Answer:
- CD = DE = DF = 0
- BC = CE = 15 N tension
- FA = 15 N compression
- CF = 15√2 N compression
- BF = 25 N tension
- BG = 55/2 N tension
- AB = (25√5)/2 N compression
Explanation:
The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.
At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.
Then link CE has 15 N tension.
The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.
In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.
At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.
At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.
The <em>horizontal</em> forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.
In summary, the link forces are ...
- (25√5)/2 N compression in AB
- 15 N tension in BC
- 25 N tension in BF
- 0 N in CD, DE, and EF
- 15 N tension in CE
- 15√2 compression in CF
- 15 N compression in FA
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Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.
