A232-kg crate is being pushed across a horizontal floor by a force p that makes an angle of 31.8 ° below the horizontal. the coefficient of kinetic friction is 0.269. what should be the magnitude of p, so that the net work done by it and the kinetic frictional force is zero?

Suppose the crate is pushed a distance x. Only 2 forces do any work in the process:

• the horizontal component of p, with magnitude p cos(-31.8°) = p cos(31.8°)

• kinetic friction, with magnitude f = µn and where n is the magnitude of the normal force of the floor pushing upward on the crate and µ = 0.269.

If the net work done by these forces is 0, then

fx = p cos(31.8°) x   ==>   f = p cos(31.8°)

which in turn means the crate is pushed with constant speed.

By Newton's second law, we have

• net force acting on the crate in the horizontal:

∑ F = p cos(31.8°) - 0.269n = 0

• net force acting in the vertical:

∑ F = p sin(-31.8°) + n - mg = 0

Solve this system for p :

p cos(31.8°) - 0.269n = 0

-p sin(31.8°) + n = (232 kg) g = 2273.6 N

Multiply through the second equation by 0.269, then eliminate n :

p cos(31.8°) - 0.269n = 0

-0.269 p sin(31.8°) + 0.269n = 2273.6 N

==>   p (cos(31.8°) - 0.269 sin(31.8°)) = 2273.6 N

==>   p ≈ 3210 N

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.