# Problem: Block pushed over the floor with a downward and forward force

Frank is pushing a block of 50 kg over the floor, with a force of 600 N downward and forward, making a 20° angle with the horizontal. The coefficient of sliding friction between the block and the floor is 0.39.

Find the acceleration of the block and the friction force acting on the block.

## Solving the problem

Let's try to sketch what is happening.

We need to draw a horizontal surface, a block on that surface, and indicate that the block is pushed by a downward-forward force that makes 20° with the horizontal. We also need to indicate that there is friction between the block and the floor:

Next, we look at our sketch, and try to list all the forces that act on the block.

There are 4 forces acting:

• Frank's pushing force, which we'll indicate with F
• the friction force, Ff
• the force of gravity, mg
• and the normal force N, which the surface exerts on the block

Knowing the forces, let's draw a free-body diagram for the block:

Let's take a step back and examine what we know, and what the need to find:

We know the mass of the block (50 kg), the push (600 N at 20° downward-forward), and the coefficient of friction (0.39).

We need to find the acceleration of the block and the force of friction.

m = 50 kg
F = 600 N
θ = 20°
μ = 0.39

### We want to know

a = ?
Ff = ?

Since we know the mass of the block, if we first find the resultant force, we can then apply Newton's 2nd Law to get the block's acceleration.

Let's start by drawing the coordinate axes on our free-body diagram, and determine the x and y components of all the forces acting on the block.

Fx = F cos 20°
Ffx = −Ff
Nx = 0
mgx = 0
Fy = −F sin 20°
Ffy = 0
Ny = N
mgy = −mg

We can now find the x and y components of the resultant force:

x:

Rx = Fx + Ffx + Nx + mgx
Rx = F cos 20° + (−Ff) + 0 + 0
Rx = F cos 20°Ff (1)

y:

Ry = Fy + Ffy + Ny + mgy
Ry = (−F sin 20°) + 0 + N + (−mg)
Ry = NF sin 20°mg (2)

For Eq. (1) we need to find the friction force Ff.

By definition, the direction of the sliding friction force will be opposite to the motion of the block, and its magnitude will be equal to the product between the coefficient of sliding friction and the normal force:

Ff = μN

We know μ, but we don't know N.

Where did we see N before?

We saw N in Eq. (2):

Ry = NF sin 20°mg

Since the block is moving along the floor, Ry must be zero (otherwise there would be a vertical acceleration):

Ry = 0

Therefore:

0 = NF sin 20°mg

The only unknown in this equation is N, so we can solve it:

0 = NF sin 20°mg
NF sin 20°mg = 0
N = F sin 20° + mg

Having found N, we can calculate Ff:

Ff = μN
Ff = μ (F sin 20° + mg)
Ff = (0.39) [(600 N) (sin 20°) + (50 kg) (9.8 N/kg)]
Ff = (0.39) (205 N + 490 N)
Ff = 270 N

Now that we know Ff, we can find Rx from Eq. (1):

Rx = F cos 20°Ff

Rx is positive (since the block accelerates in the positive x direction), and we have shown that Ry is zero. This means that R will have direction in the positive x axis, and its magnitude will be equal to Rx:

R = Rx
R = F cos 20°Ff

Now that we finally found the resultant force, let's apply Newton's 2nd Law to find the acceleration of the block:

R = ma
 a = R m
 a = R m
 a = F cos 20° − Ff m
 a = (600 N) (cos 20°) − 270 N 50 kg
 a = 294 N 50 kg
a = 5.9 N/kg = 5.9 m/s2

And with that we have found everything we needed:

a = 5.9 m/s2
Ff = 270 N

## Tips & Tricks

• As you may have noticed, we didn't find the value of every unknown, but only of those that were required by the problem. This is generally a good practice. The obvious advantage of this is that you use your calculator less often and you don't spend your time finding unknowns you're not asked to find. Another advantage is that you can see what relationships exist between different variables under particular circumstances. This is useful because whenever you will encounter a similar problem, when you are in the middle of solving it, you will be able to recognize equations, and thus know that you're solving it correctly.

## Exercises

### #1

Robert is pushing a block of 61 kg over a horizontal surface. He pushes downward and forward, making an angle of 30° with the horizontal. Assuming that the push has magnitude 490 N, and that the coefficient of sliding friction between the block and the surface is 0.44, find the acceleration of the block, and the force of friction acting on the block.

a = 0.89 m/s2
Ff = 370 N
Phyley © 2019
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