00:01
In the question we have given radius of the pulley is 3 .24 inches, force f is 1600 lbf, length l is 36 inches, height h is 32 .4 inches.
00:15
Now we'll find this distance by under root of 36 plus 36 square plus 32 .4 square equals to 78 .95 inches.
00:28
This theta is cos inverse 36 plus 36 divided by 78 .95 which is 24 .23 radian.
00:42
This is also theta, this is 90 minus theta.
00:47
We will find the force component from the moment.
00:51
Taking the x component of the moment at point d equals to 0, the force f ax multiplied by perpendicular distance 32 .4 minus 1600 multiplied by perpendicular distance 36 plus 36 plus 3 .24 equals to 0.
01:13
So f ax equals to 3715 .55 lbf.
01:22
Since the system is in equilibrium x component of a and x component of d are same.
01:30
So f dx equals to 3715 .55 lbf...