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Calculating net force

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate the net force acting on each object indicated in the diagrams. Show your work.

    2. Relevant equations

    a2 = b2 + c2

    tanO = opposite/adjacent

    C2 = a2 + b2 – 2abCosC

    sina/A=sinb/B=sinc/C



    3. The attempt at a solution

    I attached the diagram below, it is the same as in my learning materials but I cannot share the actual course files so I recreated it just to show where the numbers are.

    I am asked to find the net force acting on the object, here are my calculations:

    a2 = 102 + 82
    a = 12.8 N
    tanO = opposite/adjacent
    tanO = 10/8
    51.34 or 51⁰
    180 – (51 + 45)
    84

    C2 = a2 + b2 – 2abCosC
    C = [ (a)2 + (b)2 – 2abCosC]1/2
    C = [(12.8)2 + (17)2 – 2(12.8) * (17)cos84)]1/2
    C =20.18

    sina/A=sinb/B=sinc/C

    sinb/17=sin84/20.18
    =56.9
    51⁰ - 56.9 = -5.9

    I am confused because my result is negative. If it is not correct, can someone show me where to plug the numbers in for the equations, and if I am using the appropriate numbers? In the first equation, should I be using 10N and 17N instead, or are the numbers I have in the correct place?

    Please let me know where the numbers are input for the equations to be done correctly. I would appreciate the help in laymen's terms, I understand everyone here is smart but I am not well versed in the language of physics yet.

    Thanks for helping if you do :)
     
  2. jcsd
  3. Mar 17, 2014 #2
    I just realised that upon copying and pasting my work, the exponents were just copied as normal numbers. The first equation is actually a^2 = 10^2 + 8^2, sorry for the confusion if caused.
     
  4. Mar 17, 2014 #3
    I forgot the attachment also ... sorry!
     

    Attached Files:

  5. Mar 18, 2014 #4

    PhanthomJay

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    1. You do not show the direction (arrowhead) of the applied forces, do they all pull away from the object?
    2. Don't use Pythagorus at the onset. You should first break up each force into its x and y components, add the x components, then the y components, then use Pythagorus and trig to get the resultant force.
     
  6. Mar 18, 2014 #5
    The most common way to tackle this type of problem, and mechanics in general is by resolving each force acting on the particle. For example, Assuming that your 8N is an upward force and you 10N is pulling downward, and If we're taking down and left to be positive:

    ##R(\downarrow): 10 N + 17cos(45°) N - 8N = 14.021 N ##
    ##R(→): 17sin(45°)N = 12.021 N##
    Now you can make a right triangle and solve for the resultant force, which will be in the ##\searrow## direction, using ##a^2 + b^2 = c^2##.
     
    Last edited: Mar 18, 2014
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