A 21.0-kg packing case is initially at rest on the floor of a 1440-kg pickup tru

Question

A 21.0-kg packing case is initially at rest on the floor of a 1440-kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed.

A) Find the magnitude of the friction force acting on the case when the truck accelerates at 2.28m/s2 northward. f=?

B) Find the direction of the friction force acting on the case when the truck accelerates at 2.28m/s2 northward. NORTH, SOUTH, WEST or EAST?

C) Find the magnitude of the friction force acting on the case when it accelerates at 4.49m/s2 southward. f=?

D) Find the direction of the friction force acting on the case when it accelerates at 4.49m/s2 southward. NORTH, SOUTH, WEST or EAST?

Explanation / Answer

Let:
m be the mass of the case,
F be the friction force,
u[s] be the coefficient of static friction,
u[k] be the coefficient of kinetic friction,
a be the acceleration of the truck,
g be the acceleration due to gravity.

(a)
ma = 21 * 2.28
= 47.88N

f=u[s]mg = 0.30 * 21 * 9.81
= 61.803N

b)As limiting static friction has not been reached, the case does not slide.
F = 61.803N northward.

(c)
ma = 21* 4.49
= 94.29N

f=u[k]mg = 0.20 * 21 * 9.81
= 41.202N

d)As limiting static friction has been exceeded, the case slides.
F = 41.202N southward.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.