A horse drags a harrow of mass 50 kg across a field by a rope at a constant speed. f inhe rope is kept horizontal, what is the coefficient of friction if the force of the horen
the rope is 350 N.

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.

ok then what happened.