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Taking Moments About a Point and Resolving Forces

  1. May 7, 2005 #1
    Hi, I have my first year university exam coming up soon and this type of question seems to feature quite prominently in the past paper questions we are given where I have to take moments about a point and also resolve forces horizontally and vertically. I suppose the best way to show you would be with a question and the solution (which we were given by the tutor).

    Question:

    13(a) A uniform rectangular lamina ABCD is of mass 2m, with AB = DC = 4
    cm and BC = AD = 6 cm. Particles, each of mass m, are attached to
    the lamina at B, C and D. Calculate the distance of the centre of
    mass of the loaded lamina from the sides AB and BC respectively.

    Solution:

    Diagram(a)

    Solution(a)

    I guess this part involves centre of gravity more than anything but I still don't understand how the moments are taken.

    (b) A uniform rod AB, of length 4a and weight W, is hinged smoothly to a
    fixed point at A. The rod is held at 60 to the horizontal, with B
    above A, by a horizontal force F acting at B. Calculate in terms of W:

    Diagram(b)

    (i) the magnitude of F;

    Solution(i)

    Where do these moments come from exactly? I have a vague idea from the solution and my knowledge of trigonometry but I'm not entirely sure.

    (ii) the magnitude of the reaction force at the hinge and its direction to
    the horizontal.

    Solution(ii)

    Not sure here how the forces are resolved, a diagram of where the horizontal and vertical forces are in relation to R would really help.

    If anyone could show me where these answers are coming from I would be really gratefull since this just doesn't seem to click with me and I have no more classes left to speak with the tutor. Thanks.
     
  2. jcsd
  3. May 7, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The first thing you need to review is how to find the torque (or moment) generated by a force about a point. The torque depends on three things:
    the amount of the force, [itex]\vec{F}[/itex]
    the point of application of the force measured from the pivot point, [itex]\vec{r}[/itex]
    the angle between [itex]\vec{r}[/itex] and [itex]\vec{F}[/itex]​
    The magnitude of the torque is given by [itex]r F \sin\theta[/itex].

    Regarding 13a:

    It may be easier to think of the center of mass as the weighted average position of the mass elements:
    [tex]x_{cm} = \frac{1}{M}\Sigma x_i m_i[/tex]
    [tex]y_{cm} = \frac{1}{M}\Sigma y_i m_i[/tex]

    Regarding 13b:

    Start by identifying all the forces acting on the rod. Then apply the conditions for equilibrium: (1) the net horizontal and vertical forces on the rod must add to zero, and (2) the net torque about any point must be zero.

    Hints: (a) Treat the force R that the hinge exerts on the rod in terms of its components [itex]R_x[/itex] and [itex]R_y[/itex]. (b) Use the hinge as your pivot point when calculating torques.
     
  4. May 7, 2005 #3
    Thanks for your help. The solution is a lot clearer now. I'll have a go at some similar questions later and see if I can do them now.
     
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