the crate is not accelerating
... so the frictional force equals the forces of the workers
m * g * μ = 380 N + 280 N ... μ = 660 / (360 * 9.8)
Two workers are sliding 360 kg crate across the floor. One worker pushes forward on the crate with a force of 380 N while the other pulls in the same direction with a force of 280 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
2 answers
Mg = 360*9.8 = 3528 N. = Wt. of crate = Normal force, Fn.
380+280 = 660 N. = Force applied.
u*Fn = u*3528 = Force of kinetic friction.
660-3528u = M*a
660-3528u = M*0 = 0
u = 660/3528.
380+280 = 660 N. = Force applied.
u*Fn = u*3528 = Force of kinetic friction.
660-3528u = M*a
660-3528u = M*0 = 0
u = 660/3528.
2 answers