An 18 kg sled is being pulled by a child holding a rope with an applied force at an angle of 45 degrees with the ground. The normal force is 50N and the sled is accelerating at 6 m/s2

Question

An 18 kg sled is being pulled by a child holding a rope with an applied force at an angle of 45 degrees with the ground. The normal force is 50N and the sled is accelerating at 6 m/s2. What is the force applied to the sled? what is the coefficient of friction?

Answer 1

Given:

m = 18 kg

θ = 45°

N = 50 N

a = 6 m / s²

Find:

F = ? N

μ = ?

Solution:

+→ΣF_x = ma ma = F * cos( 45° ) - f

+↑ΣF_y = 0 0 = F * sin( 45° ) + N - mg

mg - N = F * sin( 45° )

F = ( mg - N ) * csc( 45° ) = ( 9.81 * 18 - 50 ) * √( 2 )

∴ F ≈ 179.01 N

μN = ( mg ‐ N ) cot( 45° ) - ma

μN = m * ( g - a ) - N

μ = m / N * ( g - a ) - 1 = ( 18 / 50 ) * ( 9.81 - 6 ) - 1

∴ μ ≈ 0.372

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Answer 2

Hi Jane J.,

Why are you asking this question? You can either work forwards from the conditions given, or backwards from the result obtained. I suggest the former. Note the angle of the rope: the pull force applied in the 45 degree upwards direction must decompose into equal horizontal and vertical (normal) components. That's in the nature of a 45-45-90 degree triangle, and that is your necessary entry into this problem!

That will give you the horizontal APPLIED force vector. But from the acceleration and the mass you may calculate the NET horizontal force. The APPLIED minus the NET horizontal gives the horizontal FRICTION force. Relate that to the normal force, to find the dynamic friction coefficient.

By the way, I'm a little doubtful that the child can keep this up for a very long time. 5 seconds and the sled would be travelling 30 m s^-1, about 60 miles hr^-1 . A very fast sports car might be able to do this. Bet your instructor's car can't. Just sayin'

-- Cheers, --Mr. d.