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woleraymond

  • 3 years ago

a 150kg ladder leans against a smooth wall,making an angle 30 degrees with the floor.The center of gravity of the ladder is one third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

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  1. ujjwal
    • 3 years ago
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    Do you have a diagram? the floor provides horizontal force to what? the ladder or wall.. I guess you are talking about ladder.. But are you sure it is the center of gravity of wall whose position is given? I guess it should be of ladder..

  2. woleraymond
    • 3 years ago
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    yes the question modified....

  3. woleraymond
    • 3 years ago
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    guys i need help @experimentX @Vincent-Lyon.Fr

  4. Vincent-Lyon.Fr
    • 3 years ago
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    Are you sure the 30° angle is with the floor and not with the wall?

  5. Vincent-Lyon.Fr
    • 3 years ago
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    Since the ladder is in equilibrium, net torque about any point must be zero. Chose point A where ladder is touching the floor.|dw:1337890332208:dw| It will allow you to find the value of \(N_B\). Then write that net force is also zero. Finding horizontal component of \(\vec R_A\) is obvious.

  6. woleraymond
    • 3 years ago
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    the 30degrees is with the floor

  7. Vincent-Lyon.Fr
    • 3 years ago
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    Ok, now, have you tried applying the method I described earlier?

  8. woleraymond
    • 3 years ago
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    no idea @Vincent-Lyon.Fr

  9. Vincent-Lyon.Fr
    • 3 years ago
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    Do you know how to express the moment (torque) of a force about a certain point?

  10. woleraymond
    • 3 years ago
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    yep i do but get confused with the center of gravity constraint

  11. Vincent-Lyon.Fr
    • 3 years ago
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    What do you mean "constraint"?

  12. woleraymond
    • 3 years ago
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    the fact that the center of gravity of the ladder is one third the way up from the bottom.

  13. Vincent-Lyon.Fr
    • 3 years ago
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    Well, as you know where it is, you know its distance from A is l/3. That enables you to work out moment of weight about point A.

  14. woleraymond
    • 3 years ago
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    pls i need help its not clear..since the lenght of AB is not given..aw do we work out moment about point A

  15. woleraymond
    • 3 years ago
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    |dw:1340887287738:dw| 30cos50 X 150 = R X bc

  16. woleraymond
    • 3 years ago
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    @Vincent-Lyon.Fr u dig me?

  17. theyatin
    • 3 years ago
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    according to me no matter where the center of gravity lies. horizontal force applied must be 150cos30' .

  18. Vincent-Lyon.Fr
    • 3 years ago
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    Here is the answer to your problem (see attached file):

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  19. woleraymond
    • 3 years ago
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    Thanks for the lecture @Vincent-Lyon.Fr

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