component of weight down truck bed = m g sin 24.1= 20*9.81*sin24.1
= 80.1 N
friction force up truck bed = 114.1 N
component of acceleration up truck bed= a cos 24.1
so
-80.1 + 114.1 = m a = 20 cos 24.1 * a
a = 34 / 20 cos 24.1
A box of mass 20 kg is in the bed of a truck on a hill that is inclined at an angle of 24.1° above the horizontal. The tailgate of the truck is open. The truck is currently accelerating up the hill with a magnitude of 1.7 m/s2. themagnitude of the friction force from the truck on the box in this situation is 114.1N.
If the coeffcient of static friction between the box and the truck is 0.78, then what is the maximum acceleration that the truck can have if the box is not to slip
1 answer
1 answer