00:01
For this problem, on the topic of equilibrium of rigid bodies, we are told that a horizontal platform labeled a, b, c, d, with a given weight, supports a given load at its center.
00:13
Now, the platform is normally held in position by hinges a and b that we can see in the diagram, and by braces c -e.
00:22
If we were to remove the brace, de, we want to calculate the reactions at the hinges, as well as the force exerted by the remaining brace c -e.
00:32
So here we have our free body diagram from which we can see that the brace de has been removed.
00:41
So we'll express our forces and our weight in terms of rectangular components.
00:46
And we'll first write vector ec as 3 feet along unit vector i plus 4 feet times unit vector j plus 2 feet along unit vector k.
01:16
Now we can find the force due to the brace ce.
01:21
So fce is equal to the magnitude of this force fcee along direction vector ec over the magnitude of ec so along the direction of ec.
01:42
And so this becomes the magnitude of this force fce multiplied by 3i plus 4j plus 2k divided by the magnitude of this vector which is the square root of 3 squared plus 4 squared plus 2 squared and so we get the force of the brace c to be 0 .55709 times whatever the magnitude of this force is fce along i plus 0 .7428 times the magnitude of this force fce along j plus 0 .371339 times fce along k.
02:54
And so we have an expression therefore force fce the other force in question is the weight and the weight we know is simply minus the mass of this platform times the acceleration to gravity m times g and that's along the minus j direction and so this is simply minus 300 pounds times unit vector j or 300 pounds in the direction of minus j.
03:34
Now we also know that the sum of the moments of all the forces about b must equal to zero.
03:44
And this is due to the fact that this platform is in equilibrium.
03:49
So there's no motion here.
03:51
So if we take the moments about b, we get this to be four feet along the k direction cross with the reaction a plus plus 1 .5 feet i plus 2 feet times k crossed with the weight which is minus 300 pounds along j plus 3 feet times i plus 4 feet times k cross with vector f c e the force due to the brace and the sum of these moments must be zero.
05:08
So we can simplify this, and we can write this as minus four feet, and we can resolve our reaction a into its x, y, x, y and x, y, plus four feet, a z, j minus 1 .5 feet, times 300 pounds along k plus two feet times the magnitude of the weight 300 pounds along i plus and now we can do use the method of core factors to find the last cross product and this is the 3 .0 and 4 crossed with fce which is 0 .55 which is 0 .55 709, 0 .748, and 0 .3719.
06:55
And this is multiplied by the magnitude of fce, and this must equal to 0.
07:05
Now if we set the coefficients of the unit vector is equal to 0, we'll first look at the k coefficients.
07:13
We get 0 .7428 times fce times 3 feet minus 300 pounds times 1 .5 feet must equal to 0.
07:45
Which means from here we can find the magnitude of fce immediately.
07:52
So the magnitude of this force is to 2001 .94 pounds.
08:03
Or in keeping our significant figures it's 2002 pounds.
08:09
So we have the magnitude of the force at the brace c .e.
08:17
Now we need to continue to find the reactions at a and b...
