00:01
In this following question, we have given an opening in a floor which is covered by an area 1 into 1 .2 and mass of that sheet is 30 kz.
00:15
As shown in the figure here and it is said that the seat is hinged at point a and point b and is maintained in a position slightly above the floor by a small block c.
00:26
So we have to determine the vertical component at point a, b and c.
00:32
So for this, the position of b with respect to a is 0 .6 i -cap and the position of c with respect to a is 0 .8 i -cap plus 1 .05 k -cap.
00:52
And the position of z with respect to a is 0 .3i -car plus 0 .6 k -cam.
01:05
And the weight of the sheet will be equal to m -g.
01:10
This will be equal to 30 into 9 .8.
01:15
This will give 294 .3 newton.
01:19
So from the equation, m .a.
01:24
Equals 0.
01:26
B with a cross b j vector plus rc with respect to a into c z vector plus rg with respect to a into minus wj equals 0.
01:47
So, 0 .6 i .5.
01:54
Cross bj plus icap plus 1 .05 k -caf into c .j.
02:11
Plus 0 .3 i -cap plus 0 .6 k cap cross minus omega -j will be equal to 0.
02:24
So, 0 .613 plus 0 .8c k minus 1 .05c i .5c...