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Forces with angles, Find the acceleration

  1. Dec 17, 2011 #1
    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

    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
     

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  3. Dec 17, 2011 #2

    PeterO

    User Avatar
    Homework Helper



    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 ??
     
    Last edited: Dec 18, 2011
  4. Dec 17, 2011 #3
    I said the diagrams are not accurate, I just drew them so it would be visually easier to do the calculations. And I added the y components because they were both going in the same direction (west) and subtracted the x components because they were going opposite directions (south and north).
     
  5. Dec 18, 2011 #4

    PeterO

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    Homework Helper

    But your diagram clearly shows the x components both going west, and the y-components going in opposite directions - North and South.

    You need to draw a more accurate diagram and stick to the conventions you state on that diagram if you want to get anywhere with this problem.
     
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