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

1 Answer
Sep 10, 2016

Answer:

#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.