(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° r...
Question 1 Report
(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° respectively to the horizontal and the tension in AB is 48N. If the system is in equilibrium :
(i) sketch a diagram to represent the information ; (ii) calculate the tension in DB ;
Answer Details
(a)(i) Diagram
The body \(P\) hangs from the junction \(B\). String \(BA\) rises to the left at \(30^{\circ}\) to the horizontal and carries the tension \(T_{AB}=48\,\text{N}\); string \(BD\) rises to the right at \(60^{\circ}\) to the horizontal and carries the tension \(T_{DB}\); the weight \(W=qg\) acts vertically downward. The three forces meet at \(B\).
(a)(ii) Tension in DB
The three concurrent forces at \(B\) are \(T_{AB}=48\,\text{N}\), \(T_{DB}\) and the weight \(W\). The angles between successive forces are:
| Angle between | Size | Opposite force |
|---|---|---|
| \(T_{AB}\) and \(T_{DB}\) | \(90^{\circ}\) | \(W\) |
| \(T_{DB}\) and \(W\) | \(150^{\circ}\) | \(T_{AB}\) |
| \(W\) and \(T_{AB}\) | \(120^{\circ}\) | \(T_{DB}\) |
Applying Lami's theorem at \(B\):
\[\frac{T_{AB}}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}=\frac{W}{\sin 90^{\circ}}\]From the first two ratios:
\[\frac{48}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}\]\[T_{DB}=\frac{48\sin 120^{\circ}}{\sin 150^{\circ}}=\frac{48\left(\tfrac{\sqrt3}{2}\right)}{\tfrac12}=\frac{48(0.8660)}{0.5}=83.14\,\text{N}\]The tension in DB is \(83.14\,\text{N}\).
(a)(iii) Mass \(q\) of the body
Taking \(g=10\,\text{m s}^{-2}\), the weight is \(W=10q\). Using the first and third ratios of Lami's theorem:
\[\frac{10q}{\sin 90^{\circ}}=\frac{48}{\sin 150^{\circ}}\]\[10q=\frac{48\sin 90^{\circ}}{\sin 150^{\circ}}=\frac{48(1)}{0.5}=96\]\[q=\frac{96}{10}=9.6\,\text{kg}\]The mass of the body is \(9.6\,\text{kg}\).
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