00:01
Hi, here in this given problem, the force acting on this pipe in vector form, the force is given as 300 i -cap minus 200 j -cap and plus 150 k -cap newton.
00:21
In the first question, that is f313, we have to find moment of the force about x -x is so for the moment of the moment of.
00:35
Of this force about x axis, the position vector, it will be taken as icap, unit vector.
01:03
So, no, sorry, it will be equal to 0 .2 icap.
01:18
So the position vector is 0 .2 icap.
01:21
So moment of the force means torque, it will be given by r cross f, cross, pro.
01:31
Of position vector with a force vector means this is 0 .2 icap cross 300 icap minus 200 j cap and plus 150 k cap so when we expand the bracket i cap cross i cap that term will become 0 so 0 .2 multiplied by minus 200 i cap i cap cross j cap the first term then second term 0 .2 multiplied by 150 i cap cross k cap now here it becomes 40 minus 40 and i cross j that is k cap plus 30 i cross k that is minus j cap so finally moment of the force that is calculated to be equal to that is given to be equal to minus 30 j cap and minus 40 k cap newton meter that is the answer for the first part of the problem that is f313 then in the next one f314 here we have to find the moment about o a axis and for this o a axis now now position factor will be given as 0 .3 icap plus 0 .4 j cap meter.
03:31
So the moment means the torque again using r cross f for r 0 .3 icap plus 0 .4 j cap cross force 300 i cap minus 200 j cap plus 150 k -cap.
03:57
Using matrix method, we use a matrix of the order 3 by 3...