Calculating Friction Force in a Truck Accelerating at 2.27 m/s^2

[azila]

Homework Statement

A 20.0 kg packing case is initially at rest on the floor of a 1400-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 2.27 m/s^2 northward.

B. Find the direction of the friction force acting on the case when the truck accelerates at 2.27 m/s^2.

Homework Equations

Fk = (coefficient of kinectic)N
Fs = (coefficient of static)N
F = ma

The Attempt at a Solution

Ok, first of all when I calculate the weight, I would combine the masses of the truck and the case right? should I use the F=ma formula and plug in the masses and acceleration. Find the force. Now, what do I do?? This is where I am confused. If you can help, I would appreciate it. thanks in advance.

 

[learningphysics]

What are the forces acting on the case? Is the case sliding in the truck... or is it not sliding?

 

[azila]

the case is not sliding...it is just sitting in the truck;

 

[learningphysics]

the case is not sliding...it is just sitting in the truck;

Ok. What are the forces acting on the case? Write the \Sigma\vec{F} = m\vec{a} equation for the case.