Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?
As seen below...
So if I replace L with 2 I get T=600/sin50 meaning T=783.
Now im stuck with the following question http://img63.imageshack.us/img63/5889/klrk3.th.jpg [Broken].
I cant seem to understand how to work the answer out.
Ah yes I see if we replace L with 1 we get 1.5=1.5 meaning the formula becomes 300*L=L*T*Sin30 meaning T=600N.
The next question im stuck on is http://img63.imageshack.us/img63/4524/yhse1.th.jpg [Broken]
I would think the answer is 300*L=L*T*Sin50 .
How did you get "3L/2" on the right hand side and why isit L/2 on the left side ? Furthure more once I calcualted this euqation L=200 where do I go from here?
Right to in problem 1 we basically made it perpendicular correct ? So for now lets call the lenght L meaning that 200*L=200*T*sin30, sin30 because that will give use the hypotenues which is T but the problem with the formula is that it gives the same answer as we got in question 1 meaning we...
I cant seem to work it out correctly the force is L/2 away from the pivot and in order to work out the tension I must do something must I make T perpendicular ?
Alright so the perpendicular distance is changing its now at the mid point the half of 0.5 is 0.25 and due to the fact that the force has moved half way I assume the distance is now half as well thus making it 0.25.
Right, so I am correct when 200*L=L*T*Sin30 the "L"'s cancel out leaving 200=T*Sin30
We then Divide sin30 on each side leaving 200/sin30=T meaning 400=T is this correct ?
This is for the second problem though, wouldnt the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.