1. A 40-kg. crate is at rest on a level surface. If the coefficient of static friction between the crate and the surface is 0.69, what horizontal force is required to get the crate moving?
2007-03-03 14:40:59 · 3 answers · asked by cj440288 1 in Science & Mathematics ➔ Physics
Force resisting motion is normal force * coefficient of friction.
F = 40kg * 9.8m/s^2 * .69. (answer in Newtons) So a little more than this force is needed to move the block.
2007-03-03 15:00:42 · answer #1 · answered by David S 4 · 0⤊ 0⤋
It would take the force of .7 to get it moving. By newtons first law that any object at rest stays at rest unless any force is acting upon it. This is also true if the object is moving at a constand speed. The oblect would need a net force(an unballanced force) inorder to accelerate.
2007-03-03 22:55:23 · answer #2 · answered by fatcat988 2 · 0⤊ 0⤋
f=uN
N=mg
f=(.69)(40)(9.8)
f=270.48N
2007-03-03 23:01:50 · answer #3 · answered by climberguy12 7 · 0⤊ 0⤋