
#1
May2609, 04:46 PM

P: 49

1. The problem statement, all variables and given/known data
"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 nonzero number, but there are too many variables to just solve for "a". 2. Relevant equations Fnet=ma 3. 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? Thank you! 



#2
May2609, 05:42 PM

P: 1,623

Assume that the block starts at rest, so we'll consider the coefficient of static friction. As a hint, the product of the normal force and coefficient of static friction only gives the maximum value of force possible, not necessarily the frictional force that acts. For example suppose I apply a force of 15 N in the x direction on a block of mass 300 kg. If we take the coefficient of friction to be 0.4 then F_f = (0.4)(300 kg)(9.8 m/s/s) = 1176 N. If this frictional force were to act the block would be accelerating; however, we know that it stays at rest. Therefore, we know the frictional force must be equal to the 15 N I apply. Hopefully this makes sense and helps you solve the problem.



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