A metal rod of length `50cm` having mass `2kg` is supported on two edges placed `10cm` from each end. A `3kg` load is suspended at `20cm` from one end. Find the reactions at the edges (take `g=10m//s^(2)`)

Explanation

`AB` is the rod, `C` is the centre of gravity and `W` is the weight of the rod acting downward and `W_(1)` is the weight of the load suspended at point `D`. Rod is supported at two edges `E`and `F` as shown in the figure.`R_(1)` and `R_(2)` are the reaction force at `E` and `F` respectively.

As the rod is uniform and homogenous, therefore `G` is at the centre
`AB=50cm`, `AC=25cm`, `AD=20cm`, `CD=5cm`, `AE=BF=10cm`, `ED=10cm` , `EC=FC=15cm`
`W=mg=2xx10=20N`, `W_(1)=3xx10=30N`
For translational equilibrium `sum_(i)vecFi=0`
`R_(1)+R_(2)-W-W_(1)=0`
{`W_(1)` & `W` act in the downward direction and `R_(1)` & `R_(2)` act in the vertically upward direction}
`R_(1)+R_(2)-20-30=0`
`impliesR_(1)+R_(2)=50`...........`(i)`
For rotational equilibrium `sum_(i)vectau i=0`
`=-R_(1)(EC)+W_(1)(CD)+R_(2)(FC)=0`
`implies-R_(2)(15)+30(5)+R_(2)(15)=0`
`impliesR_(1)-R_(2)=10`..........`(ii)`
adding `(i)` and `(ii)`, we get
`2R_(1)=60`, `R_(1)=30N` and `R_(2)=50-30=20N`
`R_(1)=30N`, `R_(2)=20N`