00:01
We need to find the forces exerted at the point d and the point f whenever this rail is lifted by the tongs.
00:13
So we have three different free body diagrams to represent the different points on the machine.
00:21
So we have a free body diagram for the rail, a diagram for the a junction, and also a diagram for the bdf tongue.
00:34
So to start solving this problem, we're going to first find our weight w of the rail.
00:42
And we're given the length of the rail and also its unit weight.
00:50
So to find the weight w, w is going to be equal to 39 times 44.
01:11
And that'll give us a weight of 17, 16 pounds.
01:22
So now that we have the weight, we can now take the moment, about this point e on this free body diagram.
01:38
So you can say that the sum of the moments about the point e is equal to zero, and the formula for that is going to be negative w times 0 .8 plus f of y, the y component at the point of the force, times 1 .6 equal to 0.
02:15
Solving this for f of y and plugging in w, we get f of y is equal to 1716 times 0 .8 divided by 1 .6, and this will give us a force of 8 and 58 pounds.
02:47
So next we're going to calculate an angle theta.
02:52
So we're going to calculate angle theta from this free body diagram and this angle theta also appears on the tonne free body diagram.
03:07
So we can solve for that angle theta by taking the tangent inverse of 6 over 9 .6 and this will go as an angle of 32 degrees.
03:33
So we can now say that using the free body diagram in the middle we can say that the sum of all forces in the x direction is equal to zero.
03:46
So we have that sum of all forces in the x direction is equal to zero, which in terms of that free body diagram, we can write as force of ac times a cosine of theta minus the force ab times the cosine of theta is equal to zero.
04:22
And from this equation, we can rewrite this equation as f of ac is equal.
04:32
To f of ab.
04:41
And next we're going to use the same free body diagram.
04:46
We're now going to say that some of all forces in the y direction is equal to zero, which we can write as w minus the force ac times the sine of theta minus the force av times the sign of theta is equal to zero.
05:23
So we can substitute the known terms.
05:28
And solve for the unknown f of ab.
05:41
Since in this equation we can replace the f of ac with f of a b from the previous expression.
05:53
So after plugging in the known terms, we get 1716 minus f of ab times 2 times the sine of 32 degrees is equal to 0.
06:18
And solving for the unknown f of a b, we get a force that is equal to 0...