Solve the Mystery of Force 2 on a 2.43 kg Box

[copypacer]

Homework Statement

There are two forces on the 2.43 kg box in the overhead view of the figure but only one is shown. For F1 = 20.0 N, a = 11.4 m/s2, and θ = 33.0°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)The acceleration is in the third quadrant.

Homework Equations

So, I used F2 = m(a) - F1. To try and find Force 2 and used cos and sin to fine it in vector notation. Apparently I got that wrong, HELP!

The Attempt at a Solution

I used cos and sin to fine it in vector notation. Apparently I got that wrong, HELP!

So for a I got F2 = 2.43 * 11.4 - 20 which resulted in 7.03 or something and pluged that into (7.03 sin 33 degrees) and to find (j) and did the same with cos to find (i)

It didn't work, and magnitude and direction ended up being wrong because a) was wrong HEEEELP

 

[haruspex]

With no diagram, the set-up is far from clear. Is the given angle the angle between F1 and the acceleration?
If so, it makes no sense to subtract F1 from ma as mere numbers. You can only do scalar addition and subtraction for vectors in the same direction.
Think about the net force in the direction of acceleration and the net force at right angles to that.

 

[copypacer]

With no diagram, the set-up is far from clear. Is the given angle the angle between F1 and the acceleration?
If so, it makes no sense to subtract F1 from ma as mere numbers. You can only do scalar addition and subtraction for vectors in the same direction.
Think about the net force in the direction of acceleration and the net force at right angles to that.

here it is, sorry

 

[haruspex]

here it is, sorry

Ok, my guess was almost right, and my comments stand.

 

[copypacer]

here it is, sorry

Wait I'm confused D:

Both the acceleration and force 2 are in the third quadrant, so its negative, wouldn't I still use F2 = ma - F1

so what do I do, since the two forces are not in the same direction?

I don't get it DX

 

[copypacer]

Ok, my guess was almost right, and my comments stand.

I'm trying to find the second force, so how would I solve that?

 

[haruspex]

Wait I'm confused D:

Both the acceleration and force 2 are in the third quadrant, so its negative, wouldn't I still use F2 = ma - F1

so what do I do, since the two forces are not in the same direction?

I don't get it DX

Do you know how to resolve a force into components?

 

[copypacer]

Do you know how to resolve a force into components?

...no...wait...no I don't

 

[haruspex]

...no...wait...no I don't

Well, that's rather basic, and I don't know how you would be expected to solve this problem without having been taught that.
There's a lots of stuff on the net. Try one of these:
http://www.physicsclassroom.com/class/vectors/Lesson-3/Resolution-of-Forces

http://www.s-cool.co.uk/a-level/phy...n/revise-it/resolving-vectors-into-components
As I said, in the present problem you need to resolve the given force into a component in the direction of acceleration and another at right angles to that.