00:02
So here in this way the same tower trust in problem 6 .24 but assuming that the cable is hanging from right side of the tower has fallen to the ground.
00:15
Now here all the dimensions are given in meter and here only this 1 .2 kiloctom will be the external force and we have to find between h .j and bd.
00:30
So first to calculate the forces in member h .j and in each of the members located between h .j and this bd so we need to draw a free body diagram of the entire trust and apply the equilibrium equations so first let's assume let's assume this angle this is theta 1 so this will be this is theta one if this is theta one this is also the turn and here here also this will be the down and this also this also will be theta and let's say this is theta 2 so now let's let's find the angles so for finding angle theta 1 will be equal to tiniverse of opposite by adjacent and inverse of 0 .6 by 2 .21 that should be equal to 15 .1192 degree and the angle theta 2 will be equal to the inverse of 1 .2 the point 6 plus point 6 that is 1 .2 1 .2 divided by 1 .6 so 1 .2 divided by 1 .6 .6.
02:56
And this will be equal to 36 .86 .98 degree.
03:12
Now we found the angles.
03:16
Next, let's draw the free body diagrams.
03:20
First, let's take joint f.
03:33
So for joined f, if this is joined f, then consider there will be a y here.
03:45
And this will be f of df and here says way okay is the access way so and here this will be f of ef now for joined f now use the join method to calculate the member forces so from observation from observation we can say this is theta 1 so this will also be theta 1 so from observation we can say the f f f f f f f f f f f f f f equal to 0 kiloton because there is no external force acting on the joint and the two forces are not corlinear so these two forces are not corinia so the fef equal to fdf will be equal to zero billion atom and next so let's go to join the d so for join d yeah same same like this itself so let's find join d here yeah if this is join d there will be a force fdf and fd e and fd and fd so and the angle this angle this will be teta 1 now let's solve for joint d from a observation we can say like the fbd will be equal to fde fd will be equal to fde equal to fd will be equal to fde 0 kilo tonne just because there is no external force acting on the joint and the two forces are not fully here so this will also be equal to zero and fdf we are already got it's zero so now let's go to the next turn that is joint a for joint d a this is joined a so if this is joined a there will be f a b and f a c and here here there is 1 .2 kilo and here there is 1 .2 kilo notar so and the angles here this angles this is theta one and this is also theta one so let's find i'll join a i'll join there let's take summation along y equal to zero so that will be equal to minus of fac sine teta 1 plus f .a .b.
09:45
Sine of teta 1 minus the 1 .2 will be equal to 0.
09:56
So let's take summation along h .c.
10:01
Equal to 0.
10:03
That will be equal to fac, c, cost, theta 1 plus f.
10:13
A, b, cos, beta 1 equal to 0.
10:24
So let's name these two equations as a and b...