A 3 kg box rests on a ramp with a slope of 60 degrees. Find the normal force acting on the box.

Answer 1

So, what it the first thing you want to do?

Draw a diagram, so you can see the concepts you want to use. And, what is the normal force and what does it oppose?

Because our object is not crashing through the triangle our normal for is equal to mgcos(theta) that is to say:

N = mgcosθ so, all you have to do is plug and chug.

Depending on what you use for g you should get about -15 N. My professor kept it simple and used -10 m/s² for the acceleration of gravity.

Answer 2

Hi Hadassah!

The normal force is the force from the surface against the object resting on it.

When the surface is horizontal, it is the same as the weight (gravitational force) of the box.

As the surface becomes slanted after being horizontal, the gravitational force (although still straight down) is no longer completely perpendicular to the surface. Much of the gravitational force is still perpendicular, but some of it is parallel to the surface. The amount of the gravitational force that is parallel to the surface is proportional to the sine of the angle the surface makes with the horizontal: that is, smaller angles provide smaller forces, just as smaller angles provide smaller values of sine. So . . .

F(parallel) = F(gravity) sin (theta)

That means that the perpendicular force is no longer equal to the force of gravity on the box: it decreases as the angle increases, just like the value of cosine, and

F(perpendicular) = F(gravity)cos(theta)

Since the box's position relative to the ramp's vertical level does not change, F(normal) = F(perpendicular), and

F(normal) = F(gravity)sin(theta) = (3kg)(9.8 m/s²)(cos 60°) = 14.7 Newtons