00:01
Hello students in this question a shaft is supported by a thrust bearing.
00:05
This is the shaft system with p acting downwards and q acting downwards.
00:12
This is c and d and there is a and b in here.
00:17
So these are all one meter lengths.
00:20
So the center of mass of the system will be this is totally three meters.
00:23
So center of mass will be 1 .5 meters from the either ends.
00:29
So with that in place we can find out the reaction forces.
00:32
So here the force here p is equal to 10 kilo newton and q is equal to 20 kilo newton.
00:41
So we can find out the force from the center of mass.
00:45
So we can write out the equation force will be equal to 1 .5 times 10 in the 10 kilo newton in the anti clockwise direction anti clockwise direction and we have 20 times 1 .5 in the clockwise direction.
01:04
So two rotational torques are there.
01:09
So with that in place we can calculate the balancing force.
01:13
Now what is the balancing force required to maintain this thing right.
01:18
There is one kilo newton in this direction.
01:20
So what is the balancing force towards the upside in a and b.
01:25
So a is half meter from the center.
01:29
So the balancing force the total force is 10 kilo newton and 20 kilo newton.
01:35
So to balance 10 kilo newton at 15.
01:39
So that is equal to 15 kilo newton meter for the ac part.
01:47
So or the c part anti clockwise part and for clockwise part it is 3 1 .5 times 20.
01:59
So that is 30 kilo newton meter.
02:01
So that is the torque...