A 1,440-N crate is being pushed across a level floor at a constant speed by a force of 250 N at an angle of 20.0° below the horizontal;

Question

(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 250-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

Answer 1

Fc = 1440N @ 0 Deg. = Force of crate.
Fp = 1440*sin(0) = 0 = Force parallel to floor.
Fv = 1440*cos(0) = 1440 N. = Force perpendicular to floor.

Fv' = Fv + Fap*sin20.
Fv'=1440 + 250*sin20 = 1526 N=Normal.

a. Fn = Fap - Fp - u*Fv' = 0.
250*cos20 - 0 - u*1526 = 0.
1526u = 250*cos20 = 235.
u = 235 / 1526 = 0.154 = Coefficient of kinetic friction.

b. mg = 1440 N.
m = 1440/g = 1440 / 9.8 = 147 kg = Mass
of crate.

Fv' = 1440 - 250*sin20 = 1354 N. =[Normal.

Fn = Fap - Fp - u*Fv' = ma.
250*cos20 - 0 - 0.154*1354 = 147a.
235 - 209 = 147a.
147a = 26.
a = 26 / 147 = 0.177 m/s^2.