# Problem: Object moving at constant velocity over a horizontal surface

Hanna is pulling an object of 20 kg over a horizontal plane. The force Hanna is exerting makes an angle of 30° with the horizontal. The coefficient of sliding friction μ, between the object and the plane, is 0.57.

If the object is moving at constant velocity, what is the magnitude of the force provided by Hanna?

## Solving the problem

First of all, let's represent what is happening in the problem with a simple sketch.

We will need to represent a horizontal surface, an object on it, and indicate that the object is moving at constant velocity, as well as the fact that the object is being pulled by a force that makes 30° with the horizontal, and is subject to the force of friction:

By observing the sketch, we notice that the object is subject to 4 forces:

• Hanna's pulling force, F
• the sliding friction force, Ff
• the gravitational force, mg
• and the the normal force, N

So, the free-body diagram of the object will look something like this:

We know the angle that the pulling force makes with the horizontal (30°), the mass of the object (20 kg), the coefficient of sliding friction (0.57), and that the object is moving at constant velocity.

We need to find the magnitude of the pulling force exerted by Hanna.

θ = 30°
m = 20 kg
μ = 0.57
v is constant

### We want to know

F = ?

The fact that the object is moving at constant velocity tells us that the object has no acceleration (because if it had, it wouldn't be moving at constant velocity):

a = 0

And because the acceleration is zero, the resultant force acting on the object must also be zero (for Newton's 2nd Law):

R = ma
R = m × 0
R = 0

Knowing the resultant force, we can use the following strategy to find the magnitude of Hanna's pull:

1. We find the x and y components of the resultant force, as a sum of the x and y components of all the forces that act on the object.
2. Because we already know that R is zero, Rx and Ry must also be zero, so we substitute their values in the equations that we found in step 1.
3. We then use those equations to find F (the pulling force).

We begin by drawing coordinate axes on our free-body diagram and finding the components of all the forces that act on the object:

Fx = F cos 30°
Ffx = −Ff
Nx = 0
mgx = 0
Fy = F sin 30°
Ffy = 0
Ny = N
mgy = −mg

Next, we find the x and y components of the resultant force by adding all the x and y components:

x:

Rx = Fx + Ffx + Nx + mgx
Rx = F cos 30° + (−Ff) + 0 + 0
Rx = F cos 30°Ff

y:

Ry = Fy + Ffy + Ny + mgy
Ry = F sin 30° + 0 + N + (−mg)
Ry = F sin 30° + Nmg

And since

Rx = 0
Ry = 0

we substitute their values and get

0 = F cos 30°Ff (1)
0 = F sin 30° + Nmg (2)

These two equations have three unknowns (F, Ff and N).

The number of equations has to be equal to the number of unknowns in order to solve them. Therefore, we should somehow reduce the number of unknowns to two.

We know the coefficient of sliding friction μ, and the sliding friction force has by definition the magnitude equal to μ multiplied by N:

Ff = μN

Which means that we have reduced the number of unknowns to two, and we can now solve the two equations.

By exchanging Ff with μN in Eq. (1), we get

0 = F cos 30°μN (3)

So, we now have two independent equations (Eq. (2) and Eq. (3)).

First, we solve one of them for one unknown: let's solve Eq. (2) for N:

0 = F sin 30° + Nmg
F sin 30° + Nmg = 0
N = mgF sin 30°

And substitute N in Eq. (3):

0 = F cos 30°μN
0 = F cos 30°μ (mgF sin 30°)

Finally, we solve this equation for F:

0 = F cos 30°μ (mgF sin 30°)
F cos 30°μ (mgF sin 30°) = 0
F cos 30°μmg + μF sin 30° = 0
F cos 30° + μF sin 30° = μmg
F (cos 30° + μ sin 30°) = μmg
 F = μmg cos 30° + μ sin 30°
 F = (0.57) (20 kg) (9.8 N/kg) cos 30° + (0.57) (sin 30°)
 F = 112 N 1.15
F = 97 N

Hence, Hanna's pulling force has a magnitude of 97 N.

## Tips & Tricks

• Remember that whenever an object is either at rest or moving at constant velocity, the resultant force acting on the object is 0.
• When you have two independent equations with two unknowns, you can solve them in different ways. Usually we tend to solve one of the two equations for one unknown, and substitute the result in the second equation, so that we end up with an equation that has only one unknown, which we can solve. Once we solved that equation and found the first unknown, we can substitute the found value in the other equation, and find the second unknown.

## Exercises

### #1

A light box of 1.3 kg is pulled over a horizontal table with a force that makes an angle of 45° with the horizontal. Knowing that the box is moving at constant velocity and that the coefficient of sliding friction is 0.80, find the magnitude of the pulling force.

F = 8.0 N