(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward ...
Question 1 Report
(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward vertical.
(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.
Answer Details
(a) Weight \(W=6.5\times 10=65\ \text{N}\). Let string \(AP\) (\(50^\circ\) to the downward vertical) carry tension \(T_1\) up to the left and \(BP\) (\(60^\circ\)) carry \(T_2\) up to the right.
(i) Component form:
\[\vec{T_1}=\begin{pmatrix}-T_1\sin 50^\circ\\ T_1\cos 50^\circ\end{pmatrix},\quad \vec{T_2}=\begin{pmatrix}T_2\sin 60^\circ\\ T_2\cos 60^\circ\end{pmatrix},\quad \vec{W}=\begin{pmatrix}0\\-65\end{pmatrix}.\](ii) Equilibrium of P: \(\vec{T_1}+\vec{T_2}+\vec{W}=\vec{0}\).
(iii) Horizontal: \(T_1\sin 50^\circ=T_2\sin 60^\circ\). Using Lami's theorem (angles \(110^\circ,120^\circ,130^\circ\)):
\[T_1=\frac{65\sin 120^\circ}{\sin 110^\circ}=\frac{65(0.8660)}{0.9397}=59.9\ \text{N},\quad T_2=\frac{65\sin 130^\circ}{\sin 110^\circ}=\frac{65(0.7660)}{0.9397}=53.0\ \text{N}.\]Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).
(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).
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