 1.46 \times {10}^{4} N$, rounded to two decimal places.