Question

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.

Answer 1

Answer:

50 hope this helps you

Explanation:

hope this helps you

Answer 2

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.