# Magnitude of acceleration with friction

1. Oct 6, 2009

### bolivartech

1. The problem statement, all variables and given/known data

A block is pushed across a horizontal surface by the force shown (The force is from the left at an angle 30 degrees above the horizontal). If the coefficient of kinetic friction between the block and the surface is 0.30, F=20 N, theta=30 degrees, and M=3.0kg, what is the magnitude of the acceleration of the block.

2. Relevant equations

Fk=(.3)mg

F-Fk=ma

3. The attempt at a solution

(20 N)(cos 30) - (.3)(3.0 kg)(9.8 m/s2) = (3.0 kg)a

a = 2.83 m/s2

This is incorrect though, and some how should be 1.8. I have tried separating it into X and Y forces and taking the magnitude (square root of x2+y2, but that is even farther away)

2. Oct 6, 2009

### rl.bhat

Vertical component of the applied force contributes to the normal reaction. So you to recalculate the frictional force.

3. Oct 6, 2009

### bolivartech

Alright I've thrown this in a few varying ways, but let me just say what I'm thinking so you can let me know where I've gone wrong. The frictional force is the coefficient * mass * gravity, and the Normal is mass*gravity. Since the force has a vertical element then it is also contributing to the downward force, 20N*sin30. Do I just add the result of that downward force to the frictional force?

4. Oct 6, 2009

### rl.bhat

No. downward force 20Nsin30 adds to mg. These two constitute the normal reaction. From that find the frictional force.

5. Oct 6, 2009

### bolivartech

I've never seen it that way before but I think I'm starting to get it. In my free form diagram I drew the NL as mg + Fsin30. So Fk is not Ukmg, but UkNL. Which comes out to 11.82 in this scenario, and put in exactly as I had it setup before makes the acceleration 1.8. Thanks so much, not only did I get the right answer, I understand why!