Asked by casey
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??
Answers
Answered by
Henry
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.
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.
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