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    Hi
    when doing the classic 'particle on a slope', and solving perpendicular/parallel to the slope, struggle see whether the force associated with the mass is MgSin(x) or MgCos(x). Should i draw a right angled question to help me visualise this. Im not sure where the angle-x (angle slop is inclined at) would go in this triangle???

    thanks for any help
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    You should always draw a diagram
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    (Original post by starwarsjedi123)
    Hi
    when doing the classic 'particle on a slope', and solving perpendicular/parallel to the slope, struggle see whether the force associated with the mass is MgSin(x) or MgCos(x). Should i draw a right angled question to help me visualise this. Im not sure where the angle-x (angle slop is inclined at) would go in this triangle???

    thanks for any help
    You should always draw diagrams for this sort of question. You'll find that in the classical scenario, the force parallel is given by mg sin theta - draw it out and see for yourself.

    The angle inclined would be the angle between the line of slope and the horizontal.
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    No not the diagram.
    I mean i try to draw a right-angle triangle to figure out whether the force with Mg is MgSin(x) or MgCos(x). i cant figure out where the right angle and angle (x) goes.
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    (Original post by starwarsjedi123)
    Hi
    when doing the classic 'particle on a slope', and solving perpendicular/parallel to the slope, struggle see whether the force associated with the mass is MgSin(x) or MgCos(x). Should i draw a right angled question to help me visualise this. Im not sure where the angle-x (angle slop is inclined at) would go in this triangle???

    thanks for any help
    draw a diagram and start labelling angles until you find the relevant ones
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    (Original post by starwarsjedi123)
    No not the diagram.
    I mean i try to draw a right-angle triangle to figure out whether the force with Mg is MgSin(x) or MgCos(x). i cant figure out where the right angle and angle (x) goes.


    This is a slope inclined at 30 degrees to the horizontal, clear enough?
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    (Original post by Zacken)


    This is a slope inclined at 30 degrees to the horizontal, clear enough?
    I believe so. Just to check, if the slope was incline the opposite way, i.e., from the right hand side, and i drew the same right-angled triangle, would the 30 degrees(or the angle of inclination) always be at the 'top'?

    thanks for your help!!! Been looking for a picture like this for ages
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    (Original post by starwarsjedi123)
    I believe so. Just to check, if the slope was incline the opposite way, i.e., from the right hand side, and i drew the same right-angled triangle, would the 30 degrees(or the angle of inclination) always be at the 'top'?

    thanks for your help!!! Been looking for a picture like this for ages
    It would look like this:



    This is a plane inclined at 20 degrees to the horizontal.
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    yes but if i drew a line, W, down from the box,and tried to Split the force into its perpendicular/parallel component, would W still be the hypotenuse of the triangle, with WCos20 the perpendicular component of the force??
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    (Original post by starwarsjedi123)
    yes but if i drew a line, W, down from the box,and tried to Split the force into its perpendicular/parallel component, would W still be the hypotenuse of the triangle, with WCos20 the perpendicular component of the force??
    Yes and yes.
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    (Original post by starwarsjedi123)
    I believe so. Just to check, if the slope was incline the opposite way, i.e., from the right hand side, and i drew the same right-angled triangle, would the 30 degrees(or the angle of inclination) always be at the 'top'?

    thanks for your help!!! Been looking for a picture like this for ages
    If you get a slope question that is opposite to the direction you are used to, there's no harm in redrawing the diagram the other way around. The answers you get will be the same.

    Some students find this makes them more comfortable with the question
 
 
 
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