A 24 kg crate initially at rest on a horizontal surface requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.
2007-11-13 13:56:02 · 3 answers · asked by mega4life2000 1 in Science & Mathematics ➔ Physics
M=24
F=75
U=?
W=MA
(24)(9.8)
W=235.2
N=235.2
What i did was find the force on its Y-axis to give me the normal(N). REmember Weight=Normal.
F=uN
75=u(235.2)
75/235.2=u
.318=u
This could be right im 98% sure.
2007-11-13 14:12:56 · answer #1 · answered by JEREMY L 1 · 0⤊ 1⤋
F = uN, N = mg
u = F/(mg) = 75 N/ (24 kg*9.8 m/s^2) do the math
2007-11-13 22:13:33 · answer #2 · answered by Snowflake 7 · 0⤊ 0⤋
Force to move equal to force of friction
F=f
f=u mg so
F=umg and
u= F/mg
u=75/(24 x 9.81)=
u=0.318
2007-11-13 22:13:25 · answer #3 · answered by Edward 7 · 1⤊ 0⤋