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Problem: Two blocks connected by a string are pulled horizontally

Two wooden blocks, connected by a massless string, are moving at constant velocity over a floor with friction. The smaller of the two blocks is pulled by a horizontal force of 500 N.

Knowing that the mass of the smaller block is 30 kg and the mass of the larger block is 40 kg, find the tension in the string and the coefficient of sliding friction between the blocks and the floor.

Solving the problem

Let's begin by drawing a sketch that represents what is happening in the problem.

In the sketch, we draw a horizontal floor with two blocks that are connected by a string sitting on it. Then, we indicate that the smaller of the two blocks is pulled by a horizontal force and that both the blocks are subject to the force of friction. It's also a good idea to write that the two blocks move at constant velocity. The final sketch looks something like this:

The smaller block, connected to the larger block by a string, is pulled horizontally.Velocity is constantPullFrictionMmFriction

Since the string connecting the two blocks is massless, the tension forces exerted by the ends of the string have the same magnitude, which we will indicate with T.

Also, since the two blocks are made of the same material (wood) and they move on the same floor, the coefficient of sliding friction is the same for both blocks, and we will indicate it with μ.

In this problem, we are dealing with two blocks, so it's a good idea to analyze them separately.

Looking carefully at our sketch, we can see that the smaller block is subject to 5 forces:

  • The pulling force, F
  • The tension force, T
  • The friction force, ff
  • The gravitational force, mg
  • The normal force, n

Here's the free-body diagram of the smaller block:

The smaller block with the forces that act on it: pulling force to the right, tension and friction to the left, gravity down, and normal force up.FffnmgT

The larger block, on the other hand, is subject to 4 forces:

  • The tension force, T
  • The friction force, Ff
  • The gravitational force, Mg
  • The normal force, N

Here's the free-body diagram of the larger block:

The larger block with the forces that act on it: tension to the right, friction to the left, gravity down, and normal force up.FfNTMg

We know the masses of the two blocks (30 kg and 40 kg), the magnitude of the pulling force (500 N), and that the two blocks are moving at constant velocity.

We need to find the magnitude of the tension in the string (T), and the coefficient of sliding friction (μ).

We know

m = 30 kg
M = 40 kg
F = 500 N
v is constant

We want to know

T = ?
μ = ?

Since the blocks are moving at constant velocity, they have zero acceleration, so the net force on the smaller block (fnet) is zero, and the net force on the larger block (Fnet) is also zero:

fnet = ma = m × 0 = 0
Fnet = Ma = M × 0 = 0

We can find the tension in the string and the coefficient of sliding friction following these steps:

  1. First, we find the x and y components of the net force that acts on the smaller block in terms of the x and y components of all the forces that act on the smaller block.
  2. Then, since fnet is zero, fnetx and fnety must also be zero, so we replace them with 0 in the equations that we found in step 1, and arrive at a final equation.
  3. We repeat the equivalent of the previous steps for the larger block and arrive at a final equation for the larger block.
  4. Lastly, we solve the two final equations and find the tension T and the coefficient of sliding friction μ.

Let's begin with step 1.

To find the components of the net force that acts on the smaller block, we first draw the coordinate axes on the free-body diagram of the smaller block and enumerate the components of all the forces that act on it:

The smaller block, centered on the origin of the coordinate axes, with the forces that act on it.xyFffnmgT
Fx = F
Tx = −T
ffx = −ff
nx = 0
mgx = 0
Fy = 0
Ty = 0
ffy = 0
ny = n
mgy = −mg

We can now find the x and y components of the net force by adding the x and y components of all the forces:

x:

fnetx = Fx + Tx + ffx + nx + mgx
fnetx = F + (−T) + (−ff) + 0 + 0
fnetx = FTff

y:

fnety = Fy + Ty + ffy + ny + mgy
fnety = 0 + 0 + 0 + n + (−mg)
fnety = nmg

And we replace fnetx and fnety with 0:

0 = FTff (1)
0 = nmg (2)

Additionally, we know that the magnitude of the friction force ff is the product between the coefficient of friction μ and the magnitude of the normal force n:

ff = μn

Therefore, Eq. (1) becomes:

0 = FTμn (3)

Next, let's solve Eq. (2) for n:

0 = nmg
nmg = 0
n = mg

And replace n with mg in Eq. (3):

0 = FTμmg (4)

This final equation has 2 unknowns (T and μ).

Let's now repeat the same process on the larger block in order to find the second equation.

Again, we draw the coordinate axes on the free-body diagram and determine the x and y components of each force:

The larger block, centered on the origin of the coordinate axes, with the forces that act on it.xyFfNMgT
Tx = T
Ffx = −Ff
Nx = 0
Mgx = 0
Ty = 0
Ffy = 0
Ny = N
Mgy = −Mg

Then, we calculate the x and y components of the net force Fnet:

x:

Fnetx = Tx + Ffx + Nx + Mgx
Fnetx = T + (−Ff) + 0 + 0
Fnetx = TFf

y:

Fnety = Ty + Ffy + Ny + Mgy
Fnety = 0 + 0 + N + (−Mg)
Fnety = NMg

Again, because Fnet is zero, the components are zero as well, so we can replace them in the above equations with 0:

0 = TFf (5)
0 = NMg (6)

We can write the magnitude of the friction force as:

Ff = μN

Therefore, Eq. (5) becomes:

0 = TμN (7)

Next, let's solve Eq. (6) for N:

0 = NMg
NMg = 0
N = Mg

And swap N with Mg in Eq. (7):

0 = TμMg (8)

This is the final equation for the larger block.

Thus, the two final equations are Eq. (4) and Eq. (8):

0 = FTμmg
0 = TμMg

These are two equations with two unknowns (T and μ), so we can solve them.

Let's solve Eq. (8) for T:

0 = TμMg
TμMg = 0
T = μMg (9)

And replace T with μMg in Eq. (4):

0 = FTμmg
0 = FμMgμmg

Now, let's solve this equation for μ:

0 = FμMgμmg
μMg + μmg = F
μg (M + m) = F
μF
g (M + m)
μ500 N
(9.8 N/kg) (40 kg + 30 kg)
μ500 N
(9.8 N/kg) (70 kg)
μ = 0.73

Lastly, we can get T from Eq. (9):

T = μMg
T = (0.73) (40 kg) (9.8 N/kg)
T = 2.9 × 102 N

Therefore, the tension in the string is 2.9 × 102 N, and the coefficient of sliding friction is 0.73.

Tips & Tricks

  • You have to have as many independent equations as there are unknowns in order to solve them. When you have more unknowns than equations, look carefully at each unknown and ask yourself whether there is a way to express that unknown quantity in terms of other known quantities or a mix of known and unknown quantities. This will help you to reduce the number of unknowns that appear in your equations.
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