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# M1 - statics and forces Watch

1. This is q10 Exercise 3B Advanced Maths for AQA Mechanics M1:
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A particle of mass 4kg is suspended from a point A on a vertical wall by means of a light inextensible string of length 130cm.
a) A horizontal force,P, is applied to the particle so that it is held in equilibrium a distance of 50cm from the wall. Find the value of P and the tension in the string.

b) By drawing a triangle of forces, or otherwise, find the magnitude and direction of the minimum force which would hold the particle in this position, and the tension in the string which would result.
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I've managed part a) but part b) has me foxed.

I've said, let the force applied to the particle be F at an angle alpha to the horizontal. Let T = tension in the string and F the force needed to keep the particle in equilibrium. From part a) the cos of angle of the string to the horizontal is 5/13 and its sin is 12/13.

So I get 5T/13 = Fcos(alpha)
and 12T/13 + Fsin(alpha)= 4x9.8

I have 2 equations with 3 unknowns. I'm missing something.
2. You should indeed have a whole choice of "F"s, now you have to pick the smallest in magnitude. I suggest eliminating alpha and seeing if the resulting expression for F in terms of T has a minimum.

I think my algebra/pure maths isn't up to eliminating alpha. I've tried rearranging to get F sin(alpha). Then I squared both sides of each equation so that I could use sin2 + cos2 =1 but now I have only 1 equation with two unknowns.

I would appreciate another clue.
4. The sin^2 + cos^2 =1 trick should help you on your way to getting F in terms of T, have you managed that? If so, see what the smallest value that F can take is.
5. Hi again

No, as of yet I've not worked out the sum of the squares of the equations because I thought that I'd get nowhere with only one equation and two unknowns.

So once I've got there I'm going to have F2 = some huge expression involving Ts and T2 by the looks of things. So then will I solve the quadratic for zero (that's a minimum F)?

Will get on to that equation, now you think I'm on the right lines, but I might be some time as I now have to go to work.
6. Yes, you should get F^2= some function of T. You can then complete the square to find the smallest value of F^2 (it shouldn't be zero, you're essentially trying to find the minimum of the graph F^2), and the value of T that will give it. Of course, the smallest value of F^2 might not always correspond to the smallest valid value of F, but plug in your found value of T and you'll get a valid F out.
7. Oh, lovely.

Yes managed to find minimum T by differentiating (T2 -96T/13 + 16) rather than completing the squares.

Once I'd found T I put that back into F2 so kept getting the wrong answer. Then I realised my mistake and found the correct F by using the correct T.

As usual these things take a while but I'm learning such a lot with TSR's help.

Thanks
8. (Original post by maggiehodgson)
Oh, lovely.

Yes managed to find minimum T by differentiating (T2 -96T/13 + 16) rather than completing the squares.

Once I'd found T I put that back into F2 so kept getting the wrong answer. Then I realised my mistake and found the correct F by using the correct T.

As usual these things take a while but I'm learning such a lot with TSR's help.

Thanks
kind of a long time ago but do u by any chance remember how you got the answer to part a
9. (Original post by girl :D)
kind of a long time ago but do u by any chance remember how you got the answer to part a
Yes, I keep all my maths.

I am attaching a picture of the diagram I used. I've hidden the workings (TSR rules - just hope the sketch isn't too much)

Let me know if you need further help.
Attached Images
10. Scan0002.pdf (364.3 KB, 44 views)

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