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Homework Help: Moments, calculating force exerted on ladder by wall

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Moment = force * perpendicular distance

    3. The attempt at a solution

    My book says the answer is 59N, I don't get why my method failed, I use the exact same thing for problems such as calculating the contact force produced by the ground at the end of a bridge and I've never had any problems.
    Last edited: Oct 18, 2014
  2. jcsd
  3. Oct 18, 2014 #2


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    Staff: Mentor

    Note that R is not perpendicular to the ladder.
  4. Oct 18, 2014 #3
    When I do these kinds of problems usually the beam/ladder are always perpendicular to the pivot/r making it hard to work out how these things work.

    Ok, does this mean that the force always needs to be perpendicular to the pivot even if it results in the force not being perpendicular to the beam/ladder?
  5. Oct 18, 2014 #4
    Step 1: Try and figure out what the moment arm is for the weight (it is easy to do so).
    Step 2: Figure out the moment arm for the normal force (this is given)
    Step 3: Equate your two moment expressions for the moment about the base of the ladder and solve for the normal force (the force the wall exerts on the ladder) (remember to declare a positive direction for moments on your diagram)
    -Remember the ladder is at rest (its not rotating)
    Last edited: Oct 18, 2014
  6. Oct 18, 2014 #5
    What value did you use for the moment arm for R, and why?

  7. Oct 19, 2014 #6
    Ok, is this the way I'm supposed to do these types of problems where the beam is not perpendicular to the pivot?
    Calculate the vector component of the force so that the force is perpendicular to the pivot even if the beam is not?


    I'm used to the following types of problems and then suddenly a new type shows up with no additional explanation from the book

  8. Oct 19, 2014 #7


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    You can always use the same approach to calculate a momentum:

    - find the angle ##\alpha## between the direction of the force F and the direction between pivot and the point where the force acts.
    - find the distance d between the point where the force acts and the pivot.
    - calculate ##F d \sin(\alpha)##
    - take care of the sign, if necessary

    There is no need to make additional lines in some diagram.
  9. Oct 19, 2014 #8
    There are two ways of getting the moment of a force around a pivot axis.
    Method 1:
    • Draw a dashed line along the line of action of the force
    • Drop a normal from the pivot point to the line of action. This is the moment arm (in your problem, this is 4 m).
    • Multiply the magnitude of the force by the length of the moment arm.
    Method 2:
    • Use as the moment arm the distance between the pivot point and the point of application of the force.
    • Take the component of the force in the direction perpendicular to this moment arm
    • Multiply this component of the force by the moment arm
    Both these methods give the same value for the moment of the force, and this value is the same as that given by mfb in the previous response.

    Here's some advice for whatever it's worth. The idea is not to learn how to solve this kind of problem or that kind of problem. The idea is to understand the fundamental concepts, and then to get practice in applying these concepts to problems.

  10. Oct 19, 2014 #9
    I think I finally understand this, can you confirm?


    I thought the arrow of force had to be perpendicular to the pivot, but it seems this is not the case. I've calculated the components of the original force to see if the different perpendicular distances give me the same moment.

    I think the contact force at R (vertically) would be 234.44/4m = 58.61N
    I think the contact force at A (horizontally) would be 234.44/1.5m = 156.29N
    Last edited: Oct 19, 2014
  11. Oct 19, 2014 #10
    Remember that a contact force always acts normal to the surface...
  12. Oct 19, 2014 #11
    This is how I would solve:
    -Say x is positive to the right and counterclockwise moments are positive, then:
    [itex] {Moment} \hspace{2 mm} {of} \hspace{2 mm} {weight:} \hspace{2 mm} -(32kg)(9.81\frac{m}{s^2})(0.75m) [/itex]
    [itex] {Moment} \hspace{2 mm} {of} \hspace{2 mm} {R:} \hspace{2 mm} R(4.0m) [/itex]
    Equate these to expressions and solve for R:
    [itex] R(4.0m) = -(32kg)(9.81\frac{m}{s^2})(0.75m) [/itex]
    [itex] R = -59N [/itex]
    Also, the contact force at the base of the ladder will not create a moment (at the base of ladder)
    Last edited: Oct 19, 2014
  13. Oct 19, 2014 #12
    This is not correct.

  14. Oct 19, 2014 #13
    youre right...I should have stated that when friction is ignored, contact forces are normal
  15. Oct 19, 2014 #14
    To the thread starter I apologize for that bit of misinformation
  16. Oct 19, 2014 #15
    I don't understand your diagrams, but I agree with your determination of the force R. This recognizes that R is oriented normal to the wall, because the problem statement implies that the frictional component of R is zero (smooth wall). I would have solved for R by using the same moment balances as those employed by N3wton in #11 (what I called Method 1).

    The contact force at the base of the ladder has two components, a normal component and a frictional component. From force balances on the ladder, the normal component is equal to the weight of the ladder, and the horizontal component is equal and opposite to R.

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