normal force = 3.6 * 9.81 * cos 35 + F
friction force = .33 ( 3.6 * 9.81 * cos 35 + F) = 9.55+.33 F
tat has to balance component of weight down slope
9.55 + .33 F = 3.6 * 9.81 * sin 35 = 20.3
.33 F = 10.7
F = 32.4 N
www.webassign.net/sercp9/4-p-047-alt.gifThe coefficient of static friction between the m = 3.60−kg
crate and the 35.0° incline of the figure below is 0.330. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
1 answer