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julz127
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Homework Statement
Find the force acting at pin E
[PLAIN]http://img841.imageshack.us/img841/4983/piczt.png
a = 53.0 mm, b = 212 mm, c = 318 mm, d = 106 mm, and e = 371 mm.
A mass of 515kg acts through H.
Homework Equations
The Attempt at a Solution
By finding the moment about point D, we can find the reaction at G, and since GI is a two force member the same force (opposite sign) acts at the other end, on pin I.
[tex] F_G = \frac {(F_H \times 2b) }{d_1}[/tex]
where [tex]d_1 = (2c+d) sin (180 - Atan (\frac{c}{d}))[/tex]
Solving this part I get 5205N, I know this is correct.
To find the forces at E isolate the IKE member. Sum the forces about E, to find the reaction JK then use equilibrium equations to find the x and y forces at E.
[tex]
F_{GIx} = 5205 cos (56.31)
[/tex]
[tex]
F_{GIy} = 5205 sin (56.31)
[/tex]
[tex]
M_E = (F_{GIy} \times 2a) + (F_{GIx} \times \frac {2a}{cos(30)})
[/tex]
This next part I'm having trouble with, can I simply break up the components of [tex]F_{IG}[/tex] and then have them act at I? With opposite sign of course.
I've bee working with the assumption that JK acts perpendicular to EI, is this correct?
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