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Problem 3 Find the forces in the horizontal member of the frame due to the 100 lb weight. Free body diagrams are required. in 12 in 3 in 8 in

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Answer #1

The FBD's of the two members are as shown below:

ー11 in 12 in Ax 3 in Cx 8 in Ay W Cy FBD of member AD W 1001b S in -1 1 in -12 in CX 8 in Ex MCy ЕУ FBD of member EC

1) We first isolate member AD at joint C

applying c.o.e we get,

?MA = 0

100 x (11-3) - Cy x 16 + 100(28+3) = 0

Cy = 243.75 lbs

?Fx = 0

Ax = Cx ...........(i)

?Fy = 0

Ay -100 + 243.75 -100 = 0

Ay = 43.75 lbs

2) Now, we isolate member EC at pin C

again applying c.o.e we get,

?ME= 0

-100(11-3) + Cy*16 + Cx * 8 = 0

-100*8 + 243.75 x 16 + Cx * 8 = 0

Cx = -387.5 lbs

?Fx = 0

Ex = Cx = -387.5 lbs

?Fy = 0

-Ey +100 -243.75 = 0

Ey = -143.75 lbs

also, from (i)

Ax = -387.5 lbs

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