CAIE A-Level Mathematics A2 4.1 Forces and Equilibrium Questions

Practise drawing force diagrams, resolving coplanar forces and applying equilibrium or limiting-friction equations to particles, strings, pulleys and inclined planes.

Exam points

draw every force with its correct direction before resolving along chosen axes
resolve horizontally and vertically or parallel and perpendicular to an inclined plane
apply resultant zero and F = μR at limiting equilibrium to solve tensions, angles or masses

Covered syllabus points

4.1.1—Forces 4.1.2—The vector nature of force 4.1.3—Forces 4.1.4—Friction 4.1.5—The model of a 'smooth' contact 4.1.6—Equilibrium 4.1.7—Newton's laws

[Maximum number: 6]

A particle of mass 12 kg is stationary on a rough plane inclined at an angle of $25^{\circ}$ to the horizontal. A force of magnitude $P \mathrm{~N}$ acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35 .

(a)

Draw a sketch showing the forces acting on the particle.

[ 1 ]

The force diagram should show the weight vertically downward, the normal reaction perpendicular to the plane, the force P parallel to the line of greatest slope, and the friction force F along the plane opposing the tendency to slide. The angle of inclination is $25^\circ$ .

Figure for Answer to Question (a) — CAIE A-Level Mathematics A2

(b)

Find the least possible value of P.

[ 5 ]

Resolving parallel to the plane, with the friction opposing the tendency to slide down,
$P+F=12g\sin25^\circ$ .
The normal reaction is
$R=12g\cos25^\circ$ ,
and the limiting friction is F=0.35R. Using $g=10\text{ m s}^{-2}$ ,
$F=0.35(120\cos25^\circ)=38.1\text{ N}$ .
Therefore,
$P=120\sin25^\circ-38.1=12.6\text{ N}$ .

[Maximum number: 8]

Figure for Question 2 — CAIE A-Level Mathematics A2

Coplanar forces, of magnitudes $F \mathrm{~N}, 3 \mathrm{~N}, 6 \mathrm{~N}$ and 4 N , act at a point P, as shown in the diagram.

(a)

Given that $\alpha=60$ , and that the resultant of the four forces is in the direction of the 3 N force, find F.

[ 3 ]

Resolve perpendicular to the direction of the 3 N force:
$4\sin30^\circ+F\sin60^\circ-6=0.$ $F=4.62\text{ N}.$
Accept
$F=\frac{8}{\sqrt3}=\frac{8\sqrt3}{3}.$

(b)

Given instead that the four forces are in equilibrium, find the values of F and $\alpha$ .

[ 5 ]

Resolve vertically and horizontally:
$F\sin\alpha+4\sin30^\circ-6=0,$
and
$F\cos\alpha+3-4\cos30^\circ=0.$
Thus
$F\sin\alpha=4,$
and
$F\cos\alpha=0.464102\ldots.$

Then
$F^2=4^2+0.464102^2,$
so
$F=4.03\text{ N}.$
Also
$\alpha=\tan^{-1}\left(\frac4{0.464102\ldots}\right)=83.4^\circ.$

[Maximum number: 3]

Figure for Question 3 — CAIE A-Level Mathematics A2

Two particles A and B, of masses $3 m \mathrm{~kg}$ and $2 m \mathrm{~kg}$ respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a plane. The plane is inclined at an angle $\theta$ to the horizontal. A lies on the plane and B hangs vertically, 0.8 m above the floor, which is horizontal. The string between A and the pulley is parallel to a line of greatest slope of the plane (see diagram). Initially A and B are at rest.

Given that the plane is smooth, find the value of $\theta$ for which A remains at rest.

It is given instead that the plane is rough, $\theta=30^{\circ}$ and the acceleration of A up the plane is $0.1 \mathrm{~m} \mathrm{~s}^{-2}$ .

T-2 m g=0
3 m g \sin \theta-T=0
M1 for resolving forces parallel to the plane and solving for $\theta$ $\theta=41.8(41.810 \ldots)$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3