00:01
Here, in this question, based on the given conditions of a system, we must draw a free body diagram of the whole system.
00:10
So let's first draw a free body diagram of this system.
00:17
So this is the free body diagram which we draw for the given system.
00:24
Now in the question, paul is in equilibrium condition and net torque will be equal to zero.
00:33
So here, l is the length of the pole, h is the height, capital m is the mass of the pole, ft is the tension, fpx is horizontal force, fpi is vertical force and small m is the mass of the traffic light pole.
00:57
Now when a pushing force capital f is acting at some point with post, position vector r, then the resulting torque is the product of position vector and the force f and the angle between them.
01:13
Or we can write this mathematically that tau is equal to f r sine theta.
01:24
So this is the resulting torque which is the product of position vector are force f and the angle theta between them.
01:38
Now comes to part a, so here we consider that the equilibrium of torque at vertical.
01:49
So due to equilibrium of torque at vertical, summation tau is equal.
02:09
To 0.
02:11
Therefore, ph minus 1 divided by 2mg cos theta minus mggl pose theta must be equal to 0.
02:35
Now here we rewrite this equation for f of t or we can say f t.
02:43
Therefore, f t is equal to 1 divided by 2 m g cos theta plus m g l cos theta divided by h now here small m is the mass of the pole capital m is the mass of the traffic light h is the height and l is the length so in this we substitute 12 kg for small m 7 .20 meter for l 21 .5 kg for capital m 9 .8 for g and 37 degree for theta and also 3 .8 meter for h therefore we get f t is equal to minus 12 multiply by 9 .8 multiplied by 7 .20 divided by 2, pose 37 degree minus 21 .5 multiplied by 9 .80 multiply by 7 .20 pose 37 degrees...