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    Can anyone help me with part c of the first question? It's fairly simple but I haven't been taught how to do it.
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    (Original post by pippabethan)
    Can anyone help me with part c of the first question? It's fairly simple but I haven't been taught how to do it.
    It's the same as in the first two parts, except the directions are reversed, so tack on a negative sign or resolve in the opposite direction (horizontally), preferably the latter. Your reaction force thingy will also change because the force is pushing in the same direction (vertically-wise) as the weight.
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    (Original post by Zacken)
    It's the same as in the first two parts, except the directions are reversed, so tack on a negative sign or resolve in the opposite direction (horizontally), preferably the latter. Your reaction force thingy will also change because the force is pushing in the same direction (vertically-wise) as the weight.
    I thought I did that but didn't get the correct answer.
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    (Original post by pippabethan)
    I thought I did that but didn't get the correct answer.
    Show us your working.
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    (Original post by pippabethan)
    I thought I did that but didn't get the correct answer.
    When you're given a force pointing inwards into an object, then just extend that line to it points outward from the other side of the object to make it easier. Then resolve vertically and horizantally for both R and the Net Force, use F=ma to find a, then use s = ut + .5at^2 to find s. Then just use Wd = F*d.

    What was your answer? I'll cross check.
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    (Original post by Zacken)
    Show us your working.
    I did it the way I did part b) which got the answer in the book. I think there shouldn't be the vertical component this time, but I tried the numbers without it and still got it wrong.
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    (Original post by KyleBroflovski)
    When you're given a force pointing inwards into an object, then just extend that line to it points outward from the other side of the object to make it easier. Then resolve vertically and horizantally for both R and the Net Force, use F=ma to find a, then use s = ut + .5at^2 to find s. Then just use Wd = F*d.

    What was your answer? I'll cross check.
    Thank you! I got 2760m and 262kJ I think.
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    (Original post by pippabethan)
    I did it the way I did part b) which got the answer in the book. I think there shouldn't be the vertical component this time, but I tried the numbers without it and still got it wrong.
    There is a vertical component, in this case, if you resolve downwards, you get: 100 \sin 20^{\circ} + mg - R = 0

    You also seem to have done horizontally to the right, but it's obvious that the force is acting to the left, not that it should make much of a difference, but, y'know.

    Also, the vertical component has no effect here, given that the thingy only moves horizontally.
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    (Original post by KyleBroflovski)
    When you're given a force pointing inwards into an object, then just extend that line to it points outward from the other side of the object to make it easier.
    I always do that
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    (Original post by Zacken)
    There is a vertical component, in this case, if you resolve downwards, you get: 100 \sin 20^{\circ} + mg - R = 0
    Okay so since mg=R, that would mean the resultant vertical force is 100sin20? Thank you
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    (Original post by pippabethan)
    Okay so since mg=R, that would mean the resultant vertical force is 100sin20? Thank you
    Noooo! mg=R is not true. In either case- the vertical component is useless here, the box can only move sideways, use only horizontal components of force.
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    (Original post by Zacken)
    Noooo! mg=R is not true. In either case- the vertical component is useless here, the box can only move sideways, use only horizontal components of force.
    Why doesn't mg=R? Also by ignoring the vertical component, it would give 900m for the answer to the distance part but the book says 557m?
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    (Original post by pippabethan)
    Why doesn't mg=R? Also by ignoring the vertical component, it would give 900m for the answer to the distance part but the book says 557m?
    Done. Check out my working.
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    (Original post by pippabethan)
    Why doesn't mg=R? Also by ignoring the vertical component, it would give 900m for the answer to the distance part but the book says 557m?
    Ah shizzle, i just read the question properly. You need to find the reaction force R which is mg + 100 sin 20. And then uR is the frictional force.

    So your resultant horizontal force is 45 (don't round off btw) - uR.
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    (Original post by KyleBroflovski)
    Done. Check out my working.
    Name:  d6f003be-b0b5-424d-ba80-324531b2421a.jpg
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    Thank you! I understand it now
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    (Original post by Zacken)
    Ah shizzle, i just read the question properly. You need to find the reaction force R which is mg + 100 sin 20. And then uR is the frictional force.

    So your resultant horizontal force is 45 (don't round off btw) - uR.
    Thank you!!
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    (Original post by KyleBroflovski)
    Done. Check out my working.
    Name:  d6f003be-b0b5-424d-ba80-324531b2421a.jpg
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    you shouldn't leave your answers to such an odd number of sig figs - 2 SF or 3 SF is ideal
 
 
 
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