Exam-Style Problem

9231 P23 - Jun 2018 - Q4 - 10 marks

6191

A uniform \(\operatorname{rod} A B\) has length \(2 a\) and weight \(W\). The end \(A\) rests on rough horizontal ground and the end \(B\) rests against a smooth vertical wall. The angle between the rod and the horizontal is \(\theta\), where \(\tan \theta=\frac{4}{3}\). One end of a light inextensible rope is attached to a point \(C\) on the rod. The other end is attached to a point where the vertical wall and the horizontal ground meet. The rope is taut and perpendicular to the rod. The rope and rod are in a vertical plane perpendicular to the wall.
(i) Show that \(A C=\frac{18}{25} a\).

The magnitude of the frictional force at \(A\) is equal to one quarter of the magnitude of the normal reaction force at \(A\).
(ii) Show that the tension in the rope is \(\frac{1}{4} W\).
(iii) Find expressions, in terms of \(W\), for the magnitudes of the normal reaction forces at \(A\) and \(B\).

Solution Checked by expert

Answer: \(AC=\dfrac{18}{25}a\), the tension is \(\dfrac14W\), and the normal reactions are \(\boxed{R_A=\dfrac{23}{20}W}\) and \(\boxed{R_B=\dfrac{39}{80}W}\).

Since \(\tan\theta=\frac43\), use the \(3\)-\(4\)-\(5\) triangle:

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

(i) Let \(D\) be the corner where the wall meets the ground. Since the rope is perpendicular to the rod, triangle \(ACD\) gives

\[AC=AD\cos\theta.\]

The horizontal distance from \(A\) to the wall is

\[AD=2a\cos\theta.\]

Therefore

\[AC=2a\cos^2\theta=2a\left(\frac35\right)^2=\frac{18}{25}a,\]

as required.

(ii) Let the normal reactions at \(A\) and \(B\) be \(R_A\) and \(R_B\), and let the frictional force at \(A\) be \(F_A\). We are given

\[F_A=\frac14R_A.\]

Resolve forces vertically. The rope pulls downwards with vertical component \(T\cos\theta\), so

\[R_A-W=T\cos\theta=\frac35T.\]

Hence

\[R_A=W+\frac35T.\]

Resolve forces horizontally:

\[R_B-F_A=T\sin\theta=\frac45T.\]

Thus

\[R_B=F_A+\frac45T=\frac14R_A+\frac45T.\]

Substitute \(R_A=W+\frac35T\):

\[R_B=\frac14\left(W+\frac35T\right)+\frac45T=\frac14W+\frac{19}{20}T.\]

Now take moments about \(A\). The reaction and friction at \(A\) have no moment about \(A\). Hence

\[R_B(2a\sin\theta)=W(a\cos\theta)+T(AC).\]

Substitute \(\sin\theta=\frac45\), \(\cos\theta=\frac35\), and \(AC=\frac{18}{25}a\):

\[R_B\cdot\frac{8a}{5}=W\cdot\frac{3a}{5}+T\cdot\frac{18a}{25}.\]

Cancel \(a\) and multiply by \(25\):

\[40R_B=15W+18T.\]

Use \(R_B=\frac14W+\frac{19}{20}T\):

\[40\left(\frac14W+\frac{19}{20}T\right)=15W+18T.\]

\[10W+38T=15W+18T.\]

Therefore

\[20T=5W,\qquad T=\frac14W,\]

as required.

(iii) From \(R_A=W+\frac35T\),

\[R_A=W+\frac35\cdot\frac14W=\boxed{\frac{23}{20}W}.\]

Then

\[F_A=\frac14R_A=\frac{23}{80}W.\]

Using \(R_B-F_A=\frac45T\),

\[R_B=\frac{23}{80}W+\frac45\cdot\frac14W=\frac{23}{80}W+\frac{16}{80}W=\boxed{\frac{39}{80}W}.\]