Asked by laven
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).
(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).
Answers
Answered by
Henry
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.
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.
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