What is the coefficient of kinetic friction?

Question

A 22 kg suitcase is dragged in a straight line at a constant speed of 1.10 m/s across a level airport floor by a student on the way to Mexico. The individual pulls with a 1.0x10^2 N force along a handle which makes an upward angle of 30 degrees with respect to the horizontal. What is the coefficient of kinetic friction between the suitcase and the floor?

 

Answer: μ_k=0.509

Answer 1

Apply Newton's second law in horizontal direction.

Fcos30 - μN = ma = 0, since a = 0

But N = mg-Fsin30 (balance in vertical direction)

So, μ = Fcos30/(mg-Fsin30) = 100.cos30/(22g-100.sin30) = 0.509, if g = 10 m/sec^2

Answer 2

Hey Sun -- 220N suitcase does get "lighter" by 50N with the 30 deg angle tug of 100N ...

Friction from the Fn of 170N must equate the horiz. tug of 85N ... Ksliding ~ half or 0.5 :)