plzz answerrr?

Question

to move a large crate across a rough floor, you push on it with a force F at an angle of 21 degree below the horizontal. Find the force necesssary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57

Answer 1

The forces on the crate are

mg, down
N, up
Friction, back
F, at 21 degrees

Vertical forces: there is no acceleration so the net force must be zero.

0 = N + F sin 21 - mg

which rearranges to

N = mg - F sin 21

Horizontal forces: set them equal to zero to get the minimum F that get things moving

0 = F cos 21 - μ N

where μ is the coefficient of static friction.

Plug in μ and the expression for N found above and solve for F.

Answer 2

F is into the ground
so F cos 21 (=0.934 F) will move it forward
-----------------------
static friction will be on normal weight= (mg) + F sin 21
Force needed to just move it = f = u*[(mg) + F sin 21]
f = 0.57[32*9.8+ 0.358 F]
f = 178.75 + 0.2 F
f has to be overcome before it starts moving