**1. The problem statement, all variables and given/known data**

There are two forces on the 2.80 kg box in the overhead view of the figure but only one is shown. For F1 = 10.9 N, a = 10.2 m/s2, and θ = 34.9°, 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.)

I’m not sure if this link will work, but here’s the picture: http://ift.tt/1bYSsVd

**2. Relevant equations**

∑F=ma

∑F=F_{1}+F_{2}…+

**3. The attempt at a solution**

Ok so I’ve been stuck on this one for a while. I started out finding the sum of the forces:

∑F=(2.8kg)(10.2m/s^{2})=28.56 N

Then I tried solving for F_{2}:

28.56N=10.9N+F_{2}

F_{2}=17.66

Then I found the x and y components of the acceleration:

a_{x}=10.2*sin(34.9)=5.84m/s^{2}

a_{y}=10.2*cos(34.9)=8.37m/s^{2}

Could you multiply a_{x} and a_{y} by the mass and get the F_{1x} and F_{1y}?

I’m so confused, could someone please help guide me through this problem?

http://ift.tt/1lF7zGf