# Homework Help: AP Test on Force Corrections (3 of them, but only one at a time)

1. Oct 11, 2009

### RustyComputer

1. The problem statement, all variables and given/known data
A 2-kilogram block slides down a 30 degree incline as shown above with an acceleration of 2 meters per second squared. The magnitude of the frictional force along the plane in most nearly:

2. Relevant equations
F=ma

3. The attempt at a solution
So the first thing I did was draw a free body diagram, with all three forces present. Force of friction going opposite of the direction the box is going, normal force perpendicular from the surface, and mass times gravity going straight down.

Then I tilted the diagram 30 degrees, to make the force of friction and normal force straight vertical and horizontal and put mg at an angle of 30 degrees to the left from it's initial point.

After that, I made a right triangle to find the horizontal and vertical components of mg, and wrote mg as the hypotenuse, mg_x as the horizontal component, and mg_y as the horizontal component.

Then here are the equation stuff I did.

F_f - mg = ma
F_f = ma + mg_x
F_f = ma + mgsin30
F_f = m (a + gsin30)

Then I plugged everything in,

F_f = 2(2+9.8sin30)

And came up with the answer

F_f = 13.8

But the problem is that there are no answers close to my solution with the given. So what I would like to know for this problem is, where did I go wrong?

2. Oct 12, 2009

### rl.bhat

F_f = m (a + gsin30)
This expression is wrong. Downward force in mgsinθ. Frictional force acts in the upward direction. So the net force
mgsin30 - F_f = ma

3. Oct 14, 2009

### RustyComputer

So the correct way to go about that is:

mgsin30 - F_f = ma
F_f = - ma + mgsin30
F_f = m (- a + gsin30)

Then plugging in:

F_f = 2 (- 2 + 9.8sin30)
F_f = 5.8

And as it turns out, that is a multiple choice answer, and it is correct.

I'll have the next question up in a bit...