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Dec17-11, 10:45 PM
HW Helper
P: 2,318
Quote Quote by plumeria28 View Post
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:

sin = o/h sin35 = o/6.5N o = 3.7N [S]

sin = o/h sin25 = o/7N o = 3N [N]

Fapx = Fapx1 + Fapx2
Fapx = 3.7N - 3N
Fapx = 0.7N

cos = a/h cos35 = a/6.5N a = 5.3N [W]

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
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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

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 ??