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Size:  113.7 KBProbably being a complete idiot as I'm exhausted but I can't do part 5(a). How do you work out the perpendicular distance from weight to D. Thank you in advance. Much appreciated 😊
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    (Original post by josephinemar25)
    Probably being a complete idiot as I'm exhausted but I can't do part 5(a). How do you work out the perpendicular distance from weight to D. Thank you in advance. Much appreciated 😊
    http://www.thestudentroom.co.uk/forumdisplay.php?f=38

    Have asked for this to be moved for you (when it is there, you get an answer much more quickly if someone can do it).

    Also, you don't have an attached question! :hide:
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    (Original post by josephinemar25)
    Name:  image.png
Views: 49
Size:  113.7 KBProbably being a complete idiot as I'm exhausted but I can't do part 5(a). How do you work out the perpendicular distance from weight to D. Thank you in advance. Much appreciated 😊
    This seems to have caught a few people out, as it's the third time I've seen this question in the last week.

    Uniform rod, so weight acts halfway along - a distance "a" from B.

    See attached. Line of action of force is in orange, and perpendicular distance is either of the green lines, whose lengths are "a sin(theta)".


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    (Original post by ghostwalker)
    This seems to have caught a few people out, as it's the third time I've seen this question in the last week.

    Uniform rod, so weight acts halfway along - a distance "a" from B.

    See attached. Line of action of force is in orange, and perpendicular distance is either of the green lines, whose lengths are "a sin(theta)".


    Name:  Untitled2.jpg
Views: 36
Size:  24.3 KB
    THHANNKK YOOUU that makes perfect sense
 
 
 
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