00:01
Here we have to determine that the force in each member of the load is truss here.
00:05
So here this is point b, this is point a, let's say here is point a, point c, point d.
00:18
This angle is given that is equal to 40 degree, this again is given that is 40 degree, this is 34 meter, this is 40 degree, this is 6 .8 meter and the load of 300 newton is acting in the downward direction from here.
00:32
So this term is given, so we are considering the joint d from here.
00:39
So we are considering the joint d, so from joint d this angle is given that is equal to 40 degree.
00:45
So f of ad is acting from here, a load that is equals to 3060, it's 3060 newton.
00:55
So 3060 newton is acting in the downward direction and from here f of cd is acting.
01:01
So what we have to do is we have to equate this vertical force.
01:07
So equating this vertical forces from here, so f of cd multiplied by the sine of 40 degree is equals to 3060.
01:17
Solving the term we get the value of fcd that become equals to 4760 .5149 newton.
01:23
So this is the value of f of cd.
01:26
Now we are considering the force in cd.
01:29
So equating the horizontal force here that is f of ad plus f of cd multiplied by the cos of 40 degree that is equals to 0.
01:38
So f of ad becomes equals to minus of f of cd multiplied by the cos of 40 degree.
01:43
So this from here becomes equals to minus 4760 .514 g cos of 40 degree.
01:51
So solving the term from here we get the value of f of ad that become minus 3646 .766 newton.
01:59
This is the force in ad.
02:01
Now we are considering the joint c.
02:04
So from the joint c this is somewhat like this that is force of bc is acting in this direction.
02:09
Here the force of ac which is making an angle of 40 degree...