A woman pushes a crate along the floor of a warehouse with a force of 31.9 N directed downward at an angle of 23.6° from the horizontal. If the crate has a mass of 11.9 kg and the coefficient of friction is 0.0975, find...
(a) the force of friction
b. (b) theacceleration
Wc = mg = 11.9 kg * 9.8 N/kg = 116.6 N=
Weight of crate.
Fc = 116.6 N @ (0) Deg.=Force of crate.
Fp = 116.6*sin(0) = 0 = Force parallel to floor.
Fv = 116.6*cos(0) = 116.6 N. = Force
perpendicular to floor.
a. Ff = u*(Fv+31.9*sin23.6),
Ff = 0.0975*(116.6+12.77) = 12.61 N.
Fn = Fap - Fp - Ff,
Fn = 31.9*cos23.6 - 0 - 12.61=16.62 N.
= Net force.
b. a = Fn/m = 16.62 / 11.9 = 1.397.