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    I can't figure out the distance of Q away from AB and AC.
    8. The diagram shows three particles P, Q, R attached to light rods AB, BC, CA respectively. The rods are rigidly jointed together so that ABC is a right-angled triangle with AB = 8 cm, AC = 6cm and CAB = 90°. The masses (in kg) of P, Q, R, are 2m, 3m, 5m respectively and AP = 7·5cm, BQ = 7cm, AR = 3·5cm.
    (a) Find the distance of the centre of mass of the system from
    (i) AC, (ii) AB. [9]
    (b) The system is freely suspended from B and hangs in equilibrium. Calculate the angle that AB makes with the vertical.
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    (Original post by tequila27)
    I can't figure out the distance of Q away from AB and AC.
    8. The diagram shows three particles P, Q, R attached to light rods AB, BC, CA respectively. The rods are rigidly jointed together so that ABC is a right-angled triangle with AB = 8 cm, AC = 6cm and CAB = 90°. The masses (in kg) of P, Q, R, are 2m, 3m, 5m respectively and AP = 75cm, BQ = 7cm, AR = 35cm.
    (a) Find the distance of the centre of mass of the system from
    (i) AC, (ii) AB. [9]
    (b) The system is freely suspended from B and hangs in equilibrium. Calculate the angle that AB makes with the vertical.
    Post the diagram.
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    (Original post by RDKGames)
    Post the diagram.
    Name:  捕获.PNG
Views: 6
Size:  7.4 KBThanks for helping
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    (Original post by tequila27)
    Thanks for helping
    It would probably be easier to break down this problem down in terms of xy-coordinates. If you place this diagram on the xy-place, with AC being on the y axis and AB being on the x-axis, then A would be the origin.

    Then you can just consider calculating the horizontal distance of the centre of mass from AC, and then separately calculating the vertical distance of it from AB.

    To begin, you'd want to place down the coordinates of R, P, Q. Clearly, we have R(0,3.5), P(7.5,0). The coordinates of Q can be worked out by saying that the angle at B is \theta and hence Q(8-7\cos \theta,7\sin \theta).
    We know that \cos \theta = \frac{AB}{BC} and \sin \theta = \frac{AC}{BC}. We can work out BC by Pythagoras since we know lengths AB, AC. This is sufficient to find Q.

    Carry from from there to determine the centre of mass.
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    Thank you very much!!


    (Original post by rdkgames)
    it would probably be easier to break down this problem down in terms of xy-coordinates. If you place this diagram on the xy-place, with ac being on the y axis and ab being on the x-axis, then a would be the origin.

    Then you can just consider calculating the horizontal distance of the centre of mass from ac, and then separately calculating the vertical distance of it from ab.

    To begin, you'd want to place down the coordinates of r, p, q. Clearly, we have r(0,3.5), p(7.5,0). The coordinates of q can be worked out by saying that the angle at b is \theta and hence q(8-7\cos \theta,7\sin \theta).
    We know that \cos \theta = \frac{ab}{bc} and \sin \theta = \frac{ac}{bc}. We can work out bc by pythagoras since we know lengths ab, ac. This is sufficient to find q.

    Carry from from there to determine the centre of mass.
 
 
 
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