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    I'm really stuck with this question:

    A sledge is held at rest on a smooth slope which makes an angle of 25 with the horizontal. The rope holding the sledge is at an angle of 20 above the slope. The normal reaction acting on the sledge due to contact with the surface is 80N

    Find the tension in the rope and the weight of the sledge.
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    (Original post by SebastianTindall)
    I'm really stuck with this question:

    A sledge is held at rest on a smooth slope which makes an angle of 25 with the horizontal. The rope holding the sledge is at an angle of 20 above the slope. The normal reaction acting on the sledge due to contact with the surface is 80N

    Find the tension in the rope and the weight of the sledge.
    How about a diagram? Mark on the forces and angles.
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    1. Draw a diagram labelled with the forces and angles.
    2. Resolve forces parallel to the slope.
    3. Resolve forces perpendicular to the slope.
    4. Solve simultaneous equations. You have two equations and two unknowns.

    Advice
    Use substitution rather than trying to add the equations.
    Just write W for weight instead of mg, it looks neater and you are asked to find weight anyway.
    You will get terms involving sin(20), cos(20), sin(25) and cos(25). Don't use your calculator to evaluate them until you have rearranged sufficiently to isolate T/W.
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    I have drawn a diagram and resolved perpendicularly and parallel, however the two equations i have got both have variables in them and i can't seem to find a way to fix it :s
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    (Original post by SebastianTindall)
    I have drawn a diagram and resolved perpendicularly and parallel, however the two equations i have got both have variables in them and i can't seem to find a way to fix it :s
    Then as Brister says, solve the equations simultaneously.
    Unless, of course, there are 3 variables and 2 equations in which case check your equations for a mistake
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    The two equations i have got are Tcos20 = wsin25 and 80N + Tsin20 - wcos25=0 ?
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    (Original post by SebastianTindall)
    I have drawn a diagram and resolved perpendicularly and parallel, however the two equations i have got both have variables in them and i can't seem to find a way to fix it :s
    When you resolve parallel to the slope, you should be able to write T in terms of W, or W in terms of T.
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    (Original post by SebastianTindall)
    The two equations i have got are Tcos20 = wsin25 and 80N + Tsin20 - wcos25=0 ?
    Yeah, so from the first equation you can, for example, go with T = \frac{Wsin(25)}{cos(20)} and substitute into the second equation.
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    "Yeah, so from the first equation you can, for example, go with and substitute into the second equation."

    Yes i have done this but the result is very awkward :/
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    Anyone who reads this it turns out the solution is found using triangle of forces :/
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    (Original post by SebastianTindall)
    The two equations i have got are Tcos20 = wsin25 and 80N + Tsin20 - wcos25=0 ?
    T = \frac{Wsin(25)}{cos(20)}

    Then 80 + Wsin(25)tan(20) - Wcos(25) = 0

    So W = \frac{80}{cos(25) - sin(25)tan(20)}

    Finally T = \frac{80sin(25)}{cos(20) \times (cos(25) - sin(25)tan(20))}

    You get the values using a calculator, so it is not that awkward.

    The triangle of forces or equally Lami's theorem will of course give the same result, and yes you will avoid having as many sines and cosines. However, the angles for the triangle can be a bit fiddly and it is easy to get it wrong. It really depends on your preference.

    Just out of interest, if you multiply the equation for W on top and bottom by cos(20) then you get

    W = \frac{80cos(20)}{cos(25)cos(20) - sin(25)sin(20)} = \frac{80cos(20)}{cos(45)} = \frac{80sin(70)}{sin(45)} = \frac{80sin(70)}{sin(135)}

    which is what you get from using the sine rule with the triangle of forces or Lami's theorem.

    Similarly
     T = \frac{80sin(25)}{cos(25)cos(20) - sin(25)sin(20)} = \frac{80sin(155)}{sin(135)}

    Now isn't that neat?
 
 
 
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