A uniform beam, XY, 4m long and weighing 350N rests on two pivots P and Q. It is kept in equilibrium by weights of 80N attached at X and 1000N attached at a...

Answer Details

Measure positions from \(X\): \(X=0,\ P=0.8\text{m},\ Q=0.8+2.2=3.0\text{m},\ Y=4.0\text{m}\). The uniform weight \(350\text{N}\) acts at the centre, \(2.0\text{m}\). The \(80\text{N}\) acts at \(X=0\); the \(1000\text{N}\) is \(0.6\text{m}\) from \(Q\), i.e. at \(2.4\text{m}\).

(a) Vertical equilibrium: \(R_P+R_Q=80+350+1000=1430\).

Take moments about \(P\) (anticlockwise positive):

\[80(0.8)-350(1.2)-1000(1.6)+R_Q(2.2)=0\]

\[64-420-1600+2.2R_Q=0\Rightarrow R_Q=\frac{1956}{2.2}=889.09\text{N}\]

Then \(R_P=1430-889.09=540.91\text{N}\).

(b) Replace with \(1200\text{N}\) at distance \(s\) from \(Q\) (position \(3.0-s\)). Keeping the reaction at \(P\) unchanged at \(540.91\text{N}\), take moments about \(Q\):

\[R_P(2.2)=80(3.0)+350(1.0)+1200\,s\]

\[540.91(2.2)=240+350+1200s\Rightarrow 1190=590+1200s\Rightarrow s=0.5\text{m}\]

So the \(1200\text{N}\) weight should be placed \(0.5\text{m}\) from \(Q\) (between \(P\) and \(Q\)).