00:01
Okay, in this problem we see a hiker going across a bridge.
00:05
And the hiker is stopping a fifth of the length across the bridge.
00:10
The bridge is supported on each end by two forces.
00:16
Those are f1 and f2.
00:18
These forces are counteracting the weight of the hiker and the weight of the bridge.
00:23
On the right side of the screen here, you'll see that i've written the weight of the bridge as wb and the weight of the hiker as wh.
00:31
For part a of this problem, we want to figure out what f2 or the force of the support on the near end is equal to.
00:41
And we know, or at least we would hope, that this bridge is not rotating.
00:47
It's staying still.
00:49
It's stable.
00:50
That means our torques have to all add up to zero.
00:57
And we're going to call the far end our center of rotation.
01:10
And since the bridge is not rotating, the torque at that far end adds up to zero.
01:21
And torque is the force applied times the lever arm.
01:26
So we have three forces that would be creating a torque around that far end.
01:32
The first is the weight of the bridge times our lever arm.
01:37
We're going to assume this bridge is approximately uniform and that the center of gravity of the bridge is at the middle.
01:47
So we're going to call that one half of the length of the bridge.
01:52
This would create a negative torque as it would push the bridge clockwise around that center of rotation.
02:01
And we have the weight of the hiker.
02:02
Again, this is a negative torque pushing the bridge clockwise.
02:10
And the hiker is four -fifths of the length from that far end...