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Physics Question

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  • 30-12-2006 7:42pm
    #1
    madgal
    Closed Accounts Posts: 158 ✭✭


    I've revised all the mechanics expirement questions bar the one about the moments.
    I'm getting really lost with the whole anti-clockwise and clockwise moments stuff. The book said something about postive and negative, but all the figures appear to be positive... anyone have a clue? (if your stuck for an example 2002 - Q1.)


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    #2
    Thomas_S_Hunterson
    Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    Haven't looked at the question, but here's an explanation of the mathematical theory behind the positive/negative thing.

    Add up all the clockwise moments and add up all of the anti-clockwise moments seperately.

    For a body in equilibrium, they should be equal. Therefore if you subtract one from the other, you will get zero

    i.e. [clockwise] minus [anticlockwise] = 0
    or
    [anticlockwise] minus [clockwise] = 0

    You can let either the clockwise or the anticlockwise be a negative number and the other one be a positive and then by adding, you will get zero.

    Sorry if this is unclear, I'm just finding it hard to expain such a fundamental concept any more clearly.


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    #3
    poobum
    Closed Accounts Posts: 592 ✭✭✭poobum


    what he said and remember to apply the whole distance thing to, distance of each force from turning point(hope i phrased that right)


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    #4
    madgal
    Closed Accounts Posts: 158 ✭✭madgal


    yes... but i don't understand the whole clockwise or anticlockwise thing.. i get they equal zero but don't understand what figures...


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    #5
    Schlemm
    Closed Accounts Posts: 799 ✭✭✭Schlemm


    don't forget about the force of gravity, which acts downwards and F=mg!!! (Where m is the mass of the meter stick or an object.) Forgetting this is a classic mistake!

    For the 2002 q.1, the first thing you should do is draw a diagram showing all the forces and the direction of these forces acting on the stick. The 2 spring balances act upwards. The other forces are acting downwards as shown. What is not shown is that the weight of the stick is acting downwards, so draw this in too. The force of the stick is its weight in Newtons, which is given (1N). If the weight of the stick were given in kg or grams, convert to SI units and use F=mg to work out its weight in newtons (where m is the mass of the stick).

    The stick is in equilibrium, so the vector sum of the forces in any direction is= 0. So the forces acting down on the stick (2N plus 1.8N plus 1.2N PLUS THE WEIGHT OF THE STICK, 1N) are equal to the forces going up (the spring balances). The law of equilibrium is obeyed because the vector sum of the forces is zero, ie, the forces up are equal to the forces down. Make sure to answer all parts of the quesiton here! They are asking you to do both the calculation and the explanation. I think this is ok but double check with ur teacher.

    The next bit of the question is where you work out the clockwise and anticlockwise moments around the 10cm mark. In this case, the clockwise moments are the forces which will cause the stick to turn in a clockwise direction. DON'T FORGET THE WEIGHT OF THE STICK!!! Look at your diagram and imagine a 2D image of yourself turning the stick around the 10cm mark using each of the forces shown. For example, turning the stick with the spring balances will move it in an anticlockwise direction...so these are your anticlockwise forces! Now imagine pulling on the stick with the downward forces, and imagine allowing the stick to rotate around the 10cm point using only its weight......In either case, it will rotate in a clockwise direction, so these are your clockwise forces!
    State the formula that u will use (moment=force x distance)
    The clockwise moments are:
    (0.05mx2N) + (0.405mx1N) + (0.7mx1.8N) + (0.86mx1.2N)
    NB USE S.I. UNITS HERE! METERS!!! This whole equation is equal to +2.797Nm (NB AGAIN! UNITS! THE UNIT OF A MOMENT IS A NEWTON METER OR Nm) (+ because we will say that the clockwise moments are +ve).

    (don't forget moment=force x perpendicular distance to fulcrum.....so this is why we work out the distance of each force from the 10cm mark.)

    Now work out the anticlockwise moments:
    (0.1mx2N) + (0.65mx4N) = - 2.8Nm.

    So if in equilibrium, clockwise forces=anticlockwise forces
    +2.797=-2.8 (roughly...there should be a squiggle over the = sign). So the sum of the moments is zero around this point...

    ..They are asking 2 questions here again in the 2nd part, and the reason that the second law of equilibrium is obeyed is because the sum of the moments around any point is zero.

    Remember if a body is in equilibrium, the vector sum of the forces in any direction is zero (part 1 of this q) and the sum of the moments around any point is zero (part 2 of this question). Know this!!!!

    Sorry it's so long. Hope that helps.

    www.examinations.ie has past papers and marking schemes, both of which are worth a look.
    Other links:
    http://www.technologystudent.com/forcmom/force2.htm
    http://www.skoool.ie/skoool/homeworkzone.asp?id=1946


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