1. Force problem.

Force questions always confuse me because of the N. (newtons?)

Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 55.5 N and F2 = 42.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

http://img171.imageshack.us/img171/3181/0411vf7.gif

2. Originally Posted by rcmango
Force questions always confuse me because of the N. (newtons?)

Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 55.5 N and F2 = 42.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

http://img171.imageshack.us/img171/3181/0411vf7.gif
You want to start with a Free-Body Diagram. I'm defining a +x axis to the right and a +y axis upward.

You have two given forces, $F_1$ and $F_2$. There are two implied forces, the weight w acting downward, and the normal force N acting upward.

So take components of each force in the x and y directions and use Newton's 2nd in each component direction. (We automatically know that there's no acceleration in the y direction, right? So all the acceleration must be in the x direction.)

-Dan

3. okay, so you already have the x component of force 2 right, its the same as the force?

also using trig i believe the x component of the force 1 angle is 23.5

and i know accleration is net force / mass

so couldn't i just add the x components and divide by the mass?

i'm confused here.

4. Originally Posted by rcmango
okay, so you already have the x component of force 2 right, its the same as the force?

also using trig i believe the x component of the force 1 angle is 23.5

and i know accleration is net force / mass

so couldn't i just add the x components and divide by the mass?

i'm confused here.
You aren't confused, you have it right! (Just make sure that, in my coordinate system at least, that you have the x component of F2 as negative since F2 is pointing in the negative x direction.

-Dan

Edit: You should probably be carrying more digits in your intermediate answers, though, in case you aren't. For example $F_{1x} \approx 23.4553~N$ not 23.5. Using more decimals gives you a slightly better answer at the end, where you can then round off properly.

5. i think i'm close but not correct just yet.

okay so the x0 of force 2 is: -42.5

and x0 of force 1 is: 23.4553

so if i add these together i get: -19.0447

so then i divided -19.0447 by the mass

nevermind i just realised that i can't have a negative acceleration. its right

6. Originally Posted by rcmango
i think i'm close but not correct just yet.

okay so the x0 of force 2 is: -42.5

and x0 of force 1 is: 23.4553

so if i add these together i get: -19.0447

so then i divided -19.0447 by the mass

nevermind i just realised that i can't have a negative acceleration. its right
A detail that many people tend to forget is that these are vector equations. What we are doing is taking components along one direction or another. For simplicity we define a positive component to be in one direction and a negative component to be in the other.

Your first answer is correct: The acceleration comes out to be $-3.81~m/s^2$.

What does this mean? It simply means that the acceleration is in the -x direction, or as we defined it, off to the left of the diagram.

-Dan