# More Forces with some trig; I don't know where the mistake is

1. Feb 20, 2012

### iJamJL

1. The problem statement, all variables and given/known data
Mass m sits on a frictionless, horizontal table. Assume:
- All vertical forces acting on m (such as gravity and the normal force) sum to zero.
- All applied forces act parallel to the table
When a force of magnitude F is applied to mass m, it accelerate at magnitude a = 4.49 m/s2.

NOTE: The axes lie along the table.

Suppose force F1 has magnitude 2F and points at 45 degrees to the x-axis. If forces F1 and F2, with a magnitude F, act on mass m at the same time, what will the magnitude of its acceleration be now?

2. Relevant equations
Components

3. The attempt at a solution
I attempted to solve this by drawing a diagram, and F1 is pointing out at 45 degrees relative to the x-axis. I found its components, with the x-direction vector being 8.98cos(45) = 6.3498. F2 is already in that direction as well, so I just added the two vectors to get a sum of 10.8398. When I entered this answer as 10.84 into Webassign (online homework - I don't know who is familiar with it), it said that the answer is incorrect. Is it that I didn't put enough digits, or did I solve the problem incorrectly?

2. Feb 20, 2012

### tiny-tim

welcome to pf!

hi iJamJL! welcome to pf!
but that's only the x-component of the total force …

3. Feb 20, 2012

### Staff: Mentor

your description is a bit confusing you probably need to draw a force diagram to show what you mean. You mention one force F that accelerates the m horizontally and then shift
into F1 and F2 which I thought were components of a second force but then I got lost when you said F1 has a magnitude of 2F and that F1 and F2 have a magnitude of F?

4. Feb 20, 2012

### iJamJL

Re: welcome to pf!

Thanks for the welcome!

Hm, I see what you mean, tiny-tim. Let me see if I can get this right.

You're saying that I've got the x-component down for the F1 where it is 8.98 as the resultant. The x-component becomes 6.3498, and the y-component becomes 6.348. I can then carry these both on with the 4.49 from F2, making the vectors like this:

-> 10.8398
^ 6.348

Then the resultant vectors, with the Pythagorean theorem, results to 12.5618. Is this correct?

5. Feb 20, 2012

### iJamJL

They do not give any values for the forces. That is how the problem is given.

6. Feb 20, 2012

### tiny-tim

yes, that's correct

(though it would be better, and easier, if you added the forces

2F/√2 + F and 2F/√2 )​

7. Feb 20, 2012

### iJamJL

You lost me there, tiny-tim. LOL but I understand it the other way, so I guess we could let it rest. Thank you!