posted by jay on .
A 22 kg box rests on the back of a truck. The
coefficient of static friction between the box
and the truck bed is 0.369.
The acceleration of gravity is 9.81 m/s2 .
What maximum acceleration can the truck
have before the box slides backward
Wb = mg = 22kg * 9.8N/kg = 215.6N. =
Weight of box.
Fb = (215.6N,0 deg).
Fp = Fh = 215.6sin(0)= 0 = Force paral-
lel to plane = Hor. force.
Fv = 215.6cos(0) = 215.6N. = Force perpendicular to plane = Normal.
Ff = u*Fv = 0.369 * 215.6 = 79.56N. =
Force due to friction.
Fn = Fp - Ff = 22a,
0 - 79.56 = 22a,
a = -3.62m/s^2.