00:01
Hello students, in the first part of this question we need to find out the torque produced by each forces about an axis.
00:08
First let us found it out for point a.
00:16
So the equation for finding out torque tau is equal to r into f where f is the magnitude of force and r is the position of the vector.
00:25
So first let's find it out the torque produced by force f1 about point a.
00:36
So we know that the position vector from a to the point of application of the force that is f1 is equal to zero because f1 passes through point a.
00:58
Therefore we can say that the torque at this point will be zero newton per meter.
01:06
Now the second one is the torque by force f2 at point a.
01:14
So this will be also position vector from a to point of application of f2 is l by 2.
01:23
So in the clockwise direction.
01:29
So we can write the torque at point a due to force 2 as tau 2a is equal to l by 2 into f2.
01:39
Substituting the values that is 2 by 2 into 12 we can write it as 12 newton meter.
01:48
Now we can find out the torque by force f3 at point a.
01:57
Finding it out the position vector from a to the point of application of f3 is l in the clockwise direction that is tau is equal to l into f3 where l will be equal to 2 into 16 which is equal to 32 nanometer.
02:15
So this is the torque at point a by the third force.
02:20
Now let's find out the net torque at point a.
02:25
Net torque will be equal to the sum of torque produced by each forces.
02:34
So adding all of it at different at point a by different forces that is tau 2a plus tau 3a.
02:46
Tau 1a is already zero therefore 12 plus 32 we get 44 nanometer is newton meter is a sum of torque that is a net torque...