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 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 ************************************************ 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 _______________________________________________________________________ ______ 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" 55o angle is smaller than your 25o angle, so when you got two x-components of 3N and 3.7N you did not recognise that at least one of them was wrong, as the x-component 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 ??