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Force at point A and E (Static Equilibrium) 
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#1
Jan314, 10:43 PM

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a man is trying to fix the light bulb at the top of his house ceiling.to reach the bulb,he is using a stepladder that is in equilibrium state.sides AC and CE are each 9.0 m long and hinged at C.Bar BD is attached with a rod that is 3.0 m long,halfway up.The man who is climbing 5.0 m along the ladder (AC) is 67 kg in weight.Assume that the floor is frictionless and the mass of the ladder is negligible.
Find the forces at points A and E. 


#2
Jan314, 10:54 PM

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Welcome to PF;
Good question  I take it this is an Aframe ladder? What have you tried so far? 


#3
Jan314, 11:08 PM

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yes with bar in the middle. I get 1220.38 N at point A and 928.58 N at point E.I also tried other calculation which i get 129.6 N at point A



#4
Jan414, 01:08 AM

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Force at point A and E (Static Equilibrium)
That's neet  so which one is right?
I cannot tell unless you show me your working + your reasoning. 


#5
Jan414, 01:16 AM

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first draw free body diagram and find angle and stuff so
at point A 9(Fna sin 19.47)=4(mg cos 70.53) + 4.5(Fb cos 19.47) 9(Fna sin 19.47)=875.4 N + 2785.74 N Fna=1220.38 N but I'm not sure.the other calculation is pretty much the same but with different angle and get 129.6 N so what did you get at point A? overall i did 3 calculations all with different answers.please help me 


#6
Jan414, 07:05 PM

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Is Fna is the magnitude of the force at point A, normal to side AC
Is Fb the total force at point B? What happened to the force at point C? Which angle is 19.47deg? ... I cannot see your reasoning. Note: you don't need to calculate any angles to do this problem  and your end result will be more accurate if you don't. eg. lets call the angle ∠ACE is β, then cos(β)=1/3, sin(β)=(√2)/3 Each time you need a cosine of that angle, just write in 1/3. 


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