A2 June 2022 Q3

EdexcelCurrent spec7 marksCentres of Mass

Figure 1: framework: A above B above C on a vertical line, with AB = BC = 3a; C, D and E on a horizontal line with CD = DE = 4a; F above D with DF = 3a and BF = 4a; AF = FE = 5a along the straight line AE, and CF = 5a — Figure 1

Nine uniform rods are joined together to form the rigid framework \(ABCDEFA\), with \(AB = BC = DF = 3a\), \(BF = CD = DE = 4a\) and \(AF = FE = CF = 5a\), as shown in Figure 1. All nine rods lie in the same plane.

The mass per unit length of each of the rods \(BF\), \(CF\) and \(DF\) is twice the mass per unit length of each of the other six rods.

(a) Find the distance of the centre of mass of the framework from \(AC\) (4)

The mass of the framework is \(M\). A particle of mass \(kM\) is attached to the framework at \(E\) to form a loaded framework.

When the loaded framework is freely suspended from \(F\), it hangs in equilibrium with \(CE\) horizontal.

(b) Find the exact value of \(k\) (3)

Mark scheme (a)

Scheme

Marks

Moments about \(AC\):

rod	\(CD\)	\(DE\)	\(EF\)	\(FA\)	\(AB\)	\(BC\)	\(BF\)	\(DF\)	\(CF\)
Mass ratio	4	4	5	5	3	3	8	6	10
From \(AC\)	\(2a\)	\(6a\)	\(6a\)	\(2a\)	0	0	\(2a\)	\(4a\)	\(2a\)

3.1a

\(8a\times 4a + 2\times 3a\times 4a + 2\times 5a\times 2a + 10a\times 4a + 2\times 4a\times 2a = 48a\bar{x}\)

A1
A1

1.1b
1.1b

\((132a = 48\bar{x}) \Rightarrow \bar{x} = \dfrac{11}{4}a\)

1.1b

(4)

Notes

M1: Dimensionally correct equation for moments about \(AC\) or a parallel axis. All terms needed and horizontal distances
Must be using the mass ratio. Allow slips, but not consistently density and not consistently lengths. Condone without \(a\)

A1: One side of the equation correct

A1: Both sides of the equation correct

A1: Or equivalent single term
Condone if \(a\) is missing in the working and appears at the end

Mark scheme (b)

Scheme	Marks	AO
Moments about \(F\):	M1	3.1a
\(Mg(4a - \bar{x}) = kMg\times 4a\)	A1ft	1.1b
\(\Rightarrow k = \dfrac{5}{16}\)	A1	1.1b
	(3)
	(7 marks)

Notes

M1: Dimensionally correct moments equation. Accept any complete alternative method using \(M\) and \(kM\) to obtain an equation in \(k\) only.
Condone if g and / or M cancelled throughout
Condone incorrect distances
Condone if use \(M = 48\) throughout

A1ft: Correct unsimplified equation (accept without \(g\) and/or \(M\)) Correct mass and distance combination for their \(\bar{x}\)

A1: Or 0.3125 Condone 0.31 or 0.313

Alternative

Scheme	Marks	AO
Moments about \(C\):	M1
\(4a(M + kM) = M\bar{x} + 8akM\)	A1ft
\(\Rightarrow k = \dfrac{5}{16}\)	A1	1.1b
	(3)