a man is pulling a 100kg crate with a force of 250 newtons. the rope makes a downward angle of 30.0 degrees with the horizontal. if the crate is sliding at a constant rate, calulate the coefficient. ?? how do i do this??

Answer 1

External Force:

F = 250N @ 30 deg.

X = hor. = 250*cos30 = 216.51N.

Y = ver. = 250*sin30 = 125N.

The Crate Force:

100kg * 9.80 = 980N @ 0 deg.

Fh = 216.51.

Fv = 980 + 125 = 1105N.

Fh - u*Fv = ma,
a = 0 @ constant rate of speed.
Therefore, Fh - u*Fv = 0,
u*Fv = Fh,
1105u = 216.51,

u = 216.51 / 1105 = 0.196.