- #1

netrunnr

- 15

- 0

## 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?

## Homework Equations

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:

NW

_{x}= 8.2 cos 141º = -6.37

_{i}

NW

_{y}= 8.2 sin 141º = 5.16

_{j}

E

_{x}= 9 cos 0º = 9

_{i}

E

_{y}= 9 sin 0º = 0

_{j}

subtract to get

-6.37

_{i}+ 5.16

_{j}

+

9

_{i}+ 0

_{j}

______________________________

3.37

_{i}+ 5.16

_{j}= R

now find R in N

3.37

^{2}+5.16

^{2}= 37.98

[itex]\sqrt{37.8}[/itex] = 6.16N = F

6.16N = 3kg * a

6.16N / 3 kg = 2.05m/s

^{2}= a

my answer is supposed to be 2.48m/s

^{2}

where did I lose it?