00:01
Hello, here a shop sign weights 245 newtons and it's supported by a uniform beam which is 110 newtons heavy.
00:22
We have to calculate the tension in the wire.
00:28
Let's do this.
00:30
So first let's us illustrate it.
00:39
So the weight of the sign is applied here at the distance of 1 .70 meters from the point of rotation from the pivot and and the weight of the bar is applied in the middle in the center of the bar.
01:20
Meanwhile tension produces a torque which is in the opposite direction and that's dictated by the angle of 35 degree.
01:40
Now let's write down the equilibrium condition.
01:43
So here the torque created by w1 plus torque created by w2 equals the torque created by 4.
01:58
So the first one equals to w1 times l, the second one.
02:10
Equals to w2 times l over 2 and the torque created by tension equals to tension times distance let's call it x times sine of 35 degree and here let's show this distance x x that is this distance from the pivot so therefore now we can rewrite this condition for the balance.
02:49
Therefore, t equals to w1l plus w2l divided by x sign of 35 degree.
03:07
Let's calculate it, that is 245 times 1 .70 meters plus 1 .10 newtons times 1 .17 meters divided by 2 divided by x which is 1 .35 meters times sign of 35 degree...