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Forces component question Watch

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    http://www.mediafire.com/view/?40faentnevqki77#

    I need help on the question in the link above. It's annoying me so much! The answer's D, but how? Any ideas on how to arrive at D??

    Thanks a lot.
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    W = mass

    You use sin theta when angle isn't included between the forces you want.

    So if they asked for the weight acting perpendicular to plane then it'd be Wg cos theta

    Its just basic trig but you'd know when its included it is cos theta and when it's excluded its sin theta
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    (Original post by L'Evil Fish)
    W = mass

    You use sin theta when angle isn't included between the forces you want.

    So if they asked for the weight acting perpendicular to plane then it'd be W cos theta

    Its just basic trig but you'd know when its included it is cos theta and when it's excluded its sin theta
    But weight=mass X gravitational field strength (9.81)
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    (Original post by krisshP)
    But weight=mass X gravitational field strength (9.81)
    That too

    The mass isn't all acting perpendicular to a plane.

    So it's

    Wg Cos Theta
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    and can you please explain really slowly as I am not so good when it comes to these inclined surface situations
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    I can't upload photos as the app won't let me
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    ?????:confused:
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    I do not understand at all really
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    Okay draw me a right angled triangle.

    Now draw a box on the hypotenuse.

    Draw a line vertically down from the centre of the box down through the plane and triangle.

    Now draw two dotted lines, one parallel to the plane (hypotenuse) from the centre of the box.

    And one perpendicular to the one you just drew.

    Tell me when that's done.
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    (Original post by krisshP)
    http://www.mediafire.com/view/?40faentnevqki77#

    I need help on the question in the link above. It's annoying me so much! The answer's D, but how? Any ideas on how to arrive at D??

    Thanks a lot.
    You just need to find the Parallel and Perpendicular (of the slope) components of W, and the angle between W-Perpendicular and W is the same as Theta.

    With this in mind, make a triangle, with W as the hypotenuse, W-Parallel as the opposite, and W-Perpendicular as the adjacent side, and it's just simple trigonometry from there.

    The picture I attached should hopefully help you.
    Attached Images
     
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    (Original post by L'Evil Fish)
    Okay draw me a right angled triangle.

    Now draw a box on the hypotenuse.

    Draw a line vertically down from the centre of the box down through the plane and triangle.

    Now draw two dotted lines, one parallel to the plane (hypotenuse) from the centre of the box.

    And one perpendicular to the one you just drew.

    Tell me when that's done.
    Is the second dotted line suppose to through the centre of the box on hypotenuse??

    thanks
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    (Original post by krisshP)
    Is the second dotted line suppose to through the centre of the box on hypotenuse??

    thanks
    Yeah towards the plane...

    The post above is a good way too
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    (Original post by L'Evil Fish)
    Yeah towards the plane...

    The post above is a good way too
    i'm done!
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    (Original post by krisshP)
    i'm done!
    Okay, now the angle in between your vertical line and your perpendicular line is the same as the angle of your original right angled triangle (inclined from horizontal) label them both with theta
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    (Original post by L'Evil Fish)
    Okay, now the angle in between your vertical line and your perpendicular line is the same as the angle of your original right angled triangle (inclined from horizontal) label them both with theta
    done.
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    Oh I get

    thanks so much L'Evil Fish and FaisalTreShah. Also if you use your hands and twist both a bit, it makes even more sense ! You two are so smart to recognise the angles and stuff so quickly and fast! Can't believe the question was in the multiple question section A side of the paper. It's got some difficulty though.

    THANKS AGAIN !
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    (Original post by krisshP)
    done.
    Now if you join up the dotted perpendicular line with the vertical line to create a right angle triangle. How would you find the length of the dotted perpendicular line?

    Cos theta = adjacent / hypotenuse

    Hypotenuse = Vertical Line (which has a magnitude of mass x gravity) label this line w.

    So the dotted line is found with:

    Wg Cos Theta
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    my teacher once tried explaining something similar, but she failed and struggled so I couldn't understand
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    (Original post by krisshP)
    my teacher once tried explaining something similar, but she failed and struggled so I couldn't understand
    Look at the post up there, with the picture!
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    (Original post by L'Evil Fish)
    Now if you join up the dotted perpendicular line with the vertical line to create a right angle triangle. How would you find the length of the dotted perpendicular line?

    Cos theta = adjacent / hypotenuse

    Hypotenuse = Vertical Line (which has a magnitude of mass x gravity) label this line w.

    So the dotted line is found with:

    Wg Cos Theta
    which dotted line? The one perpendicular to the one parallel to the inclined surface?

    I understand now, but still.
 
 
 
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