00:01
Hello everyone in the present problem it is given that force of the cable is 500 newton.
00:08
Coming to the a part of the problem we have here the coordinate of point b is 2 .5 comma 1 .2 .0.
00:37
Hence, position vector is given as rb equal to 2 .5 icap plus 1 .2j cap plus 0 kill cap.
01:03
Now coordinate of c is 0 .2 .2 .2 .0 .0.
01:16
And r .c is 0 icap plus 2 .0 j cap plus 0 .9 kilcap.
01:32
Next rbc equal to rc minus rb which is equal to 0 ic which is equal to 0 i cap plus 2 .0 .0 .j cap plus 2 .0 j cap plus 0 .9 k cap minus 2 .5 i cap plus 1 .2 j cap plus 0k cap.
02:07
Simplifying we have rbc is equal to minus 2 .5 i cap plus 0 .8 j cap plus 0 .9 k cap.
02:24
Now magnitude of rpc is root of minus 2 .5 square plus 0 .8 square plus 0 .9 square.
02:45
So magnitude of rpc is equal to 2 .775.
02:56
Meter.
02:58
Now, force f in cable in vector form is given as f vector is equal to 500 divided by magnitude of rbc multiplied by rbc vector vector...