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Can anyone check my working on the resultant force ,test question

  1. Nov 22, 2013 #1
    1. The problem statement, all variables and given/known data

    [URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps268ea812.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps268ea812.png[/URL][/PLAIN]

    [URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zpsbae78257.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zpsbae78257.png[/URL][/PLAIN]

    2. Relevant equations

    ƩMA = Ʃ r x F
    Total Force of x,y= 0

    3. The attempt at a solution

    bi)150(0.4)+(50)(0.3)-Fsin50(0.5)
    =0

    Ffb=150(0.4)+(50)(0.3)/ sin50(0.5)
    =195.81N ( answer is wrong why?)
     
    Last edited: Nov 22, 2013
  2. jcsd
  3. Nov 22, 2013 #2

    TSny

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    Hello, cracktheegg.

    You have not calculated the cross products correctly for the torques at G and E. It's more complicated than just multiplying the forces by the given distances. Did you draw the vectors ##\vec{r}## and ##\vec{F}## for each force?
     
  4. Nov 22, 2013 #3
    bi)Moment of mass=(50)cos30 *0.3

    Moment of object=(150* cos30 * 0.4

    moment of GE= Fsin50(0.5)

    is this correct?
     
    Last edited: Nov 23, 2013
  5. Nov 23, 2013 #4

    TSny

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    The moments for the 50 N force and the unknown force look correct.

    For the 150 N force, note that the position vector extends from A to E.
     
  6. Nov 23, 2013 #5
    Sry, can you show me the working, i really can't visualize.
     
  7. Nov 23, 2013 #6

    TSny

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    Here is the position vector ##\vec{r}## for E. You'll need to work out ##\vec{r} \times \vec{F_E}##.
     

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  8. Nov 23, 2013 #7
    I use 0.05/0.4
    tan^-1(0.05/0.4)
    0.4/cos(tan^-1(0.05/0.4))=r⃗
    r⃗ * 150=60.46

    I do (60.46 + (50)cos30 *0.3)/sin50(0.5)
     
  9. Nov 23, 2013 #8

    TSny

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    OK, this gives you the magnitude of ##r##. (You could also have gotten this with the Pythagorean theorem).

    You need to take into account the angle θ between r and the force. M = r F sinθ.
     
  10. Nov 23, 2013 #9

    Thanks I finally get the answer and start to understand why

    0=150cos(67.13)+50sin30+179.4cos50-Rx
    0=150sin(67.13)+50cos30-179.4sin50+Ry
    Should I use the above to find the magitude at hinge A?

    And what does iii) mean?
     
    Last edited: Nov 23, 2013
  11. Nov 27, 2013 #10
    Can anyone help me with b iii)
    I can't get the answer, I use moment to find the force
     
  12. Nov 27, 2013 #11

    TSny

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    It looks like you chose your x axis along the table and y axis perpendicular to the table. Note that the 150 N force does not make an angle of 67.13 degrees with the table. The angle of 67.13 degrees is the angle between the 150 N force and the position vector from A to E.

    I think this part of the problem would be much easier if you choose your x axis horizontal and your y axis vertical.
     
  13. Nov 27, 2013 #12

    TSny

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    My interpretation of this part, is that you need to find the force P of the arm that would just allow the force from the strut FB to go to zero. So, it's like part (i) except you replace the force of the strut by the force P and solve for the force P.
     
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