M2 June 2005 Q6

EdexcelOld spec12 marksMoments Resolving Forces

Figure 2: horizontal pole AB of length 3 m hinged at A to a wall, rod CD from C on the wall below A to D on the pole, AD = 0.5 m, angle ACD = 30 degrees — Figure 2

A uniform pole \(AB\), of mass 30 kg and length 3 m, is smoothly hinged to a vertical wall at one end \(A\). The pole is held in equilibrium in a horizontal position by a light rod \(CD\). One end \(C\) of the rod is fixed to the wall vertically below \(A\). The other end \(D\) is freely jointed to the pole so that \(\angle ACD = 30^\circ\) and \(AD = 0.5\) m, as shown in Figure 2. Find

(a) the thrust in the rod \(CD\), (4)

(b) the magnitude of the force exerted by the wall on the pole at \(A\). (6)

The rod \(CD\) is removed and replaced by a longer light rod \(CM\), where \(M\) is the mid-point of \(AB\). The rod is freely jointed to the pole at \(M\). The pole \(AB\) remains in equilibrium in a horizontal position.

Scheme	Marks
\(M(A)\):
\(P \times 0.5\sin 60 = 30g \times 1.5\)	M1 A2
\(P = 90g.\tfrac{2}{\sqrt{3}} \approx 1020\) N (1000N)	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\rightarrow\) \(X = P\cos 60 = \tfrac{1}{2}P\)	M1 A1
\((\approx 509\ \text{N}\ \ (510\text{N}))\)
\(\uparrow\) \(Y + P\cos 30 = 30g\)	M1 A1
\((\Rightarrow Y = -588\ \text{N})\)
resultant \(= \sqrt{(X^2 + Y^2)} = \sqrt{(509^2 + 588^2)} \approx 778\) N or 780N	M1 A1
	(6)

Mark scheme (c)

Scheme	Marks
In equilibrium all forces act through a point	M1
\(P\) and weight meet at mid-point; hence reaction also acts through mid-point so reaction horizontal	A1 cso
	(2)
	(12 marks)

OR

\(M(\text{mid-point})\): \(Y \times 1.5 = 0\ \Rightarrow\ Y = 0\)	M1
Hence reaction is horizontal	A1