Question

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip. You start to wonder how heavy the crate is. You call to ask the boy his mass, and he replies “50 kg.” From your recent physics class you estimate that the static and kinetic coefficients of friction are 0.8 and 0.4 for the boy’s shoes, and 0.5 and 0.2 for the crate. Estimate the mass of the crate.

Answer 1

The figure below shows all the forces acting on the boy and on the crate while the boy is pushing the crate. The crate and the boy are drawn separately for clarity.

The red arrows indicate the forces on the crate while the blue arrow indicates the forces on the boy. , are the masses of the crate and the boy respectively. is the frictional force on the crate from the floor and is the frictional force on the boy from the floor.

The vertical forces on the crate and the boy should be balanced as there is no motion in the vertical direction. This implies,

…… (1)

…… (2)

When the boy is pushing on the crate with force, the boy feels a force on him exerted by the crate. These two forces are a Newton’s third law forces pair and hence they are of equal magnitude ().

As the boy is barely moving the crate, it should be that the net horizontal force on the crate is zero. This means that,

…… (3)

We also know that the frictional force on the crate is related to the normal reaction on the crate from the floor through.

As the boy is barely moving the crate we should use for the coefficient of friction between the floor and the crate (). Thus,

…… (4)

Here, is the kinetic coefficient of friction between the floor and the crate.

Use the equations (3) and (4), and solve for the applied force by pushing.

Substitute for.

…… (5)

As pointed out earlier. Thus,

…… (6)

Now we can concentrate on the forces acting on the boy. Since the boy is at horizontal equilibrium, the two opposing horizontal forces on the boy should cancel each other. This gives,

…… (7)

Once again, the frictional force on the boy is related to the normal force on him through where is the frictional coefficient between the floor and the boy.

Substitute for in equation (7) to solve for.

…… (8)

Substitute for to solve for.

…… (9)

From equation (8) we can see that the highest pushing force the boy can exert on the crate, (which is equal to), is determined by the frictional coefficient between the boy and the floor as the mass of the boy is fixed.

We are given that the crate is barely moving even when he is applying the highest force he can. The highest force on the crate occurs when is equal to the highest value it can take, the static coefficient of friction between the boy and the floor (). Thus, we can rewrite equation (8) by replacing with.

…… (10)

Equate the equations (6) and (9) and solve for.

…… (11)

Substitutefor, for, and forin equation (11) to solve for.

The mass of the crate can be estimated to be about.