Friction problem - box on the table
A box that weighs 10.0 N is being dragged with constant velocity along a horizontal surface of the table by a rope that is at an angle α of 45 deg with that surface. The tension in the rope is 5.0 N. What is the coefficient of friction?
Solution.
Since the box is in motion with constant velocity, the sum of all forces acting on it must be equal ZERO. For easier calculations we decompose all forces into their horizontal (x direction) and vertical (y direction) components.
The drawing below visualizes the forces acting on the box.
Given are:
Fg = 10.0 N – weight of the box,
F = 5.0 N - tension in the rope,
α = 45 deg – angle between the rope and the surface of the table.
SOLUTION:
The sum of forces in the vertical direction is zero, as the weight Fg decreased by force F2 is compensated by the force of reaction (not shown in the Figure) according to Newton’s 3rd law.
The sum of all horizontal forces acting on the box must be 0, because the box is moving with constant velocity – Newton 1st law. The details of calculation are as follows.
We replace the force F by it components F1 and F2.
From trigonometry we have
F1 = F cosα (1)
F2 = F sinα (2)
The F2 component reduces the force with which the box presses on the table. The frictional force will the be
Ff = (Fg – F2) μK (3)
where μK is the coefficient of kinetic friction, because the box in our problem is moving.
As the box is moving with constant velocity, the force F1 must be equal to the Ff. From such requirements we have
F cosα = (Fg – F2) μK (4)
Substituting F2 from Eq. 2 into Eq. 3 we get
F cosα = (Fg – F sinα) μK (5)
and solving Eq. 5 with respect to μK , we have
μK = F cosα / (Fg - F sinα) (6)
which is an answer to our problem. Inserting into Eq. 6 numbers given in the problem we will get
μK = 0.547
The frictional coefficient is dimensionless, as it is a ratio of forces.
Such a large coefficient of kinetic friction suggests a very rough surface both of the table and of the box.