Problem: Mass pulled up an incline with friction

Micheal is pulling a mass of 30 kg up an incline which makes an angle of 26° with the horizontal.

The mass has an upward acceleration of 2.5 m/s2 and the coefficient of kinetic friction between the mass and the incline is 0.58.

What is the magnitude of the pulling force Micheal is exerting?

Solving the problem

As usual, we begin by drawing a sketch of what we think is happening.

We draw an incline that makes an angle of 26° with the horizontal, a mass on it, and indicate that the mass is pulled upwards, has an acceleration directed upwards, and is subject to the force of friction:

The mass being pulled up the inclineFrictionAccelerationPull26°

Looking at the sketch, we can infer that 4 forces are acting on the mass:

  • Micheal's pull, F
  • the friction force, Ff
  • the force due to gravity, mg
  • and the normal force, N

Let's draw a free body diagram of the mass:

The free-body diagram of the massNmgfFF

This is what we know: the mass (30 kg), the angle that the incline makes with the horizontal (26°), the acceleration of the mass (2.5 m/s2), and the coefficient of kinetic friction (0.58).

We need to find the force exerted by Micheal.

We know

m = 30 kg
θ = 26°
a = 2.5 m/s2
μ = 0.58

We want to know

F = ?

Since we know the mass and the acceleration, we can find the magnitude of the resultant force by applying Newton's 2nd Law:

R = ma
R = ma

Now that we determined the resultant force acting on the mass, we can find F using the following strategy:

  1. We first find the x and y components of the resultant force, Rx and Ry, as a sum of the x and y components of all the forces that act on the mass.
  2. We then find the values of Rx and Ry by decomposing R, and substitute them in place of Rx and Ry in the equations that we found in the previous step.
  3. Finally, we use the resulting equations to find F.

Let's begin with the first step of our strategy, by drawing the coordinate axes on our free-body diagram. For convenience, we choose the x axis to be in the direction of motion. Then, we determine the x and y components of all the forces that act on the mass:

Free-body diagram with coordinate axes. The components of the gravitational force are also represented.xyNmgymgxmgf26°FF
Fx = F
Ffx = −Ff
Nx = 0
mgx = −mg sin 26°
Fy = 0
Ffy = 0
Ny = N
mgy = −mg cos 26°

And we find the components of the resultant force as a sum of the components of all the forces:


Rx = Fx + Ffx + Nx + mgx
Rx = F + (−Ff) + 0 + (−mg sin 26°)
Rx = FFfmg sin 26° (1)


Ry = Fy + Ffy + Ny + mgy
Ry = 0 + 0 + N + (−mg cos 26°)
Ry = Nmg cos 26° (2)

The next step is to determine Rx and Ry by decomposing R:

First of all, R has the same direction as the acceleration of the mass, i.e. the positive x direction:

The graphical representation of the resultant force acting on the massxyR


Rx = R = ma
Ry = 0

Now, let's substitute these values in Eq. (1) and Eq. (2), respectively:

Rx = FFfmg sin 26°
Ry = Nmg cos 26°
ma = FFfmg sin 26° (3)
0 = Nmg cos 26° (4)

We have two equations with 3 unknowns: F, Ff, N.

To find F, we need to reduce the number of unknowns to 2.

We can do that by remembering that the magnitude of the friction force is:

Ff = μN

We already know the value of the coefficient of kinetic friction μ. Therefore, Eq. (3) becomes:

ma = FFfmg sin 26°
ma = FμNmg sin 26° (5)

At this point, we have two equations, Eq. (4) and Eq. (5), with two unknowns, F and N, so we can easily solve them.

Let's solve Eq. (4) for N:

0 = Nmg cos 26°
Nmg cos 26° = 0
N = mg cos 26°

And substitute it in Eq. (5):

ma = FμNmg sin 26°
ma = Fμmg cos 26°mg sin 26°
ma + μmg cos 26° + mg sin 26° = F
F = ma + μmg cos 26° + mg sin 26°
F = m (a + μg cos 26° + g sin 26°)
F = (30 kg) [2.5 m/s2 + (0.58) (9.8 m/s2) (cos 26°) + (9.8 m/s2) (sin 26°)]
F = (30 kg) (2.5 m/s2 + 5.1 m/s2 + 4.3 m/s2)
F = (30 kg) (11.9 m/s2)
F = 360 N

Hence, Micheal is pulling the mass with a force of 360 N.

Tips & Tricks

  • Even though it is good practice to find the value of only those unknowns that are required by the problem, sometimes you have to deal with long equations that have lots of variables in them. In that case it is acceptable to find the values of some unknowns that are not asked by the problem, so that you can keep your equations simple and short (e.g. instead of substituting N with mg cos 26°, you can just calculate the value of N, and keep it in the equation as such). In this problem though, we didn't do this, and our equations got a little bit complex, but you should feel free to simplify your equations whenever you think it makes sense.
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