00:01
Hi, in this question we need to find the magnitude of the tension t and the force acting at the point a.
00:11
The mass of the eye beam is given as 95 kg per meter.
00:15
The length of the beam is given as 5 meters and height of the beam is given as 0 .5 meters.
00:22
Here what are the forces acting? there will be horizontal reaction at the point a and there will be vertical reaction at the point a.
00:30
Then the weight of the entire beam will be acting at the center of gravity of the beam that is at the point g and there is a load of 10 kiloons which is acting at the point c which is located at a distance of 1 .5 meters from the point b.
00:48
This is the cable on which there is a tension t this particular tension t is inclined to x -axis and y -axis will have two components one will be along negative x -direction and another will be in the positive y direction.
01:06
Considering the equilibrium of the system, let us take the moments about the point a.
01:11
When we take the moments about the point a, we can write sigma m .a is equal to 0.
01:20
This tension component, t .c .25, multiply with the perpendicular distance.
01:29
The perpendicular distance is height divided by 2.
01:33
That is 0 .5 meters is the height divided by 2 plus the vertical component of the tension t sine 25 multiply with the perpendicular distance.
01:51
The perpendicular distance would be 5 meters minus 0 .12 meters minus 10...