# Problem: Block pushed up a frictionless ramp

Rob is pushing a block of 12 kg up a frictionless ramp. The ramp makes an angle of 35° with the horizontal. Assuming that Rob exerts a force of 148 N, find:

• The resultant force acting on the block.
• The acceleration of the block.
• The magnitude of the normal force.

## Solving the problem

Let's represent what is happening in the problem with a simple sketch.

We need to represent a ramp that makes an angle of 35° with the horizontal, a block on it, and indicate that the block is pushed up by a force parallel to the ramp:

By looking at the sketch, we can conclude that there are three forces acting on the block:

• Rob's push, F
• the gravitational force, mg
• and the normal force N, which the ramp exerts to prevent the penetration of the block

Having figured out the forces acting on the block, let's draw a free-body diagram of the block:

Let's think about what we know, and what we're asked to find:

We know the mass of the block (12 kg), the angle the ramp makes with the horizontal (35°), and the force exerted by Rob (148 N).

We're asked to find the resultant force acting on the block, the acceleration that the block has as a result, and the normal force exerted by the ramp on the block.

m = 12 kg
θ = 35°
F = 148 N

### We want to know

R = ?
a = ?
N = ?

Let's start by finding the resultant force.

In this case we have a block moving up a ramp, so for our convenience, we will use tilted coordinate axes, with the x axis in the direction of motion (uphill).

After drawing the coordinate axes on the free-body diagram of the block, we proceed to find the components of the individual forces acting on the block:

Keep in mind that the angle that the gravitational force, mg, makes with its y component, mgy, is equal to the angle that the ramp makes with the horizontal (35° in this case).

Fx = F
Nx = 0
mgx = −mg sin 35°
Fy = 0
Ny = N
mgy = −mg cos 35°

And the components of the resultant force will be:

x:

Rx = Fx + Nx + mgx
Rx = F + 0 + (−mg sin 35°)
Rx = Fmg sin 35° (1)

y:

Ry = Fy + Ny + mgy
Ry = 0 + N + (−mg cos 35°)
Ry = Nmg cos 35° (2)

We know that the motion of the block is along the ramp, therefore Ry must be zero. Otherwise there would be an acceleration in the y direction, which is not the case.

Ry = 0

Therefore, the magnitude of R is equal to the absolute value of Rx. We know that Rx must be positive because the block is accelerating in the positive x direction. So, we can write:

R = Rx

We have already determined Rx in Eq. (1):

Rx = Fmg sin 35°

Therefore,

R = Fmg sin 35°
R = 148 N − (12 kg) (9.8 N/kg) (sin 35°)
R = 148 N67 N
R = 81 N

So, the resultant force on the block is 81 N and directed in the positive x direction:

Next we need to find the acceleration.

Since we know the resultant force and the mass, we can apply Newton's 2nd Law to get the acceleration:

R = ma
 a = R m
 a = R m
 a = 81 N 12 kg
a = 6.8 m/s2

Lastly, we need to find the magnitude of the normal force.

The normal force N appears in Eq. (2):

Ry = Nmg cos 35°

And we've shown that Ry is zero, therefore:

0 = Nmg cos 35°
N = mg cos 35°
N = (12 kg) (9.8 N/kg) (cos 35°)
N = 96 N

At this point, we've found everything we were asked to find:

R = 81 N
a = 6.8 m/s2
N = 96 N

## Tips & Tricks

• When an object is upon an incline, the angle between mg and mgy is equal to the angle that the incline makes with the horizontal.
• For objects moving along an incline, the y component of the resultant force is always zero (assuming that we take the x axis in the direction of motion, and the y axis perpendicular to it).

## Exercises

### #1

A heavy box of 135 kg is pushed up an inclined plane which makes an angle of 14.0° with the horizontal.

If the push has magnitude 550 N and the plane can be considered frictionless, what are: the resultant force acting on the box, the acceleration of the box, and the normal force?

R = 230 N
a = 1.70 m/s2
N = 1.29 × 103 N