00:01
It is said a force is directed from point a, which is given to be 4, minus 1 and 5, to point b, which is given to be minus 2, 0 and 3.
00:10
Now let's first calculate the ab vector.
00:12
Ab vector is nothing but change in x, that is minus 2, minus 4 i cap, plus change in y, 0 minus of minus 1, j cap, plus change in z, which is 3 minus 5, yeah cap.
00:24
Now this value is equal to minus 6 i cap plus j cap minus 2, a cap.
00:28
Now, the unit vector along ab is nothing but ab vector divided by its magnitude.
00:35
Now, ab vector we have already calculated.
00:37
Its magnitude is equal to 6 square, that is 36 plus 1 plus 2 square, which is 4, whole to the power of 1 by 2 or its root.
00:46
And this comes out to be 1 by root over 41 into minus 6 i cap plus j cap minus 2, a cap.
00:54
Now since force is directed along this direction, and the magnitude of force, which is given to be p, then of course b vector is going to be 500 kilo newton which is given in the equation times a b unit vector which is 1 by root over 41 minus 6 i cap plus j cap minus k cap now we need to calculate the moment of course about the origin know the force is acting from point a which is given to be 4 minus 1 and 5 and the origin we already know to be 0 0 0 so the this implies oa vector or the r vector is going to be 4 i cap minus j cap plus 5k cap.
01:31
Now torque acting would simply just be equal to oa vector cross the force vector p.
01:37
Now replacing all these things 4 icap minus j cap plus 5k cap cross the force vector we have found out to be 500 kilo newton divided by root over 41 into we have minus 6 i cap plus jcap minus 2 k cap.
01:55
Now if you compute the cross product we will get the value of the torque to be 500 divided by root over 41 into minus 3 i cap minus 22 j cap minus 2 k cap thus the magnitude of the torque is nothing but the magnitude of this root over 3 square that is 500 by root over 41 into root over 3 square plus 22 square plus 2 square if you calculate this value we'll get this approximately 1740 kilo newton meter and this is the answer of the first question in the second question it is said the axis of rotation or the point about which it should be rotated is c that is given to be 3, 4 ,5.
02:36
And the force, that is the moment, is also acting along on that point itself.
02:40
Then obviously, let's suppose this is so close, oa is so close that oa vector is equal to zero vector...