Free Body Diagrams

1. Oct 22, 2013

frownifdown

1. The question I'm struggling on has been answered on here before but with different values. Here is the link to the question https://www.physicsforums.com/showthread.php?t=344213 but on mine I have a 1.9kg object and F1=5 and F2=2.

I figured out that the Y components are 0 for both of them and I have the x component solved for F1, but I was sort of following it step by step and I'm unsure of why I did what I did. The equation used by gbednba in this example was Ax=3sin20=1.02 --> -1.02-1+5=2.97/2.6=1.14 which is the answer. I don't really understand what the 20 degree angle has to do with solving it, so if you could help me incorporate it into the problem I would appreciate that a lot. I guess I just don't really know what is going on here so if you can help me out, that would be great.

Oh and the one problem I still have to solve is for Ax of diagram B, which I don't even know what to do for.

2. Oct 22, 2013

Simon Bridge

Welcome to PF;
The values don't matter - you need to apply the method not the solution.
When you put a set of axis on your diagram, you will find that some forces are at an angle to some axis.

In the diagram on the left:

the 3N force is 20deg from the -y axis - which is why the 20deg is important.
When you sum the forces in the y direction you get: $2.82-3\cos(20^\circ)=ma_y$

See how the axis in the diagrams have been rotated to simplify the math?
In diagram A it was most convenient to rotate the axis 20deg from the usual position because that way most of the forces lined up with some axis - less trig to do.
In diagram B there was more choice - they chose a 15deg rotation to line up the force being calculated - so there are fewer steps in the calculation.

Last edited: Oct 22, 2013
3. Oct 23, 2013

frownifdown

Ah, thank you so much. So now I understand what the angle has to do with it, but I'm looking at B and am still a little stumped. When solving for the x component of the acceleration, would I take F2 (2) - 2cos(15) to get the net forces on the x axis, or do I need to figure out the x component of the 2N on the positive X axis as well and combine that (2cos15=1.93) with the F2 (-2) to get -.07 as the net forces acting on the X axis.

Sorry if my reply was confusing, it just reflects where I am right now.

Edit: Nevermind, I just figured it out! Thank you so much for all your help.

Last edited: Oct 23, 2013
4. Oct 23, 2013

Simon Bridge

Well done ... just for the next person who comes along:
All the forces have to be resolved to components along the axes. It is just that the axes have been chosen so that some forces only have one non-zero component.

i.e. for B: $\vec{F}_2 = -F_2\hat{\imath} + 0\hat{\jmath}$
Resolving to axes means that we can just use magnitudes instead of having to do a whole vector equation.
The full equation for B would go like this:

$[-F_2\hat{\imath} + 0\hat{\jmath}]+ [0\hat{\imath} + (1.414)\hat{\jmath}] + [2\cos(15^\circ)\hat{\imath} + 2\sin(15^\circ)\hat{\jmath}]+[2\sin(15^\circ)\hat{\imath} -2\cos(15^\circ) \hat{\jmath}]=m[a_x\hat{\imath} + a_y\hat{\jmath}]$

... grouping the terms in i and j gives you the familiar pair of equations.

The core trick here is to realize that the orientation of the x and y axes is totally arbitrary - y does not have to be "up" - it can point anywhere.