00:02
In this particular case, here is the length of the bridge.
00:11
The tension acts in this direction at an angle, let's say, that's 40 degree.
00:22
The weight of the car acts over here, and the distance of that is, let's say, c, that this is of the car.
00:35
And the weight of the bridge tells you act as a distance of, let's say x now we have to have some force in this direction at an angle theta so since f y equals zero we have got let's just start with f x equals zero we have f cosine theta equals t cosine 40 degree now if this point is o talk about the point o equals zero if this point is o, let's call this point to be, actually let's just choose this point to be o.
01:53
So talk about the point o is zero.
01:56
So you have weight of the car times the length of the bridge minus c plus weight of the bridge times l minus x.
02:13
These are all counterclockwise torque which is balancing the clockwise torque which is f sine theta times the length of the bridge so we find that f sign theta is weight of the car is 900 kilogram 900 g times different everything by l it because 1 minus c over l the length of the bridge is 1 .5 so basically l is 1 .5 plus 7 .5 equals 9 meter and and c is halfway that distance because c is half of l so c over l becomes half plus weight of the bridge is 2 ,500 kilograms so 2 ,500 g divided by 1 minus x over l...