00:01
Hello everyone, in this session we have asked to calculate the magnitude, direction of equilibrant vector and angle.
00:12
Now for the solution of the question on resolving components of given force along x -axis is f1 vector is equals to f1 cos 45 ik f2 vector is equals to minus f2 cos 40 ik f3 vector is equals to minus f3 cos 80 ik and along y -axis the components are f1 vector is equals to f1 sin 45 ik f2 vector is equals to f2 sin 40 ik and f3 vector is equals to f3 sin 80 ik.
01:30
Now the net force force along y -axis is f1 vector is equals to f1 sin 45 ik plus f2 sin 40 ik plus f3 sin 80 ik.
02:01
Here the value of f1 is equals to 2 .4, f2 is equals to 1 .7 and f3 is equals to 1 .6.
02:12
On putting the values and simplifying we get fy vector is equals to 1 .697 plus 1 .088 plus 1 .568 the whole j.
02:31
On simplifying we get 4 .353 jk kilo newton.
02:41
Now the net force along x -axis is fx vector is equals to f1 cos 45 jk plus f2 sorry minus f2 cos 40 ik plus sorry it's also minus f3 cos 80 ik.
03:23
On putting the values of f1, f2 and f3 and simplifying we get 0 .1158 ik plus ik kilo newton.
03:39
Therefore the net force is calculated as fnet vector is equals to fx vector plus fy vector...