# Forces, vectors and acceleration

## Homework Statement

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.2 N, acting 51° north of west. What is the magnitude of the body's acceleration?

F=ma

## The Attempt at a Solution

I have 51º NW at 8.2N and 0º E at 9N
51º is actually 141º from x-axis and I will do the calculations from there.

I am basically subtracting one vector from the other to see the resultant and I get the acceleration by using F = ma

math:
NWx = 8.2 cos 141º = -6.37i
NWy = 8.2 sin 141º = 5.16j

Ex = 9 cos 0º = 9i
Ey = 9 sin 0º = 0j

subtract to get
-6.37i+ 5.16j
+
9 i + 0j
______________________________
3.37i + 5.16j = R

now find R in N
3.372+5.162 = 37.98
$\sqrt{37.8}$ = 6.16N = F
6.16N = 3kg * a
6.16N / 3 kg = 2.05m/s2 = a

my answer is supposed to be 2.48m/s2
where did I lose it?

## The Attempt at a Solution

BruceW
Homework Helper
subtract to get
-6.37i+ 5.16j
+
9 i + 0j
______________________________
3.37i + 5.16j = R

It looks like you're doing OK until you get to this bit. You say subtract, but you are adding the two vectors (which is the correct thing to do). But then the answer you get doesn't look right. You did -6.37+9 which is not equal to 3.37

180-51= ?

BruceW
Homework Helper
I think he's doing 90+51. This is another thing to watch out for. The question is a bit vague whether the angle is to be taken from the horizontal or from the vertical.

oh , if you take the angle between the F vectors to be 129 , you get the answer correctly

51 degree north of west means 51 degree taken from x axis in II quadrant??

first thanks for pointing out the math error 9 - 6.37 is 2.63! my bad.

51 degree north of west means 51 degree taken from x axis in II quadrant??

yes, this is my problem. it was taken from the x axis in QII and I was taking it from the norther direction in Q II so I did 51º + 90º and it should have been 180º - 51º giving me an angle of 129º from the positive x axis.

thanks!