June 2022 Paper 1 Q5

OCR MEICurrent spec5 marksResolving Forces

5 A sphere of mass 3 kg hangs on a string. A horizontal force of magnitude \(F\) N acts on the sphere so that it hangs in equilibrium with the string making an angle of \(25^\circ\) to the vertical. The force diagram for the sphere is shown below.

Force diagram: the sphere with Weight acting vertically down, a horizontal force F N to the left, and Tension acting up and to the right along the string at 25° to the vertical

(a) Sketch the triangle of forces for these forces. [2]

(b) Hence or otherwise determine each of the following:

the tension in the string
the value of \(F\).

[3]

Two acceptable triangles of forces: Weight (vertical), F N (horizontal) and Tension forming a closed loop, with the 25° angle between Weight and Tension — Mark scheme (a)

Scheme	Marks	AO
Either of these is acceptable	B1 B1	1.1b 1.1b
	[2]

Mark scheme (b)

Scheme	Marks	AO
Using the triangle of forces \(F = 3g\tan 25^\circ\) or Tension \(= \dfrac{3g}{\cos 25^\circ}\)	M1	1.1a
\(F = 13.7\)	A1	1.1b
Tension \(= 32.4\) N	A1	1.1b
	[3]

Notes

M1: Allow sin/cos or \(25^\circ\)/\(65^\circ\) interchange to find \(F\) or \(T\)

A1: cao

Alternative method

Scheme	Marks
Resolve vertically \(T\cos 25^\circ = 3g\)	M1
\(T = 32.4\) N	A1
Resolve horizontally \(F = T\sin 25^\circ = 13.7\)	A1

M1: Allow sin/cos interchange or \(25^\circ\)/\(65^\circ\) interchange

A1: cao