AS June 2019 Q1

EdexcelCurrent spec9 marksCentres of Mass

Figure 1: rhombus framework ABCD with AB along the top and DC along the bottom, and a diagonal rod AC — Figure 1

Five identical uniform rods are joined together to form the rigid framework \(ABCD\) shown in Figure 1. Each rod has weight \(W\) and length \(4a\). The points \(A\), \(B\), \(C\) and \(D\) all lie in the same plane.

The centre of mass of the framework is at the point \(G\).

(a) Explain why \(G\) is the midpoint of \(AC\). (1)

The framework is suspended from the ceiling by two vertical light inextensible strings. One string is attached to the framework at \(A\) and the other string is attached to the framework at \(B\). The framework hangs freely in equilibrium with \(AB\) horizontal.

(b) Find

(i) the tension in the string attached at \(A\),

(ii) the tension in the string attached at \(B\).

(4)

A particle of weight \(kW\) is now attached to the framework at \(D\) and a particle of weight \(2kW\) is now attached to the framework at \(C\). The framework remains in equilibrium with \(AB\) horizontal and the strings vertical.

Either string will break if the tension in it exceeds \(6W\).

Mark scheme (a)

Scheme	Marks	AO
The rods are uniform and the axes of symmetry intersect at midpoint of \(AC\).	B1	2.4
	(1)

Notes

B1: Any equivalent clear justification. Needs to mention uniformity and symmetry and the midpoint of \(AC\)

Mark scheme (b)

Scheme	Marks	AO
Use moments: e.g. M(\(A\)): \(\left(2aW + aW + 3aW = 4aT_B + aW\right)\)	M1	2.1
e.g. M(\(A\)): \(5W.2a\cos 60^\circ = 4aT_B\) or M(\(B\)): \(3a \times 5W = 4aT_A\)	A1	1.1b
Resolving vertically: \(T_A + T_B = 5W\)	M1	2.1
\(\Rightarrow T_A = \dfrac{15W}{4}\), \(T_B = \dfrac{5W}{4}\)	A1	1.1b
	(4)

Notes

M1: Form ANY moments equation. Require all terms. Dimensionally correct. Condone sign errors.

A1: Correct unsimplified (including trig) equation
e.g. M(\(G\)): \(T_A.2a\cos 60^\circ = T_B.(4a - 2a\cos 60^\circ)\) or \(T_B.(4a\cos^2 30^\circ)\)

M1: Form a second equation in \(T_A\) and/or \(T_B\) e.g. by resolving vertically or a second moments equation, and solve for \(T_A\) and \(T_B\)

A1: Both tensions correct. If answers reversed, allow M marks.

Mark scheme (c)

Scheme	Marks	AO
\(T_A\) will be the larger, so the first to exceed \(6W\) so need to use \(T_A = 6W\) (e.g. by M(\(B\)) but they may use two equations) to form an equation in \(k\) only.	M1	3.1a
\(6W \times 4a = 5W \times 3a + kW \times 6a + 2kW \times 2a\)	A1	1.1b
\(24aW = 15aW + 10kaW\)	A1	1.1b
\(k = 0.9\)	A1	1.1b
	(4)
	(9 marks)

Notes

M1: Realise that the first to break will be the rope at \(A\) and complete method to form an equation in \(k\) only (allow uncancelled \(W\)’s) using \(T_A = 6W\). Require all terms (in all equations used). Dimensionally correct. Condone sign errors.
M0 if they use \(T_B = 6W\) to find \(k\) (this gives \(k = 9.5\))

A1: Unsimplified equation or inequality in \(k\) only with at most one error

A1: Correct unsimplified equation or inequality in \(k\) only

A1: Correct only. Decimal or fraction.