00:01
Hello everyone in this problem we have given a line here we have given the different points that is f here k a e g here it is j here it is j l then c after that there is d and here it is h and here it is b now we have given the different forces acting also so we have given the distance between each point is d.
00:42
So here the value of d is 8 .70 centimeter.
00:48
So if i convert into meter it will be 8 .70 into 10 days to par minus 2 meter.
00:57
So in the first part of the problem we have given at the point c there is a force that is 31 newton in the upper direction.
01:08
So here this one is point c and here the force is in the upward direction that is 31 newton.
01:20
And we have to find the value of the torque at point j.
01:25
So first of all we are considering the distance.
01:28
So it is d and here it is d.
01:31
So the total distance between j and c is 2d.
01:37
So as we know the value of the torque that is tau can be written as force in into the distance that is 2 of d.
01:46
So the value of f is 31 newton into 2 into d that is 8 .70 into 10 raised 2 minus 2.
01:54
So after multiplying we get 5 .39 newton meter.
02:00
Now for the next part that is b part we have given a force that is 22 newton at the point c that is acting in the downward direction.
02:11
So here point c is there and there is a force that is 22 newton which is acting in the downward direction.
02:23
So if i consider the torque along the point j now the direction will be in the clockwise.
02:31
So it will be in the negative sign.
02:33
So that torque can be written as minus of force into the distance that is 2 of d.
02:39
So here it becomes minus 22 into 2 into d that is 8 .70 into 10 raised 2 minus 2.
02:47
So after multiplying we get torque is equals to minus 38 .2 newton meter...