Jason works for a moving company. A 95 kg wooden crate is sitting on the wooden ramp of his truck, the ramp is angled at 11 degrees.
What is the magnitude of force, directed parallel to the ramp, that he needs to get it to start moving up the ramp?
What is the magnitude of force, directed parallel to the ramp, that he needs to get it to start moving down the ramp?
2 answers
Maria/John -- please don't bother switching names. It's silly.
a. M*g = 95 * 9.8 = 931 N. = Wt. of crate.
Fp = 931**sin11=178 N. = Force parallel to incline.
Fn = 931*Cos11 = 914 N. = Normal force.
u = 0.40? = coefficient of starting friction.
Fs = u*Fn = 0.4 * 914 = 366 N. = Force of starting friction.
F-Fp-Fs = M*a.
F-178-366 = 95*0,
F = 544 N.
b. F+Fp-Fs = M*a.
F+178-366 = 95*0.
F =
Fp = 931**sin11=178 N. = Force parallel to incline.
Fn = 931*Cos11 = 914 N. = Normal force.
u = 0.40? = coefficient of starting friction.
Fs = u*Fn = 0.4 * 914 = 366 N. = Force of starting friction.
F-Fp-Fs = M*a.
F-178-366 = 95*0,
F = 544 N.
b. F+Fp-Fs = M*a.
F+178-366 = 95*0.
F =
1 answer