netrunnr
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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:
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?