# Calculating the net force

1. Mar 15, 2016

### alexandria

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

2. Relevant equations

3. The attempt at a solution

8.0 N (north) + 10.0 N (south) = 2.0 N (south)

i used the tip-to-tail method to add the vectors
c = ([2.0 N]^2 + [17.0 N]^2 - 2 (2.0 N) (17.0 N) cos45°)1/2
c = 15.6 N

SinA/17.0 N = Sin45°/15.6 N
A = 40°

Net force = 15.6 N (S 40° W)

2. Mar 15, 2016

### haruspex

I don't get 40 degrees from that, but also you have fallen into a trap.
sin(x) increases as x goes from 0 to 90 degrees, but then decreases to 0 at 180 degrees.
sin(x) and sin(180-x) produce the same number. So when you know the sine and want to find the angle, you have to decide whether it is the angle returned by the arcsin function, or 180 degrees minus that angle.
There are deterministic processes for ensuring the right answer, but in this case just look at your triangle diagram. If you draw the 2N to scale it will be much shorter.
Like I said, it's always worth checking whether the answer looks sensible.

3. Mar 15, 2016

### alexandria

SinA/17.0 N = Sin45°/15.6 N
how do i solve this to determine SinA?

4. Mar 15, 2016

### alexandria

and does my answer look 'sensible' so far? i feel like im doing something wrong

5. Mar 15, 2016

### haruspex

What did you do to get 40 degrees? Please show your working.

6. Mar 31, 2016

### alexandria

sorry to bring back this forum again, but i re-did this question, can anyone tell me if this is correct.
39.

8.0 N (north) + 10.0 N (south) = 2.0 N (south)

c = ([17.0 N]2 + [2.0 N]2 - 2 (17.0 N) (2.0 N) cos45°) ½

c = 15.6 N

SinA/a = SinB/b = SinC/c

SinA/2.0 N = Sin45 degrees/15.6 N

SinA = (2.0 N) Sin45° / 15.6 N

A = sin^-1 (0.091)

A = 5.2 degrees

90 degrees – 5.2 degrees = 39.8 degrees

Fnet = 15.6 N [North 39.8 degrees West]
???

7. Mar 31, 2016

### haruspex

That all looks right except for a typo in making the post. You wrote 90- instead of 45- near the end.

8. Mar 31, 2016

### alexandria

so then, its 45 - 5.2 = 39.8
thanks for the help.