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Forces exerted on the ladder by the wall and ground

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A uniform ladder which is 10 m long and weights 300 N leans with its upper end against a smooth vertical wall and its lower end on rough horizontal ground. The bottom of the ladder is 6 m from the base of the wall. A man weighting 700 N stands on the ladder at a point 6 m above the ground. Calculate the magnitudes and directions of the forces exerted on the ladder by (a) the wall, (b) the ground.

    Answers: (a) 506 N at 90 degrees to the wall, (b) 1121 N at 63 degrees to the ground.

    2. The attempt at a solution
    At first I did a graph of the situation:

    f8e74195f143.jpg

    R = normal reaction to the wall, S = to the ground, F = friction (rough ground).

    Find 8 by Pythagoras. S is = 700 N + 300 N = 1000 N.

    And after this point I am a bit stuck. From a different book example I get:
    (F * 8) + (300 * 3) + (700 * 1.5) = (S * 6)
    8 F = 4050
    F = 506.25 N

    The answer looks right, but:
    As I understand, the body is in equilibrium so in that case: normal reaction S (anti-clockwise) should be multipled by the perpendicular line which is 6 m -> (S * 6), then all of other forces which are facing clockwise direction should be used: we have F and we sort of move it to the left so the perpendicular line is 8 m (F * 8), two other clockwise forces are 700 and 300. 300 * 3 and 700 * 1.5 (the vertical side of the triangle is 2 m, the hypotenuse is 2.5 m and by Pythagoras the horizontal line is 1.5). So we get F = 506 N. But how to find the angle? Because F is a force directed towards the wall that is the reason why it is equal to 90? And why don't we take into account the R force?

    And how to start with (b)?

    Any help please?
     
  2. jcsd
  3. Oct 8, 2015 #2

    BvU

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    (1) Here you calculate a contribution to the total torque ##\Sigma \vec \tau## which must be 0 for equilibrium. Apparently with the top of the ladder as axis of rotation, considering you move F to the left over 6 m. And that's why R doesn't contribute. That gives you F. And sum of forces horizontal = 0 gives you R

    (2) The wall is smooth, so it can only exercise a force perpendicular to itself.
     
  4. Oct 13, 2015 #3
    Vertically: we find S = 700 + 300 = 1000 N

    Then we take moments at the point where the ladder touches the wall: (700 * 1.5) + (300 * 3) + (F * 8) = (1000 * 6) -> F = 506 N. In the answer 506 N at 90 degrees to the wall but isn't 506 the F force? I mean the friction with the ground? Why is it 90 degrees then?

    If we take the horizontal forces R = F (similarly as the vertical forces) then R = 506 N and then ok the force is at 90 degrees to the wall. But in that case what is the 1121 N at 63 degrees to the ground answer? Shouldn't it be F? But in that case we have already found F and it is equal to 506 N.

    Really lost on this one.


    Update:

    1. First we find 8 m the vertical side by Pythagoras. Then 3 m at the 300 N because it's the midpoint of the triangle and 1.5 m at 700 N because it's also a midpoint of the smaller traingle.
    2. Vertically: S = 700 + 300 = 1000 N.
    3. Moments about ladder touching the wall: (700 * 1.5) + (300 * 3) + (F * 8) = (1000 * 6) -> F = 506 N.
    4. Horizontally: F = R -> R = 506 N.
    5. R is the normal reaction to the wall and that's why it's at 90 degrees.
    6. Now we need to find the force exerted on the ladder by the ground. We take it as RG and it's the force which on noted as "5 m" on the graph. By Pythagoras: 506.252 + 10002 = 11212 N.
    7. Now we're looking for the angle which is below the 5 m line (the angle which the line makes with the horizontal / ground). In that case: tan θ (1000 / 506.25) -> θ = 63.2 degrees.

    Looks logical to me. Could't understand that the force exerted on the ladder by the wall is not the friction (F) or the normal reaction (S) but the hypotenuse (the ladder) of these forces. And of course F = R and R is 90 degrees and it's logical.

    Is everything right now? I am a bit unsure of the first part where I find 1.5 m of 700 N. The smaller traingle has 4, 5 and 3 (ver, hyp, hor) as it's side. If the smallest triangle has a 2 m vertical side, which is the middle point of the vertical side of the small triangle, then, as I understand, I can conclude that the smallest triangle has 1.5 as a horizontal. side.
     
    Last edited: Oct 13, 2015
  5. Oct 13, 2015 #4

    BvU

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    You mean floor. That force is the vector sum of F and S. Is not in the direction of the ladder.
     
  6. Oct 13, 2015 #5
    Oh yes, I meant floor. The force exerted on the ladder by the floor is not the fliction (F) or the normal reaction (S) but the vertor of these forces.

    One quick question:

    1. First we find 8 m the vertical side by Pythagoras. Then 3 m at the 300 N because it's the midpoint of the triangle and 1.5 m at 700 N because it's also a midpoint of the smaller traingle.

    I am a bit unsure of the first part where I find 1.5 m of 700 N. The smaller traingle has 4, 5 and 3 (ver, hyp, hor) as it's side. If the smallest triangle has a 2 m vertical side, which is the middle point of the vertical side of the small triangle, then, as I understand, I can conclude that the smallest triangle has 1.5 as a horizontal side.

    This logic is correct?
     
  7. Oct 13, 2015 #6

    BvU

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    You can check for yourself: make your drawing a bit better to scale !
     
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