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# Forces on a bridge watch

1. Hey, I have no idea how to do this:

A lorry of weight 40000N crosses a 20m long bridge of weight 100000N. What are the forces supplied by each bridge support when it is 5m from one support?

Thanks
2. Total force 'down' must equal total force up

20/5 = 4 therefore 4 supports.

40000 + 100000 = 140000

140000/4 = 35000 each support.
3. Think of the principle of moments.
4. (Original post by Hancock orbital)
Total force 'down' must equal total force up

20/5 = 4 therefore 4 supports.

40000 + 100000 = 140000

140000/4 = 35000 each support.
I think you misunderstood, unless I did

Two supports, one at each end. The truck is 5 m from one end.

OP, think torque. Net torque and net force must be zero for the system to remain in equilibrium.
5. (Original post by D-Day)
I think you misunderstood, unless I did

Two supports, one at each end. The truck is 5 m from one end.

OP, think torque. Net torque and net force must be zero for the system to remain in equilibrium.
yeah - four supports does seem strange for a 20m bridge. duh.
6. I'm not sure if this is right but i got 80,000N and 87,000N.....
7. Got to add up to 140000 - so no its not right
8. Hmm....

Not very sure how to do it then...

i did 40kNx5m=200KNm+ (100knx10m)= 1200kNm= anticlockwise momentum

clockwise= 20mx= 1200kNm
clockwise=1200kNm/20m=60,000N

then i did the 140,000-60,000= 80,000N

and then i did something similar to the other side and got 87,000N....

So then is the answer 60,000N and 80,000N, if it has to add up to 140kN?
9. HELP ~ What have i done wrong?
10. Net torque is zero. Assume that the pivot point is on one of the supports. This eliminates an unknown.

Now move the pivot point to the other support.

11. Well so....
F2=600,000N
F1=800,000N

12. Check your work. Those are off by a factor of ten. Should be 80,000 and 60,000.
13. Ok, thanks alot

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