1. ## Re: Rotating shaft

mechanical engineering. I follow your moments from the point A as this is what I have. Its the next part I don't know how to get from there to my answer of F1=6 and F2 =2. This is where I become unstuck...

2. ## Re: Rotating shaft

Yeah I 100% have the same as you for 0.5F1-1.5F2-2Rb
How do I get from that though to the final answer of F1=6 and F2=2
Been at this for days and cant clear this final hurdle to finish this up.....
Currently Iam no further on than where I started....

3. ## Re: Rotating shaft

You have two equations in two unknowns, F_1 and F_2:

$\displaystyle \sum F = 0 = R_a + R_b -(F1 + F_2)$

$\displaystyle \sum T_a = 0 = 0.5F1 + 1.5F_2- 2R_b$

Using a technique you learned in the 9th grade: rearrange the first equation so that F_2 is by itself:

$\displaystyle F_2 = R_a + R_b - F1$

and subsutite for the F_2 term in the second equation:

$\displaystyle 0 = 0.5F1 + 1.5 (R_a + R_b - F1)- 2R_b = -F_1+1.5R_a-0.5R_b$

and rearrange to solve for F_1:

$\displaystyle F_1=1.5R_a-0.5R_b = 1.5(5N)-0.5(3N) = 6N$

So that's F_1. Now use this value of F_1 back in the earlier equation for F_2:

$\displaystyle F_2 = R_a + R_b - F1 = 5N + 3N - 6N = 2N$.

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