00:02
In this problem, we are asked to find out the magnitude of the force f such that the moment m equals to 900 newton meter.
00:15
So here let us consider the given information from the diagram.
00:20
We have the point a which is 0 ,0, 10 .5 and the point b which is 4, negative 3, 1 .5.
00:31
So using these two points we have the position vector of ab to be 4 minus 0 which is 4.
00:41
Negative 3 minus 0 is negative 3 and 1 .5 minus 10 .5 is negative 9.
00:48
So now let us evaluate the magnitude of this vector.
00:52
So we have the magnitude to be square root of 4 squared which is 16 plus negative 3 the whole squared which is 9 plus negative 3 the whole squared which is 9 plus negative of 9 the whole squared which is 81.
01:04
Adding all of this up we get square root of 106 which is approximately equal to 10 .30.
01:13
So now let us evaluate the force vector we have the f vector to be equal to the magnitude of f times the position vector of a, b over the magnitude of the vector ab.
01:34
So now substituting the values we have f times 4 negative 3 negative 9 divided by 10 .30.
01:44
So further simplifying this we get f vector to be equal to f times the vector 0 .39 negative 0 .29 and negative 0 .87.
02:01
So now let us consider the formula of the moment.
02:07
The moment is given by the cross product of the position vector of a and the force f...