00:01
Problem 3 .80.
00:02
In this problem we have to replace these couples that are acting on this system by a single equivalent couple and find the magnitude and direction of that equivalent couple system.
00:15
So first of all, what we will do that we will try to resolve all these forces, all these couples into its component couples and then write those couples into its vector form and then we will.
00:30
Will find the direction.
00:32
Let's say that this is moment a, this is moment b and this is moment c.
00:40
Now in the problem it is given that moment a and moment c are acting in yz plane so their components will be along y axis and z axis.
00:50
So moment a will have a component along z axis and it will be 900 cost 20 in positive z direction and and it will have another component along z -axis and that will be 900 sine 20 and that will be along positive y -axis.
01:10
So we can write moment a in vector form as 900 cost 20 and this is positive j directions so k -cap and other component is 900 sign 20 and it is along positive y xx so jcap now moment b is completely along x xx so we need not to resolve this this moment so we can write it as moment as it is as moment b is equals to 840 since it is a negative x direction so minus i cap now moment due to a moment uh due to a moment c that is mc we can write it as it will have a component along positive z direction and another component along negative y direction.
02:08
So positive z direction component will be 1200 cost 20 k cap and negative y component will be minus 1200 sine 20 jcap.
02:24
So total moment acting on the system can be calculated by adding these moments...