Posted by greg on Thursday, March 15, 2012 at 9:33am.
A 1,440N 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;
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 250N 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).

physical aspects of nature  Henry, Friday, March 16, 2012 at 8:28pm
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.