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A uniform bridge of weight 1200kN and of length 17m rests on supports at each end which are 1m wide. A stationary lorry of weight 60kN has it’s centre of mass 3.0m from the centre of the bridge.

Calculate the support force on the bridge at each end.

I got the answer to be 642N and 618N. Whereas, the answer in the book says 640 and 620.

I took the distance between the two supports to be 15, should I take it as 17m, if so why?

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  • $\begingroup$ Where did you get 15m from? Everything fits ok if you take it to be 17m $\endgroup$
    – valdo
    Commented Oct 9, 2011 at 15:04
  • $\begingroup$ Take 1m of either ends because the width of supports are 1m. Also, the diagram, shows 15m gap in between the supports. $\endgroup$
    – Deric
    Commented Oct 9, 2011 at 15:05
  • $\begingroup$ @Deric You seem confused about what "momentum" means. This question is about statics. en.wikipedia.org/wiki/Momentum $\endgroup$
    – Mark Eichenlaub
    Commented Oct 9, 2011 at 15:18
  • $\begingroup$ Oh, i meant 'moments'. Sorry. $\endgroup$
    – Deric
    Commented Oct 9, 2011 at 15:25
  • $\begingroup$ The answers are wrong by a factor of 1000 guess why. ""on supports at each end which are 1m wide."" This "wide" i do not really understand. $\endgroup$
    – Georg
    Commented Oct 9, 2011 at 15:32

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The book is correct - how many significant figures are you given the data to ?

I would probably use the middle of the supports (ie 16m) but that doesn't matter for working out the vertical forces. This question is also nothing to do with momentum, unless there is a part 2 where you work out the sideways force when the truck moves.

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  • $\begingroup$ It does have to do with momentum. You have to use the equation moment of a force = Fd. $\endgroup$
    – Deric
    Commented Oct 9, 2011 at 15:03
  • $\begingroup$ I don't think the problem is regarding s.f $\endgroup$
    – Deric
    Commented Oct 9, 2011 at 15:04
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    $\begingroup$ 642 and 618 are 640 and 620 to 2sf, you are only given the data to 2sf so 642 is wrong, whatever the value 'should' be $\endgroup$
    – Martin Beckett
    Commented Oct 9, 2011 at 15:12
  • $\begingroup$ @martin Of course it matters what the width of the bridge is. In the limit of a very long bridge, the truck is in the center and the two supports contribute equally. In the limit of a very short bridge, the truck is all the way to one side and only one support contributes. $\endgroup$
    – Mark Eichenlaub
    Commented Oct 9, 2011 at 15:20
  • $\begingroup$ @mark - sorry, I meant within the accuray given - it didn't matter if you used the inner, outer or midpoint of the mounts $\endgroup$
    – Martin Beckett
    Commented Oct 9, 2011 at 16:31
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To determine where the force acts, you need to understand the flow of stress in the supports--- the bridge is deforming the supports slightly due to its sag, and the contact is most significantly stressed at the point closest to the center of the bridge, and this is where you place the force. If the supports were metal springs instead of rigid blocks, you could calculate the deformation in each spring, and see that it is somewhat bigger nearer the center, but as the spring gets saggier, the force distributes more equally. In the limit of complete sagginess, the force is distributed over the whole meter length of the support, and in the limit of complete rigidity, it is right at the end closest to the center. Assuming complete rigidity is the closer approximation, your calculations are correct, and equal to the answer in the book to the given precision.

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