(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° r...
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(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° respectively to the downward vertical, find the tension in each of the strings. [Take \(g = 10 ms^{-2}\)].
(b) A constant force F acts on a toy car of mass 5 kg and increases its velocity from 5 ms\(^{-1}\) to 9 ms\(^{-1}\) in 2 seconds. Calculate :
(i) the magnitude of the force ; (ii) velocity of the toy car 3 seconds after attaining a velocity of 9 ms\(^{-1}\).
Answer Details
(a) Weight \(W = mg = 5\times10 = 50\,\text{N}\). Let \(T_1\) (string \(AP\)) and \(T_2\) (string \(BP\)) act at \(40^{o}\) and \(30^{o}\) to the downward vertical on opposite sides.
Horizontal: \(T_1\sin40^{o} = T_2\sin30^{o}\).
Vertical: \(T_1\cos40^{o} + T_2\cos30^{o} = 50\).
Using Lami's theorem is neat here. The angles at \(P\) are: between \(W\) and \(T_1 = 140^{o}\), between \(W\) and \(T_2 = 150^{o}\), between \(T_1\) and \(T_2 = 70^{o}\).
\[\frac{T_1}{\sin150^{o}} = \frac{T_2}{\sin140^{o}} = \frac{50}{\sin70^{o}}\] \[\frac{50}{\sin70^{o}} = \frac{50}{0.9397} = 53.21\] \[T_1 = 53.21\times\sin150^{o} = 53.21\times0.5 = 26.6\,\text{N}\] \[T_2 = 53.21\times\sin140^{o} = 53.21\times0.6428 = 34.2\,\text{N}\]Tension in \(AP \approx \mathbf{26.6\,\text{N}}\); tension in \(BP \approx \mathbf{34.2\,\text{N}}\).
(b)(i) \(a = \dfrac{9 - 5}{2} = 2\,\text{ms}^{-2}\), so \(F = ma = 5\times2 = \mathbf{10\,\text{N}}\).
(ii) Continuing at \(a = 2\,\text{ms}^{-2}\) for a further \(3\,\text{s}\) from \(9\,\text{ms}^{-1}\):
\[v = 9 + 2\times3 = \mathbf{15\,\text{ms}^{-1}}\]Download The App On Google Playstore
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