(a) A body of mass 5 kg is placed on a smooth plane inclined at an angle 30° to the horizontal. Find the magnitude of the force: (i) acting parallel to the ...

Answer Details

(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).

(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).

(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).

(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).

Take moments about A (clockwise = anticlockwise):

\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\]

(ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).

\[R_A=210-100=\textbf{110 N}.\]