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Thread: statics

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    statics

    statics-win_20170801_16_10_27_pro.jpgClick image for larger version. 

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    i am stuck on question 6(iii).
    the answer i got for part (i) was w/6 and for (ii) x=w/6 and y=w

    for part (iii) i know f=w/g again. should i take moments about a point? i tried but cant get anywhere
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    Re: statics

    (i) $\displaystyle \sum \tau = 0 \implies W \cdot L\sin{\theta} = F \cdot 2L\cos{\theta} \implies F = \dfrac{W\tan{\theta}}{2} = \dfrac{W}{6}$

    (ii) horizontally ... $H_x = F = \dfrac{W}{6}$

    vertically ... $H_y = W$

    (iii) $F = \dfrac{W}{6} \implies 2F = \dfrac{W}{3}$ ...

    $W \cdot L\sin{\theta} = \dfrac{W}{3} \cdot 2L\cos{\theta} \implies \tan{\theta} = \dfrac{2}{3}$
    Thanks from markosheehan
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    Re: statics

    statics-win_20170801_18_16_20_pro.jpg do you know how to do question 8 here

    i am not sure what it means by the supporting force at R and the couple M
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    Re: statics

    R is a vertical supporting force at point c.

    As far as I can tell, M is a moment about point c counteracting the moments induced by the beam weight of 50N and load weight of 100N at point b.

    See diagram ...
    Attached Thumbnails Attached Thumbnails statics-static_couple.jpg  
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    Re: statics

    Thanks. Yes your answers are correct.
    How did you know the direction of Rc and that it was going up vertically. It does not say the direction of it in the question.

    What are you doing in your second equation.? Are you taking moments about a point and if so where ? Why do you not include the Rc in the second equation.?
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    Re: statics

    Quote Originally Posted by markosheehan View Post
    How did you know the direction of Rc and that it was going up vertically. It does not say the direction of it in the question.
    both the 100N and 50N weights are in the downward direction ... what direction for $R_c$ is necessary for equilibrium in the y-direction?

    What are you doing in your second equation.? Are you taking moments about a point and if so where ? Why do you not include the Rc in the second equation.?
    Moments are about point $c$ ... $R_c$ does not contribute to the net torque about point $c$ since its moment arm is zero.
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    Re: statics

    oh yes i forgot that its moment is 0 as it goes through point c.
    would M also not be 0 as it acts through point c as well no?
    why did you represent Mc as a circle on the diagram?
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    Re: statics

    Quote Originally Posted by markosheehan View Post
    oh yes i forgot that its moment is 0 as it goes through point c.
    would M also not be 0 as it acts through point c as well no?
    why did you represent Mc as a circle on the diagram?
    M represents a torque in the counter-clockwise direction that balances the the two torques induced by the 50N and 100N weights.

    Have you learned any of this in class?

    Last edited by skeeter; Aug 2nd 2017 at 01:39 PM.
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    Re: statics

    but how do you know the torque force M doesnt act through point C. it says the couple M is at C in the question no? and the sum of moments through C is zero.
    thanks for the video
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    Re: statics

    M acts about point c. Torques act about a point, not at a point.
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