"A force of 300N is applied to a 70kg carton on a level floor at an angle of 15deg below the horizontal. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.28. Find the friction force acting on the carton."
I don't know which coefficient to use - static or kinetic. I know that static friction involves a stationary object, so a=0 and Fnet=0... and kinetic friction involves a moving object, so Fnet=ma. I would ideally look to solve for acceleration, to see if its a zero or non-zero number, but there are too many variables to just solve for "a".
The Attempt at a Solution
**u=coefficient of friction
I broke up the forces into their components:
F(y) = n - W - F(y) = n - mg + Fsin(theta) = 0
F(x) = Fcos(theta) - f(friction) = ma
So, F(x)--> un = Fcos(theta) - ma
I know that if its static friction, F(x)=f.. and if its kinetic friction, F(x)>f
The answer is 289.9N, which is Fcos(theta), but when I solve for kinetic and static friction seperately once I've found the normal, I don't get this answer at all. I'm assuming from the answer, that static friction is used, because that's why friction=Fcos(theta), so it doesn't move.
However, when I solve f(k)=u(k)*n = 0.28(603N)=167N
And f(s)=u(s)*n = 0.4(603N) = 241.2N
Both values of friction don't give me 289.9N,
Can someone help me out?