Question

A 45 n toolbox is dragged horizontally at constant speed by a rope making an angle of 40 with the floor the tension in the rope is 20 n, what is the coefficient of kinetic friction?

Answer 1

The freebody diagram of the considered system is shown below.

The net force equals 0 because the speed is constant. Hence, we obtain:

$$\vec{N}+\vec{T}+\vec{mg} +\vec{F}_{fr} =0$$

which can be written in terms of x- and y-projections as:

$$F_{fr} = T \cos{\alpha}\\ N + T \sin{\alpha} = mg$$

Taking into account that

$$F_{fr} = \mu N$$

one can derive:

$$\mu = \frac{T \cos{\alpha} }{N} = \frac{T \cos{\alpha}}{mg - T \sin{\alpha}}$$

Substituting the numerical values, we obtain:

$$\mu = \frac{20 \cos{40^\circ}}{40 - 20 \sin{40 ^\circ}} \approx 0.56$$

Answer: 0.56