Exam-Style Problem

9231 P21 - Nov 2017 - Q4 - 10 marks

6167

A small ring \(P\) of weight \(W\) is free to slide on a rough horizontal wire, one end of which is attached to a vertical wall at \(Q\). The end \(A\) of a thin uniform \(\operatorname{rod} A B\) of length \(2 a\) and weight \(\frac{5}{2} W\) is freely hinged to the wall at the point \(A\) which is a distance \(a\) vertically below \(Q\). A light elastic string of natural length \(2 a\) has one end attached to the ring \(P\) and the other end attached to the rod at \(B\). The string is at right angles to the rod and \(A, B, P\) and \(Q\) lie in a vertical plane. The system is in limiting equilibrium with \(A B\) making an angle \(\theta\) with the horizontal, where \(\sin \theta=\frac{3}{5}\) (see diagram).
(i) Find the tension in the string in terms of \(W\).

(ii) Find the coefficient of friction between the ring and the wire.

(iii) Find the magnitude of the resultant force on the rod at the hinge in terms of \(W\).

(iv) Find the modulus of elasticity of the string in terms of \(W\).

Solution Checked by expert

Answer: \(\boxed{T=W}\), \(\boxed{\mu=\dfrac13}\), \(\boxed{R_A=\dfrac{\sqrt{13}}2W}\), and \(\boxed{\lambda=\dfrac83W}\).

Since \(\sin\theta=\frac35\), we have

\[\cos\theta=\frac45.\]

(i) Take moments about the hinge \(A\) for the rod.

The weight of the rod is \(\frac52W\) and acts at the midpoint of the rod, a distance \(a\) from \(A\). Its perpendicular moment arm is \(a\cos\theta\).

The tension \(T\) acts at \(B\) at right angles to the rod, so its moment about \(A\) is \(T(2a)\).

Hence

\[T(2a)=\frac52W(a\cos\theta).\]

Substitute \(\cos\theta=\frac45\):

\[2aT=\frac52W\cdot a\cdot\frac45=2aW.\]

Therefore

\[\boxed{T=W}.\]

(ii) Now consider the ring \(P\). Since the system is in limiting equilibrium, the friction is \(F=\mu R\).

The horizontal component of the tension is

\[T\sin\theta=W\cdot\frac35=\frac35W.\]

The normal reaction from the wire is the weight of the ring plus the vertical component of the tension:

\[R=W+T\cos\theta=W+W\cdot\frac45=\frac95W.\]

Therefore

\[\mu R=\frac35W.\]

\[\mu\cdot\frac95W=\frac35W.\]

Cancel \(W\):

\[\mu=\frac{3/5}{9/5}=\frac13.\]

Thus

\[\boxed{\mu=\frac13}.\]

(iii) For the rod, resolve the force at the hinge into horizontal and vertical components.

The horizontal component balances the horizontal component of the tension:

\[X=T\sin\theta=W\cdot\frac35=\frac35W.\]

The vertical component balances the weight of the rod after allowing for the vertical component of tension:

\[Y=\frac52W-T\cos\theta=\frac52W-W\cdot\frac45.\]

\[Y=\frac{25}{10}W-\frac{8}{10}W=\frac{17}{10}W.\]

The resultant force at the hinge has magnitude

\[R_A=\sqrt{X^2+Y^2}.\]

Hence

\[R_A=\sqrt{\left(\frac35W\right)^2+\left(\frac{17}{10}W\right)^2}.\]

Therefore

\[R_A=W\sqrt{\frac{9}{25}+\frac{289}{100}}=W\sqrt{\frac{36+289}{100}}.\]

\[R_A=\frac{\sqrt{325}}{10}W=\frac{5\sqrt{13}}{10}W=\boxed{\frac{\sqrt{13}}{2}W}.\]

(iv) The string has natural length \(2a\).

From the geometry of the diagram, the length of the string is

\[PB=\frac{a+2a\sin\theta}{\cos\theta}.\]

Using \(\sin\theta=\frac35\) and \(\cos\theta=\frac45\),

\[PB=\frac{a+2a(3/5)}{4/5}=\frac{a+6a/5}{4/5}=\frac{11a/5}{4/5}=\frac{11a}{4}.\]

So the extension is

\[x=\frac{11a}{4}-2a=\frac{3a}{4}.\]

By Hooke's law,

\[T=\frac{\lambda x}{2a}.\]

Since \(T=W\),

\[W=\frac{\lambda(3a/4)}{2a}=\frac{3\lambda}{8}.\]

Therefore

\[\lambda=\boxed{\frac83W}.\]