00:03
We need to find the forces exerted on the cylinders, cg, ad, and ef, due to an input force p that is equal to two kips that's applied horizontally at the point j.
00:18
So i solve this problem.
00:21
We're going to first work around the bucket, taking the seven moments of the bucket, then working towards the ad cylinder, and then work towards the ef cylinder.
00:36
And these triangles here with lengths, we're going to use these to find the x and y components of the forces, cg and ad, whenever we're solving for the moments.
00:50
And these are just ratios.
00:51
These links here are just ratios of the given ones.
00:59
So i'll start the problem by taking the sum of moments about point h and setting it equal to zero.
01:08
And we can write that as 4 over 5 times f of cg times 10 plus 3 over 5 times f of cg times 10 plus force p times 16 is equal to 0 and we can plug in the known terms and solve for f of cg so we can plug in 2 for p and we take this equation software f of cg and we'll get a force of 2 .29 kips so we'll work towards point b and say that the sum of moments about point b is equal to zero and we can write that as 4 over 5 times f of 80 times 12 plus 3 over 5 times f of 80 times 10 plus p times 86 is equal to 0 and we rewrite this plugging in p so we plug in 2 for p and we take the equation solve it for f of 80 to get a force that is equal to negative 11 .03 kips and if you want the magnitude of this we can just say that the magnitude is equal to the positive value of the previous calculation...