- #1
asteg123
- 12
- 0
QUESTION: If a 60kg package rests on a level floor with coefficients of static and kinetic frictions 0.56 and 0.37 respectively, what horizontal force is required to start the package's motion? how much force is needed to slide the package across the floor at constant velocity?
using the formulas f(static*max*)=mu * N
where mu(static) = 0.56 and N=m*g = 60*9.8 = 588N
so the maximum static friction would be 329.28N... so the force needed to start the package's motion would be any value GREATER than 329.28N
and that means that the kinetic friction is
f(kinetic) = mu * N
where mu(kinetic) = 0.37
so f(kinetic) = 217.56N
and it takes 217.56N of force to slide the package across the floor at constant velocity..
__________________________________
I was just wondering if what i did was correct... and if the force needed to move an object at a constant velocity less than the force needed to move it from rest...
THX>>>>
using the formulas f(static*max*)=mu * N
where mu(static) = 0.56 and N=m*g = 60*9.8 = 588N
so the maximum static friction would be 329.28N... so the force needed to start the package's motion would be any value GREATER than 329.28N
and that means that the kinetic friction is
f(kinetic) = mu * N
where mu(kinetic) = 0.37
so f(kinetic) = 217.56N
and it takes 217.56N of force to slide the package across the floor at constant velocity..
__________________________________
I was just wondering if what i did was correct... and if the force needed to move an object at a constant velocity less than the force needed to move it from rest...
THX>>>>