M1 June 2006 Q5

EdexcelOld spec13 marksMoments

Figure 3: horizontal girder AB held by vertical cables at A and C, AC = 90 cm — Figure 3

A steel girder \(AB\) has weight 210 N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end \(A\). The other cable is attached to the point \(C\) on the girder, where \(AC = 90\) cm, as shown in Figure 3. The girder is modelled as a uniform rod, and the cables as light inextensible strings.

Given that the tension in the cable at \(C\) is twice the tension in the cable at \(A\), find

(a) the tension in the cable at \(A\), (2)

(b) show that \(AB = 120\) cm. (4)

A small load of weight \(W\) newtons is attached to the girder at \(B\). The load is modelled as a particle. The girder remains in equilibrium in a horizontal position. The tension in the cable at \(C\) is now three times the tension in the cable at \(A\).

Diagram: girder with tensions R at A and 2R at C, weight 210 at distance d from A — Mark scheme (a)

Scheme	Marks
\(R + 2R = 210 \;\Rightarrow\; R = 70\) N	M1 A1
	(2)

Mark scheme (b)

Scheme	Marks
e.g. M(\(A\)): \(140 \times 90 = 210 \times d\)	M1 A1ft
\(\Rightarrow d = 60 \;\Rightarrow\; AB = 120\) cm	M1 A1
	(4)

Notes

(b) 1st M1 for a valid moments equation

2nd M1 for complete solution to find \(AB\) (or verification)

Allow ‘verification’, e.g. showing \(140 \times 90 = 210 \times 60\) M1 A1
\(1260 = 1260\) QED M1 A1

Diagram: girder with tensions S at A and 3S at C, weight 210 and load W at B — Mark scheme (c)

Scheme	Marks
\(4S = 210 + W\)	M1 A1
e.g. M(\(B\)): \(S \times 120 + 3S \times 30 = 210 \times 60\)	M1 A2,1,0
Solve \(\rightarrow\) (\(S = 60\) and) \(W = 30\)	M1 A1
	(7)
	(13 marks)