Statics Question involving forces and moments

  • Thread starter jnbfive
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  • #1
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Just figured I'd throw this up here. I was able to solve for the reaction at E when a=2 but when I tried for a=7.5, I wound up with something completely different. I was wondering what I'm not incorporating into the steps to solving this.

Also, it says determine the reactions at A and E. The page got cut when I scanned it.
 

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  • #2
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This is my work for solving part a) when a=2

Sum of the Moment about A = 20(2)-16(1.5)+[5.5(sin(30))+2(cos(30)]E

-8=[5.5(sin(30))+2(cos(30)]E

E= -8/[5.5(sin(30))+2(cos(30)] or 3.57 lbs at a 60 degree angle, which is what the book says.

Now how come when I plug in 7.5 in a, I can't come up with 21.0 lbs? Suggestions would be appreciated
 
  • #3
PhanthomJay
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This is my work for solving part a) when a=2

Sum of the Moment about A = 20(2)-16(1.5)+[5.5(sin(30))+2(cos(30)]E
what happened to the 32 in-lb couple?
-8=[5.5(sin(30))+2(cos(30)]E

E= -8/[5.5(sin(30))+2(cos(30)] or 3.57 lbs at a 60 degree angle, which is what the book says.
correct answer (specify direction), wrong equation and math.
Now how come when I plug in 7.5 in a, I can't come up with 21.0 lbs? Suggestions would be appreciated
maybe you forgot the 32 inch pound couple?
 

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