Physics/Mechanics Equilibrium Force Question

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SUMMARY

The discussion focuses on calculating the reaction forces at points A and B for a uniform bar with a mass of 100 kg/m, supported by a pin and a roller. Using the principles of equilibrium, the calculated forces are BY = 6.389 kN and AY = 6.018 kN, with AX determined to be 1.2 kN. The equations utilized include the moment/torque equation and the sum of forces in both the x and y directions, confirming the accuracy of the calculations through multiple methods.

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bob29
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Homework Statement


Find the reaction forces of A and B.
Uniform bar with a mass of 100kg/m supported by a pin and a roller.
[tex]\Theta[/tex] = 60
g=9.81m/s2
[URL]http://img340.imageshack.us/i/mechanicsproblem.jpg/[/URL]
[URL]http://img340.imageshack.us/i/mechanicsproblem.jpg/

Homework Equations


Moment/torque = Force*Displacement
Sum of MA=0 (moment about point A)
Sum of FY=0
Sum of FX=0

The Attempt at a Solution


Taking M about point A. (Counter-clockwise as +ve)
9BY - (1500N*7m) - (2400N*sin(60) * 3.5m) - [100kg/m*9m*9.81m/s2*(1/2*9m)]=0
9BY = 57,505N
BY = 6.389kN
B = 6.389kN​

Sum of FY=0
AY + BY - 2.4kN*sin(60) - 1.5kN - (W=8.829kN) = 0
AY = 6.018kN​

Sum of FX=0
AX - 2.4kN*cos(60) = 0
AX = 1.2kN​
Would like to verify whether my attempt at the question is correct, as well as my free body diagram.
 
Last edited by a moderator:
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A diagram for the question would help.
 
Sorry, Iv'e just found the link to the figure. Your fbd is not correct at the left hand load. Something missing there. You can check your own work by taking moments about any point other than A.
 

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