Exam-Style Problem

9231 P21 - Jun 2017 - Q2 - 8 marks

6141

A uniform smooth disc with centre \(O\) and radius \(a\) is fixed at the point \(D\) on a horizontal surface. A uniform rod of length \(3 a\) and weight \(W\) rests on the disc with its end \(A\) in contact with a rough vertical wall. The rod and the disc lie in a vertical plane that is perpendicular to the wall. The wall meets the horizontal surface at the point \(E\) such that \(A E=a\) and \(E D=\frac{5}{4} a\). A particle of weight \(k W\) is hung from the rod at \(B\) (see diagram). The coefficient of friction between the rod and the wall is \(\frac{1}{8}\) and the system is in limiting equilibrium. Find the value of \(k\).

Solution Checked by expert

Answer: \(\boxed{k=2}\) or \(\boxed{k=\dfrac{4}{17}}\), depending on the direction of limiting friction at the wall.

Let \(P\) be the point where the rod touches the disc, and let \(\theta\) be the angle made by the rod with the horizontal.

Since the rod is tangent to the smooth disc, \(OP\) is perpendicular to the rod. From the geometry of triangle \(AOP\),

\[AO=\frac54a,\qquad OP=a.\]

\[AP=\sqrt{\left(\frac54a\right)^2-a^2}=\frac34a.\]

Hence

\[\sin\theta=\frac45,\qquad \cos\theta=\frac35.\]

Let \(R_P\) be the reaction from the disc, and let \(R_A\) be the normal reaction from the wall at \(A\). Resolving horizontally gives

\[R_A=R_P\sin\theta=\frac45R_P.\]

Take friction at \(A\) to be positive downwards and call it \(F_A\). Resolving vertically gives

\[F_A+(k+1)W=R_P\cos\theta=\frac35R_P.\]

Now take moments about \(A\). The reaction \(R_P\) has moment arm \(AP=\frac34a\). The weights of the rod and the particle act vertically, so their perpendicular distances from \(A\) are horizontal distances along the rod:

\[R_P\left(\frac34a\right)-kW(3a\cos\theta)=W\left(\frac32a\cos\theta\right).\]

Using \(\cos\theta=\frac35\), this becomes

\[R_P\left(\frac34a\right)-kW\left(\frac95a\right)=W\left(\frac9{10}a\right).\]

Multiplying by \(20/a\),

\[15R_P-36kW=18W.\]

Because the system is in limiting equilibrium,

\[|F_A|=\frac18R_A=\frac18\cdot\frac45R_P=\frac1{10}R_P.\]

Case 1: friction acts downwards. Then \(F_A=\frac1{10}R_P\). The vertical equation gives

\[\frac1{10}R_P+(k+1)W=\frac35R_P,\]

\[(k+1)W=\frac12R_P.\]

Thus \(R_P=2(k+1)W\). Substitute this into the moment equation:

\[15\{2(k+1)W\}-36kW=18W.\]

\[30k+30-36k=18,\]

which gives

\[k=2.\]

Case 2: friction acts upwards. Then \(F_A=-\frac1{10}R_P\). The vertical equation gives

\[-\frac1{10}R_P+(k+1)W=\frac35R_P,\]

\[(k+1)W=\frac7{10}R_P.\]

Thus \(R_P=\frac{10}{7}(k+1)W\). Substitute this into the moment equation:

\[15\left\{\frac{10}{7}(k+1)W\right\}-36kW=18W.\]

Multiplying by \(7\) gives

\[150(k+1)-252k=126.\]

Hence

\[24=102k,\]

\[k=\frac4{17}.\]

Therefore the limiting-equilibrium values are \(\boxed{k=2}\) and \(\boxed{k=\frac4{17}}\).