(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings \(\overrightarrow{XP}\) and \(\overrightarrow{YP}\). The strings are in...
Question 1 Report
(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings \(\overrightarrow{XP}\) and \(\overrightarrow{YP}\). The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensionsin the strings. [Take g = \(10 ms^{-2}\)].
(b) The height h metres, of a ball thrown into the air is \(2 + 20t + kt^{2}\), after t seconds. If it takes 2 seconds for the ball to reach its highest point, find
(i) the value of k (ii) its highest point from the point of throw.
Answer Details
(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:
\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).
(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):
\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\](ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.
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