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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:

Rob pushing the block up the ramp35°Push

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:

Free-body diagram of the blockNFmg

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.

We know

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:

Tilted coordinate axes in our free-body diagram. The x and y components of mg are also present.xyNFmgymgxmg35°

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:

The graphical representation of the resultant force which acts upon the blockxyR

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
aR
m
aR
m
a81 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?

Solution

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