A box with a mass of 50 kg is dragged across the floor by a rope which makes an angle of 30degrees with the horizontal. Determine the coefficient of kinetic friction between the box and the floor if a 250-newton force on the rope is required to move the crate at a constant speed of 20 m/s.

The coefficient of kinetic friction between the box and floor is :  0.6

Given data :

Mass of box = 50 kg

Angle = 30°

Force required on rope  ( T ) = 250 N

speed = 20 m/s

g = 9.81 m/s²

Calculate the value of the coefficient of kinetic friction

First step : determine sum of forces in the y-direction

∑ Fy = ma

N  +  T sin 30°  - mg = 0

∴ N = mg - Tsin 30° ( 1 )

Next step : Determine the sum of forces in the x-direction

∑ Fx = ma

T cos 30° - Nμ = 0

∴ μ = T cos 30° / N   ( 2 )

where ; N = mg - T sin 30°

Final step : Determine the coefficient of kinetic friction between the box and floor

Back to equation ( 2 )

μ = T cos 30° / ( mg - T sin 30° )

    = 250 cos 30° / ( ( 50 * 9.81 ) - 250 * sin 30° ))

    = 0.59 ≈ 0.6

Hence we can conclude that The coefficient of kinetic friction between the box and floor is :  0.6

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0.6

Explanation:

There are four forces on the box:

Weight force mg pulling down,

Normal force N pushing up,

Tension force T pulling 30° above the horizontal,

and friction force Nμ pushing to the left.

Sum of the forces in the y direction:

∑F = ma

N + T sin 30° − mg = 0

N = mg − T sin 30°

Sum of forces in the x direction:

∑F = ma

T cos 30° − Nμ = 0

Nμ = T cos 30°

μ = T cos 30° / N

μ = T cos 30° / (mg − T sin 30°)

Plug in values:

μ = (250 N) cos 30° / ((50 kg) (9.8 m/s²) − (250 N) sin 30°)

μ = 0.59

Rounded to one significant figure, the coefficient of friction is 0.6.

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