Wt. of crate = m * g =
3.2kg * 9.8N/kg = 31.36 N.
Fp = 31.36*sin35 = 17.99 N. = Force
parallel to the incline.
F = Fp*cos35 = 17.99*cos35 = 14.74 N.
The coefficient of static friction between the m = 3.20-kg crate and the 35.0° incline of the figure below is 0.260. What minimum force vector F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
1 answer