A uniform rod AB is resting on two supports which are at distnaces 3 m and 4 m from the ends A and B respectively. Masses of 10 kg and 8 kg are suspended from the ends A and B respectively. The weight of the rod is 45 kg and its length is 12 m. Find the reaction at the supports.
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Given:- L=length of rd = 12 m
M1= 10 kg
M2= 8kg.
M= mass of rod
To find:- Reactions R1 and R2
Solution:-
Consider the beam AB with mass 10 kg at A and mass 8k at B.
The equilibrium conditions should be applied.
ΣFy=0
R1+R2=10+8+45
All vertical upward forces(reactions) = All downward forces(rod,weights),
R1+R2=63
Now consider moment at A.
R1*3+(-W*6)+(R2*8)+(-8*12)=0
3R1+8R2=6*45+96
3R1+8R2=366
Solving both the equations,
5R1=138
R1=27.6
Similar;y solve for R2
R2=35.4
The reactions at the support are 27.6 and 35.4kg respectively.
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