What is the mass of the counterweight in a two-leaf bascule bridge?

[lofty28]

Modern day draw bridges are also known as bascule bridges. These can have one or two sides (one is called a leaf) that are drawn up from each end of the bridge so that boats can pass through water channels. In some designs, each side of the bascule bridge is held in place by a counter-weight located in a pier placed on the river bank. Imagine one side of a two-leaf bascule bridge. The length of one leaf of the bascule bridge is 25m (from pivot point on the pier to the edge of the leaf that meets the opposite leaf in the middle of the river). The mass of the leaf is 10 000 kg. If the counter weight is located one metre on the other side of the pivot point, use your knowledge of torque (moment of force) to determine the mass of the counterweight. (Hint: The mass of the leaf should be taken from the centre of the leaf and we assume the leaf is at equilibrium in the down position).

I am completely stumped.

Please help!

Thanks

 

[tiny-tim]

welcome to pf!

hi lofty28! welcome to pf!

it's exactly like a see-saw (a teeter-totter) …

how much does the fat kid have to weigh? ​

 

[lofty28]

Would I be right to say...
(f1*L)+(F2*L2)+(F3*L3)=0 (0*25)+(10,000*12.5)+(x * -1)=0 (0)+(125,000)+(x * -1)=0 (125,000)= -(x * -1) 125,000/1 = -x * 1 /1 125,000 = -x

 

[tiny-tim]

hi lofty28!

let's put some line-breaks in there …

(f1*L)+(F2*L2)+(F3*L3)=0

(0*25)+(10,000*12.5)+(x * -1)=0

(0)+(125,000)+(x * -1)=0

(125,000)= -(x * -1)

125,000/1 = -x * 1 /1

125,000 = -x

yes, that's correct until it should be …

(125,000)= -(x * -1)

125,000/1 = x * 1 /1

125,000 = x​

but why not just say "10,000*12.5 = x*1, so x = 125,000"?

 

[lofty28]

Thank you soo much IOU :D

I do have just one more question which relates to the last one...

The bridge from problem 3 needs to draw up the leaves to let a ship through. The leaves are being lifted when someone notices a truck has broken down on one leaf and cannot be moved. It takes some time to stop the motor that controls the lift of the leaf. By how much can the truck be lifted by the leaf of the bridge before it starts to roll back? Assume the coefficient of static friction between the leaf surface and truck is 0.2. In your answer report the angle of the leaf of the bridge at the point just before the truck will roll. When justifying your answer use a free body diagram and appropriate working.

Thanks

 

[tiny-tim]

I do have just one more question which relates to the last one...
…
When justifying your answer use a free body diagram and appropriate working.

this has nothing to do with the last question, this is a simple friction question!

describe (or post an image of) your free body diagram for (only) the truck ​

 

[lofty28]

I think what the question is asking is at what slope (degree) would the truck start moving backwards?
So I assume that some sort of friction equation would solve this.
All the information we were given is what I posted befor... I know how to find it if the mass was given but it is not given..

 

[tiny-tim]

... I know how to find it if the mass was given but it is not given..

ah!

standard trick …

call the mass "m", you'll find it will cancel out in the end! :tongue2:​

what do you get?