A student pulls her 22-kg suitcase through the airport at constant velocity. The pull strap makes an angle of 50 degrees above the horizontal. (a) If the frictional force between suitcase and floor is 75N, what force is the student exerting? (b)What's the coefficient of friction?
μk x N
The Attempt at a Solution
So, I have the answer to this in the back of the book, but I can't figure it out. Here is my attempt:
First I split tension in two components and tried to find the x and y Fnet,
y: -mg + N + T x sin50 = ma = 0 (constant velocity = zero acceleration)
x: T x cos50 - 75N = ma = 0
Algebraically reworking this:
(T x Cos50) / μk = N
so substituting this in, my final equation should be:
The part where I am confused is how to find the coefficient. I tried all the ways I knew and I know the hint is that the friction is 75N. I tried μk=75N/215.6N (m x g, 22kg x 9.8) = .34 but this is not correct as it isn't right in the back of the book and I know that the normal force is less because of the angle of the tension.
Ultimately, I am confused because it seems circular. I need the normal force to find the coefficient but I need the coefficient to find the normal force. Every other equation I have seen like this provides the coefficient.
Any help is greatly appreciated.