00:01
In the following question, it is said that three forces are applied to an object as shown in the figure.
00:07
Force f1 has magnitude of 24 .1 newton and is directed 30 degree to the left on y -axis.
00:16
Force f2 has magnitude of 14 .2 newton and point along positive fx.
00:21
What must be the magnitude and direction by third force f3 such that vector sum of these forces is 0 newton.
00:32
So for this resolving component of forces, vector f1 x will be equal to mod f1 vector sine 30 degree and f1 y vector is equal to mod f1 vector cost 30 degree and f2 x vector will be equal to mod f2 vector cos 0 degree which is mod f2 vector and vector f2 y is equal to mod f2 vector cos 90 degree which is equal to 0 and f3 x is equal to mod f3 vector cos theta and f3 y vector is equal to mod f3 sine theta and if we sum all the forces we get zero newton which means forces along x -axis and y -axis are balanced so y -axis forces forces f1y vector is equal to vector f3 y.
02:33
So f1, pos 30 degree is equal to mod f3, sine theta and assume this equation as first equation.
02:46
Now forces along x -axis, f1 x plus f3 xx is equal to f3x is equal to f2 x so f1 sine 30 degree plus f3 cos theta will be mod of vector f2 and by equation first f3 f3 vector will be equal to mod f1 cos 30 degree upon sine theta assume this is third equation.
03:41
Then f1, sine 30 degree plus f1, cos 30 degree upon sine theta, will be equal to mod f2 vector.
04:02
So, f1 vector, sine 30 degree plus f1 vector plus f1 vector will be equal to mod f2 vector 24 .1 sine 30 degree plus 24 .1 pos 30 degree or theta is equal to 14 .2.
04:35
So 24 .1 into 0 .5 plus 0 .86 quart theta is equal to 14 .2...