00:01
Hello students, referring to the given diagram, we are asked to determine the magnitude of moments of force about x, y and z axis using scalar product as well as vector product.
00:11
From the diagram, the vector equation can be written as at point b, it is 4i plus 3j minus 2k.
00:21
It is since z is in the negative axis and f is equal to 4i, it is given plus 12j minus 3k and we have to take the scalar approach.
00:35
Then we can write the moment of force about x axis is given as y fz minus z fy.
00:43
Then it can be written as 3 into minus 3 minus minus 2 into 12.
00:51
So on calculating this, we will be getting 15 foot pound and moment of inertia about y axis can be written as minus x fz plus z fx.
01:07
On substituting minus x is minus 4, 4 into minus 3 plus minus 2 into 4.
01:17
So on calculating this, we will be getting 4 foot pound.
01:22
Next moment of force about z axis can be given as x into fy minus y into fx.
01:31
On substituting the value 4 into 12 minus 3 into 4.
01:37
On calculating this, we will be getting 36 foot pound.
01:43
Now we have to move to vector product.
01:46
Moving on to vector approach, we can write mx is equal to moment of force about x axis is equal to ux into b cross f.
02:01
Therefore, ux can be taken as 1 0 0 and 4 3 minus 2 and f is 4 12 minus 3.
02:16
On calculating this, minus 9 minus 12 to 24 plus 24.
02:23
Therefore, the answer is 15 foot pound.
02:31
Next moment of force about y axis is is equal to u y into b cross f, which can be given as 0 1 0 4 3 minus 2 and 4 12 minus 3.
02:46
On calculating this, we will be getting 1 into 4 3 minus 12 minus of minus 8 plus 8.
02:57
Then it is minus 1 into then it is 4 foot pound.
03:06
Next is moment of force about z axis is equal to uz into b cross f.
03:13
Therefore, this is 0 0 1 into 4 3 minus 2 into 4 12 minus 3.
03:22
On calculating this, 1 into 4 12 of 48 minus 12, which can be obtained as 36 foot pound...