- Summersnow8

Please help!
A bridge with mass 9000. kg supports a truck with mass 3000. kg that is stopped in the middle of the bridge. What mass (in kg) must each pier of the bridge support?
= (9000 kg + 3000 kg) / 2
= 6000 kg
If the truck in the preceding problems stops 9.00 m from the northern end of the 32.0-m bridge, what mass (in kg) must the northern pier of the bridge support?
F1 + F2 = (9000 kg * 9.8 m/s^2) + (3000 kg * 9.8 m/s^2)
F1 + F2 = 88200 + 29400
F1 + F2 = 117,600 N
F1 (32.0 m) = (88200 N) (????? m) + (29400 N) (9.00 m)
@nikato @michele_Laino @lightgrav @radar @Greg_D @peachpi
the answers are suppose to be 5220 kg, 6580 kg

- jamiebookeater

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- anonymous

I tried this last night. Got the same numbers you did. 5220 kg doesn't even make sense for the first part because it's not enough support. The part you're missing for the 2nd part is 16 m. Weight acts at the center of the bridge, then you can solve for F1.

- Summersnow8

@peachpi, yeah the answer for the first problem is 6000 kg, but there are 2 asnwers for the second problem... I need help with the second one only

- anonymous

Is there another question for the 2nd problem?

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## More answers

- anonymous

They only ask for mass as far as I can tell

- Summersnow8

they are asking for the mass of the northern pier of the bridge.... so I don't know which answer it is or how you get the answers

- anonymous

|dw:1437603862116:dw|

- Summersnow8

how do I figure out which one is north and which one is south?

- anonymous

North is the one closest to the truck. That's why I switched to N and S as subscripts to make it a little easier
If you take the moment about the north side it's
\[\sum M_N=-(9~m)(29400~N)-(16~m)(88200~N)-(32~m)F_S=0\]
Solving that gives the force on the south support

- anonymous

That should be + 32 Fs

- anonymous

For the north you can do the same, take the moment about the south end
\[\sum M_S=(23~m)(29400~N)+(16~m)(88200~N)-(32~m)F_N=0\]

- Summersnow8

I got this:
F1 (32.0 m) = (88200 N) (16 m) + (29400 N) (9.00 m)
F1 (32.0 m) = 1675800
F1 = 1675800 / 32 m
F1 = 52368

- anonymous

yeah that's what I got, then when you divide by 9.8 you'll get 5343.75 kg

- anonymous

that's for the south pier

- anonymous

for the north pier you can use the equation above or just subtract from 12000

- Summersnow8

but the answers are 5220 kg & 6580 kg , not what we got

- anonymous

that's what I was saying above, those answers don't make sense. They don't even add up to 12,000 kg, which is the amount of mass that has to be supported

- Summersnow8

well, those are the answers in the back of the book.... are you sure our equation is right? why do we divide by gravity? I am not understanding it

- anonymous

I divided by gravity because they asked for mass, not force. The equation is right. The answer makes no sense.
The mass of the bridge and truck is 9000 + 3000 = 12000 kg.
The answers for the supports are 5220 + 6580 = 11800 kg.
This would mean the mass on the bridge is 200 kg more than it can support. Those answers cannot be right, unless something is missing from the problem

- Summersnow8

I put the wrong answer, in the back of the book the problem is asking he mass must each pier of the bridge support, which the correct answer is 6000 kg

- anonymous

I'm confused. that's for the 1st question, right?

- Summersnow8

sorry, i am confusing myself, yea that is for the 1st, but yeah the book says 5220 kg, 6580 kg for the second question...... I don't get it.....

- Summersnow8

so did you get 5543.75 (south), and how do you solve for north?

- anonymous

You can set up a moment equation from the south side, like I did above. Or you can just subtract the south from 12000

- anonymous

since we know they have to add up to 12000

- Summersnow8

so 6456.25 (north)

- anonymous

yes

- Summersnow8

okay, so the answer is 6460 ? I hope we solved this correctly :/

- anonymous

yeah ok if you round

- Summersnow8

hmmm, okay. thanks for your help

- mtimko

The correct answer (at least for my purposes) ended up being 6655 +/- 10

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