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Same force drawn twice?

by wahaj
Tags: drawn, force
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wahaj
#1
Nov27-12, 10:38 PM
P: 142
1. The problem statement, all variables and given/known data

Find reaction at A and B. A,B,C are pin joints with a pulley at C. There is a weight hanging from the pulley which is attached just above the joint at D.

2. Relevant equations



3. The attempt at a solution

This question can be solved by observation. there are two forces acting on the frame at C. A horizontal and vertical force of the same magnitude which is the tension in the string. BC is a two force member. The FBD is attached. From it I can determine that the reaction at B is tension in the string. member AC has two forces acting on it which are the two different directions of tension in the string. So the weight and the same horizontal force as the force acting on BC. So to conclude the reactions at A and B are of the same magnitude as the weight. But the book draws the horizontal tension in the string twice in the FBD for member AC which results in the horizontal reaction at A being twice as large as what I got. Why does the horizontal force C show up twice?
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haruspex
#2
Nov28-12, 01:32 AM
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I can determine that the reaction at B is tension in the string.
Total reaction or horizontal component thereof? The horizontal components of reaction at A and B will add up to string tension, no?
PhanthomJay
#3
Nov28-12, 01:13 PM
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Before spinning wheels on this, you should correct your sketches to show the joint designations A and B consistently. And dimensions?

wahaj
#4
Nov28-12, 07:11 PM
P: 142
Same force drawn twice?

I suppose providing numbers would be best. I didn't do that because I want to do my homework my self. I'm just having trouble with the concept. I didn't see this before but I labeled things wrong. Thanks for pointing that out. I'll correct that and provide a more mathematical version of my question. Sorry if I confused you. Ok so here is what I am having trouble with. at point C in the BC member's FBD I know that there is the force of the tension acting both vertically (F) and horizontally (Ax). My book draws these forces as I drew them but it also adds another force of the same magnitude as Ax acting to the left. This results in the reaction at A being 1962 N. Where does that force come from?
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haruspex
#5
Nov28-12, 08:10 PM
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Quote Quote by wahaj View Post
at point C in the BC member's FBD I know that there is the force of the tension acting both vertically (F) and horizontally (Ax). My book draws these forces as I drew them but it also adds another force of the same magnitude as Ax acting to the left.
There is also tension in the AC rod. It happens to come out equal to that in the string because of the dimensions of the rods.
wahaj
#6
Nov28-12, 08:13 PM
P: 142
but doesn't the tension in the rod cancel out as a result of the reaction at A?
haruspex
#7
Nov28-12, 08:33 PM
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but doesn't the tension in the rod cancel out as a result of the reaction at A?
Imagine taking away the rod AC. Would the system stay as is? If not, the rod is doing something important.
Allow for a force in the rod in the free body diagram for C and figure out the vertical and horizontal statics equations. You will find a nonzero force there (and as I say, it just happens to equal the string tension with this geometry).
wahaj
#8
Nov28-12, 08:48 PM
P: 142
I see, thanks. its still pretty tough for me to wrap my head around this but your explanation will help.


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