M1 June 2014 (R) Q1

Figure 1: particle P hanging from O on a string at 25 degrees to the downward vertical, with a horizontal force of 5 N — Figure 1

A particle \(P\) of weight \(W\) newtons is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(O\). A horizontal force of magnitude 5 N is applied to \(P\). The particle \(P\) is in equilibrium with the string taut and with \(OP\) making an angle of 25\(^\circ\) to the downward vertical, as shown in Figure 1.

Find

(a) the tension in the string, (3)

(b) the value of \(W\). (3)

Mark scheme (a)

Scheme	Marks
Resolving horizontally: \(5 = T\cos 65^\circ\)	M1A1
\(T =\) 12, 11.8, or better (N)	A1
	(3)

Notes

First M1 for resolving horizontally with correct no. of terms and \(T\) term resolved.

First A1 for a correct equation in \(T\) only.

Second A1 for 12 (N) or 11.8 (N) or better.

N.B. The M1 is for a complete method to find the tension so where two resolution equations, neither horizontal, are used, the usual criteria for an M mark must be applied to both equations and the first A1 is for a correct equation in \(T\) only (i.e. \(W\) eliminated correctly)

Alternatives

Lami’s Theorem: \(\dfrac{T}{\sin 90^\circ} = \dfrac{5}{\sin 155^\circ}\) (same equation as → resolution) M1A1

Mark scheme (b)

Scheme	Marks
Resolving vertically: \(W = T\cos 25^\circ\)	M1A1
\(= 11.8\cos 25^\circ =\) 11, 10.7 or better (N)	A1
	(3)
	(6 marks)

Notes

First M1 for resolving vertically with correct no. of terms and \(T\) (does not need to be substituted) term resolved.

First A1 for a correct equation in \(T\) only.

Second A1 for 11 (N), 10.7 (N) or better.

Alternatives

Triangle of forces: \(W = 5\tan 65^\circ\) M1A1

Lami’s Theorem: \(\dfrac{T}{\sin 90^\circ} = \dfrac{W}{\sin 115^\circ}\) M1A1

Or Resolution in another direction e.g. along the string M1 (usual criteria) A1 for a correct equation.