Asked by jay
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
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
Answers
Answered by
Henry
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.
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.
Answered by
Henry
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