00:01
Hi, in this question, we're given that its uniform beam weighs 500 newtons.
00:08
And since the beam is uniform, the way to be at the center is 500 newtons.
00:18
And the tension in this cable is called that ft.
00:24
We need to resolve this tension into its vertical component.
00:33
This side is going to be 35 also, alternate angles.
00:37
And so this ft cost 35 now torque is equal to force times the distance distance from this center of rotation to the force and for the tension ft it's all could be equal to f t because 35 times 0 .4 l for the width of the bar w which is 500 newton's torque torque would be equal to 500 times distance and since this is at a half point this total distances will be 1 l so this is this will be 0 .5 l now for the for the 700 newton force this is going to be 700 times times l.
02:12
Now we're going to take the clockwise as positive and anti -clockwise as negative.
02:20
So this is clockwise, 7 .00 is clockwise.
02:22
The weight is clockwise and the tension is counterclockwise.
02:27
So it's going to be negative.
02:30
And we'll add all of them together and they'll be equal to zero.
02:40
So minus fg plus 35 times 0 .4 l...