a. M*g = 1100 N. = Wt. of crate.
M = 1100/g = 1100/9.8 = 112 kg = Mass of crate.
Fn = M*g + Fap*sin20 = Normal force.
Fn = 1100 + 500*sin20 = 1271 N.
Fk = u*Fn = u*1271 = 1271u. = Force of kinetic friction.
Fap*Cos20-Fk = M*a.
500*cos20-1271u = M*0
469.8-1271u = 0,
u = 0.370.
b. Fn = M*g - Fap*sin20.
Fn = 1100-500*sin20 = 929 N.
Fk = u*Fn = 0.370 * 929 = 343.7 N.
500*cos20-343.7 = 112a.
a = 1.13 m/s^2.
A 1100 N crate is being pushed across a level floor at a constant speed by a force of 500 N at an angle of 20.0° below the horizontal. (a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 500 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in Figure P4.37b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).
1 answer
1 answer