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    For part b

    Resolve horizontally (right)

    −F cos 42◦ + Rsin 42◦ = 0

    I understand the Rsin42 but don't see how they get the Fcos42 can't see where they put the angle in.


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    (Original post by khanpatel321)
    For part b

    Resolve horizontally (right)

    −F cos 42◦ + Rsin 42◦ = 0

    I understand the Rsin42 but don't see how they get the Fcos42 can't see where they put the angle in.
    I would usually do this by resolving perpendicular and parallel to the slope, FWIW. But, going by your method:



    The blue lines are horizontal and vertical (hence perpendicular to one another). I get the 48 degree angle since the horizontal blue line is perpendicular to the vertical downwards black line, so the angle is 90 - 42. Then the angle between the the perpendicular red line and parallel up-the-slope black line is again 90 degrees, so the angle between the blue and the up-the-slope black is 90-48 = 42 again.
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    (Original post by Zacken)
    I would usually do this by resolving perpendicular and parallel to the slope, FWIW. But, going by your method:



    The blue lines are horizontal and vertical (hence perpendicular to one another). I get the 48 degree angle since the horizontal blue line is perpendicular to the vertical downwards black line, so the angle is 90 - 42. Then the angle between the the perpendicular red line and parallel up-the-slope black line is again 90 degrees, so the angle between the blue and the up-the-slope black is 90-48 = 42 again.
    Thanks

    Yeah that is how I did it but in the question it said "by resolving horizontally and vertically, or otherwise". So I thought this way may be easier but it isn't, compared to the standard way.
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    (Original post by khanpatel321)
    Thanks

    Yeah that is how I did it but in the question it said "by resolving horizontally and vertically, or otherwise". So I thought this way may be easier but it isn't, compared to the standard way.
    Ah, okay. Makes sense.
 
 
 
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