Quote by plumeria28
1. The problem statement, all variables and given/known data
Two forces act on a 4.00 kg mass which is sitting at rest on a horizontal frictionless surface. One force is 6.50 N directed 55° West of South, the other is 7.00 N directed 25° North of West. What acceleration does the mass receive?
2. Relevant equations
sin = o/h
cos = a/h
tan = o/a
a = F/m
3. The attempt at a solution
I drew forces (not accurately) on paint (attachment) to help me solve this question then did the following calculations:
Fapx1
sin = o/h sin35° = o/6.5N o = 3.7N [S]
Fapx2
sin = o/h sin25° = o/7N o = 3N [N]
Fapx = Fapx1 + Fapx2
Fapx = 3.7N  3N
Fapx = 0.7N
Fapy1
cos = a/h cos35° = a/6.5N a = 5.3N [W]
Fapy2
cos = a/h cos25° = a/7N a = 6.3N [W]
Fapy = Fapy1 + Fapy2
Fapy = 5.3N + 6.3N
Fapy = 11.6N
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tan = o/a tan = 11.6/0.7 tan = 86.5°
sin = o/h sin86.5° = 11.6/h h = 11.622N
************************************************
a = F/m a = 11.622N/4.00kg a = 2.9 m/s^2
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I was wondering if I did the question right. Did I get the right answer?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution

problems:
Firstly, you have not helped yourself with your diagram.
"Your" 55
^{o} angle is smaller than your 25
^{o} angle, so when you got two xcomponents of 3N and 3.7N you did not recognise that at least one of them was wrong, as the xcomponent of the 6.5N force is way less than the x component of the 7N force. The Y components don't llok good either. You seem to have added the y components and subtracted the x ??