(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

F=ma

3. 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?

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

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Forces, vectors and acceleration

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