A 40kg 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?

Answer 1

276 Newtons.

Coefficient of static friction is a ratio between the force it takes to budge the object over the normal force (mass times gravity).

So:

0.69 = Force / 400 N (40 kg times 10 m/s^2, the force of gravity).

Solve for 276 N.

Answer 2

Mass of the crate = 32 kg

Weight of the crate = 313.6 newtons

Virtual retarding force of friction = (313.6) x (0.57) = 178.75 newtons

That's the force it takes (acting parallel to the floor) to get it started sliding.

Answer 3

No, it won't.

A 50 kg crate with 0.42 coefficient of friction will have limiting friction of 21 kg-force. 60 Newtons is just under 7 kg-force.