Question

A block of mass 9.2kg rests on a slope an angle of 26.9∘ relative to the horizontal. What is the size of the contact force normal to the slope, in Newtons?
Use a gravitational acceleration value of g = 9.8m/s^2

Answer 1

Answer:

80.4 N

Explanation:

As the block is at rest on the slope, it means that all the forces acting on it are balanced.

We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:

- The normal reaction N (contact force), upward

- The component of the weight of the block, $mg cos \theta$, downward, where m is the mass of the block, g is the gravitational acceleration and $\theta$ is the angle of the incline

Since the block is in equilibrium along this direction, the two forces must balance each other, so they must be equal in magnitude:

$N=mg cos \theta$

And by substituting the numbers into the equation, we find the size of the contact force normal to the slope:

$N=(9.2 kg)(9.8 m/s^2)(cos 26.9^{\circ})=80.4 N$