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Edexcel M2 Moments question, can't understand something Watch

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    Hello, I'm doing question 7 in the 2011 June M2 paper, I can't seem to understand a bit of the last part (c). I don't understand why R = 5mg (you can see that in the mark scheme). I understand all the rest but just don't know why R = 5mg.

    Question Paper: http://qualifications.pearson.com/co...e_20110613.pdf
    Mark Scheme: http://qualifications.pearson.com/co...s_20110817.pdf


    Any help would be greatly appreciated.
    Thanks
    Reda
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    So when you say you don't understand why R=5mg do you mean the answer to part b) or something IN part c)?
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    (Original post by Reda2)
    Hello, I'm doing question 7 in the 2011 June M2 paper, I can't seem to understand a bit of the last part (c). I don't understand why R = 5mg (you can see that in the mark scheme). I understand all the rest but just don't know why R = 5mg.

    Question Paper: http://qualifications.pearson.com/co...e_20110613.pdf
    Mark Scheme: http://qualifications.pearson.com/co...s_20110817.pdf


    Any help would be greatly appreciated.
    Thanks
    Reda
    Hello

    First of all, you know that the normal reaction force, R, is perpendicular to this surface, so it is a horizontal force acting to the right. Now you know that the body is in equilibrium and thus, there must be no resultant force in the system.

    Now consider the horizontal forces acting in the system. A tension of magnitude \frac{25mg}{4} acts at an angle \alpha to the rod where \tan \alpha = \frac{3}{4} (figured this out using right angled trigonometry considering the triangle ABD and labelling angle ABD as \alpha). The only other horizontal force acting in the system is the component of tension parallel to the rod acting to the left - \frac{25mg \cos \alpha}{4} which is just 5 mg. As the total leftward forces = total rightward forces for horizontal equilibrium to be maintained, R = 5 mg.

    Alternatively, here is ExamSolution's video solution.
 
 
 
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