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    This is an underlined question in chapter 3 in advanced maths for AQA. I'm having trouble finding the horizontal distance from the wall to the weight.

    A particle of weight W is a attached by a light inextensible string of length a to a point A on a vertical wall. The particle is supported in equilibrium by a light rigid strut on length b attached to a point B on the wall a distance of a vertically below A. Show that the tension in the string is W and find the thrust in the strut.

    I have found that cos A (2a2 - b2)/(2a) and that cos B = b/(2a) - but I'm open to correction.

    I can't find a value for the horizontal distance from the wall to the weight that doesn't include a square root so I'm stuck for finding sin A and sin B.

    Any clues anyone?

    Thanks
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    (Original post by maggiehodgson)
    Any clues anyone?
    Your value for cosA should have 2a^2 in the denominator.

    Since you now have cosA and cosB, you can use the standard identity for cos and sin to work out the sin of each as necessary - if you need them.

    I would do a diagram to start, if you haven't already.

    You want to take moments about B to find the tension. So start by writing out that equation and see what you've got.

    If you notice the relationship between angles it's very easy, but difficult to hint at.

    You really need to put some working down for the moments before I can advise further.
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    oops

    I had actually got 2a2 as the denominator, sorry for the mis-type.

    So, I'm saying
    Let T1 = the tension in the string and T2 = the thrust in the strut.

    T1cos A + T2 cos B = W
    T1 sinA = T2 sinB

    but for that I would need the horizontal distance from the wall to the weight.

    However, you've now said "if you need sin". So I need to rethink.

    You have said "moments about B" but I don't know what that means I can only think about the forces at C (the place where T1 and T2 meet the weight in my diagram).

    But, if I draw the triangle of forces instead of the forky diagram above, I have an isosceles triangle so does that automatically make T1 = W without doing any calculations?

    I hope you can follow what I'm talking about.
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    (Original post by maggiehodgson)
    But, if I draw the triangle of forces instead of the forky diagram above, I have an isosceles triangle so does that automatically make T1 = W without doing any calculations?
    Yes. Well spotted, didn't see that one myself.

    Since you've not used moments, there's no point in cosidering them, unless you've covered them in class, and wish to look at it that way.

    Regarding your original working, if you drop a perpendicular from C to AB, meeting at D say, then you can use your second equation to work out T2.
    You don't need to know the length of CD, as it will cancel.
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    Oh thanks.

    Must try to remember to draw different diagrams in future.

    What were your original ideas then.

    By the way, I'm not attending classes I'm just self studying and using the books I've mentioned earlier. Hence why I keep popping up on TSR 'cos I've got no other personal help,

    Looking forward to seeing your original solution if you have time. The more I can see of other people's thoughts the more I'll pick up.
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    (Original post by maggiehodgson)
    ..
    Will post tomorrow - need some sleep!
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    OK.

    Refering to the attached diagram - lines in red are construction lines.

    Taking moments about B for the rod BC, we have

    W x CE = T1 x BD

    W x a x sin CAE = T1 x a x sin DAB

    But CAE and DAB are the same angle, so following cancellation we are left with

    W = T1.
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    Thanks for that. I will have to get up to the chapter on moments to fully appreciate it but I will come back to it as and when.

    It certainly seems quicker and simpler than the way I went about it when trying to use the forky diagram method rather than the triangle of forces. I took your earlier advice on not working out the horizontal distance from wall to weight as it would cancel out and sure enough it did. I ended up with SinA/SinB = b/a and substituted it into T1sinA =T2sinB and then solved simultaneous equations. I got there in the end.

    Thanks again. I can now move on to "modelling friction".
 
 
 
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