Reaction at pin support

  • #1
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Homework Statement


I'm asked to find the W and I was told that the rope at C is in tension , there is reaction at B , my question is , is there any reaction at C ?

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Answers and Replies

  • #2
haruspex
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Not sure I understand the question.
The rope AC applies a downward force at A through its tension. That is the only way that forces at C can affect forces at A. Ropes cannot transmit torque about their endpoints or forces perpendicular to the rope.
 
  • #3
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So, is there any reaction at c ?
 
  • #4
SteamKing
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So, is there any reaction at c ?
You don't care.

A free body diagram can be drawn around the beam which excludes C. All that matters, as far as the equilibrium of the beam is concerned, is that the rope is in tension, which means that Tc > 0.
 
  • #5
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You don't care.

A free body diagram can be drawn around the beam which excludes C. All that matters, as far as the equilibrium of the beam is concerned, is that the rope is in tension, which means that Tc > 0.
Why the reaction cant be drawn at c ?
 
  • #6
SteamKing
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Why the reaction cant be drawn at c ?
You don't care what the reaction at C is.

All you are interested in is finding the value of the distributed load w which keeps the rope in tension.
 
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  • #7
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You don't care what the reaction at C is.

All you are interested in is finding the value of the distributed load w which keeps the rope in tension.
i have tried to do in this way , but i do not get the ans
vertical force = 80+10-RB+TC-RC-2W=0 --------equation 1
total moment about A = 80(1) +10(3)+W(2)(5) = 0
110+10W= 2RB , RB= (110+10W) / 2 ------------equation 2
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

TC-RC = (6W-70) / 2 ----equation 3

Sub equation 2 and 3 into 1 ,
i gt 90-(110+10W) / 2 + (6W-70) / 2 -2W = 0
i gt 18W= 0
why cant i do int his way ?
 
  • #8
SteamKing
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i have tried to do in this way , but i do not get the ans
vertical force = 80+10-RB+TC-RC-2W=0 --------equation 1
total moment about A = 80(1) +10(3)+W(2)(5) = 0
110+10W= 2RB , RB= (110+10W) / 2 ------------equation 2
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

TC-RC = (6W-70) / 2 ----equation 3

Sub equation 2 and 3 into 1 ,
i gt 90-(110+10W) / 2 + (6W-70) / 2 -2W = 0
i gt 18W= 0
why cant i do int his way ?
Looks messy.

Why don't you take moments about the pin at B? This will save you some work.

Remember, the reaction at C is not a load on the beam. The only load on the beam at point A is the tension in the rope, Tc.
 
  • #9
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Looks messy.

Why don't you take moments about the pin at B? This will save you some work.

Remember, the reaction at C is not a load on the beam. The only load on the beam at point A is the tension in the rope, Tc.
see it carefully , i did take the total moment about B ,
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

if i ignore RC in my calculation , then my ans would be correct ?
 
  • #10
SteamKing
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see it carefully , i did take the total moment about B ,
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

if i ignore RC in my calculation , then my ans would be correct ?
You haven't got any reasonable answer yet that I can see. Remember, the purpose of this exercise is to find the value of W which keeps the rope in tension.

Again, for the umteenth time, RC is not a load on the beam. Like haruspex said way back, you can't push on a rope. :wink:
 
  • #11
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You haven't got any reasonable answer yet that I can see. Remember, the purpose of this exercise is to find the value of W which keeps the rope in tension.

Again, for the umteenth time, RC is not a load on the beam. Like haruspex said way back, you can't push on a rope. :wink:
so , i have redo the question , here's what i gt :

80+10+2W -RB +TC= 0

moment about A = -80(1)+10(3) +W(2)(5) -2RB = 0
110+10W-2RB= 0
RB= (-110-10W) / 2


moment about B =
-80(1)+10(1) +2W(3) - TC(2) = 0
-70+6W-2TC= 0
2TC= -70+6W
TC = (-70 + 6W) / 2

90 + 2W - ((-110-10W) / 2 ) - ( (-70 + 6W) / 2 ) = 0
W=45N/m

is it correct ?
 
  • #12
SteamKing
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so , i have redo the question , here's what i gt :

80+10+2W -RB +TC= 0

moment about A = -80(1)+10(3) +W(2)(5) -2RB = 0
110+10W-2RB= 0
RB= (-110-10W) / 2
This is a superfluous calculation.

moment about B =
-80(1)+10(1) +2W(3) - TC(2) = 0
-70+6W-2TC= 0
2TC= -70+6W
TC = (-70 + 6W) / 2
The moment calculation about point B looks good.

90 + 2W - ((-110-10W) / 2 ) - ( (-70 + 6W) / 2 ) = 0
W=45N/m

is it correct ?
Then you went and spoiled it by adding the moments summed about point A.

You can write only one moment equation. Discard the moment equation about A.
Use the moment equation about B to find W, such that TC is always in tension. (TC > 0)
 
  • #13
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This is a superfluous calculation.



The moment calculation about point B looks good.


Then you went and spoiled it by adding the moments summed about point A.

You can write only one moment equation. Discard the moment equation about A.
Use the moment equation about B to find W, such that TC is always in tension. (TC > 0)
so , the W = 60/ 7= 8.57???
 
  • #14
SteamKing
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  • #16
SteamKing
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from the moment about B above
You might want to check that original moment equation again. There's no factors of 60 or 7 contained within it.
 
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