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    I have two separate questions regarding moments changing:

    1)

    If a bus drives across a bridge supported at two points (the bus starts from the origin, O, and the bridge is supported at s1 from the origin and s2 from the origin by 2 contact forces - this is all 1D) what happens to the magnitudes of those contact forces (obviously s1 and s2 cannot change), and/or to the bridge itself, as the bus travels across the bridge? Regard the bus as a point particle with weight m2g and the bridge itself as a 1D body with weight m1g.

    2)

    This time, we've got a plank of weight m1g and a ball of weight m2g on top of it. There is one support force at s1 from the origin and another at s2 (to begin with, the ball is located between the two support forces). Now the position s2 of the second contact force is varied, all the way down to s1, and then all the way back up to the very end of the plank (i.e. the plank's length away from the origin). How do the magnitudes of the contact forces on the planks change, and/or what happens to the plank?

    I know these are broad questions but answers would be much appreciated.
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    (Original post by Big-Daddy)
    I have two separate questions regarding moments changing:

    1)

    If a bus drives across a bridge supported at two points (the bus starts from the origin, O, and the bridge is supported at s1 from the origin and s2 from the origin by 2 contact forces - this is all 1D) what happens to the magnitudes of those contact forces (obviously s1 and s2 cannot change), and/or to the bridge itself, as the bus travels across the bridge? Regard the bus as a point particle with weight m2g and the bridge itself as a 1D body with weight m1g.
    (1)
    Suppose that you have a uniform bridge of length l where the supports are at either end of the bridge - if you like, at x = 0 and x = l.

    The forces at the supports F_0 and F_l must satisfy F_0 + F_l = (m_1 + m_2)g for there to be no net force, which is the first requirement for equilibrium.

    The second requirement is for there to be no unbalanced moment. In general, if we let the position of the bus be x then taking moments about the axis through O eliminates F_0 to leave us with m_2 gx + m_1 g(\frac{l}{2}) = lF_l.

    Now clearly if x is small then F_l is small, and vice versa. The force F_0 does the opposite of F_l. By symmetry, the contact forces must be equal when the bus reaches the centre of the bridge. You can confirm this by putting x = \frac{l}{2} above.

    Of course, I have not given you exactly what you wanted; I have only considered the special and, I would think, most usual case. If you decide to put the supports elsewhere, you just need to adjust the second equation.
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    (Original post by Brister)
    (1)
    Suppose that you have a uniform bridge of length l where the supports are at either end of the bridge - if you like, at x = 0 and x = l.

    The forces at the supports F_0 and F_l must satisfy F_0 + F_l = (m_1 + m_2)g for there to be no net force, which is the first requirement for equilibrium.

    The second requirement is for there to be no unbalanced moment. In general, if we let the position of the bus be x then taking moments about the axis through O eliminates F_0 to leave us with m_2 gx + m_1 g(\frac{l}{2}) = lF_l.

    Now clearly if x is small then F_l is small, and vice versa. The force F_0 does the opposite of F_l. By symmetry, the contact forces must be equal when the bus reaches the centre of the bridge. You can confirm this by putting x = \frac{l}{2} above.

    Of course, I have not given you exactly what you wanted; I have only considered the special and, I would think, most usual case. If you decide to put the supports elsewhere, you just need to adjust the second equation.
    I see - so speaking purely qualitatively, the closer the bus is to one of the supports, the greater the proportion of the total contact force required is provided by that one. Does this apply even if the supports do not start from either end? And if we say support 1 is at x1 and 2 is at x2, then is it correct that between x=0 and x=x1, and between x=x2 and x=l, we cannot have equilibrium and both contact forces are 0 (the bridge will turn)?
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    (Original post by Big-Daddy)
    I see - so speaking purely qualitatively, the closer the bus is to one of the supports, the greater the proportion of the total contact force required is provided by that one. Does this apply even if the supports do not start from either end? And if we say support 1 is at x1 and 2 is at x2, then is it correct that between x=0 and x=x1, and between x=x2 and x=l, we cannot have equilibrium and both contact forces are 0 (the bridge will turn)?
    it could pivot about one support... Like a see saw.
    there's one contact force in a see-saw though.
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    (Original post by Big-Daddy)
    I see - so speaking purely qualitatively, the closer the bus is to one of the supports, the greater the proportion of the total contact force required is provided by that one. Does this apply even if the supports do not start from either end? And if we say support 1 is at x1 and 2 is at x2, then is it correct that between x=0 and x=x1, and between x=x2 and x=l, we cannot have equilibrium and both contact forces are 0 (the bridge will turn)?
    If you move the supports around, you can still take moments about one of them to eliminate one force. Taking moments will give a similar equation to the one I have given but with different distances from the chosen axis.

    It is possible for the bridge to turn in the case that you describe, but it does not necessarily have to. If the weight of the bridge acts between the supports and the weight of the bus does not, then the bridge may be in equilibrium. If, however, the weight of the bridge and of the bus are not between the supports, there will definitely be an unbalanced moment. Draw a diagram to see for yourself.
 
 
 
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