# Dynamics: Forces at 90 and 45 degrees

1. Apr 20, 2013

### sv_heavymetal

1. The problem statement, all variables and given/known data
Hi, I have a system of forces which is showed in the following picture:

http://imageshack.us/photo/my-images/197/fuerzas.JPG/

The exercise says: A force F produces an acceleration of 5 m/s^2 when it acts on a particle of mass m. Calculate the acceleration of the particle when subjected to the forces as shown in the image, for the following cases: alpha: 90, alpha: 45.

As you can see, I did not give the values ​​of both F or m.

Results: The acceleration must be 7.1 m/s^2 to 90 ° and 9.2 m/s^2 to 45 °, that is, the smaller the angle the bigger the acceleration.

2. Relevant equations

F= m.a

3. The attempt at a solution

I think I have to use Pitagoras, using the 2 Fs as the legs and the resultant force will be the hypotenuse?

Thanks.

2. Apr 20, 2013

### Redbelly98

Staff Emeritus
Welcome to PF!
Pythagoras only works when the two Fs are at 90°, since it only applies to a right triangle. So yes, you can use that for the alpha=90° case.

When alpha=45°, you should draw yourself a free-body diagram, and look at components of the individual forces.