CAIE A-Level Further Math AS 3.2 Equilibrium of a Rigid Body Questions

Practise locating centres of mass, resolving coplanar forces and taking moments, then applying limiting friction or contact-edge conditions to sliding and toppling.

Exam points

take moments of component areas or volumes to locate the combined centre of mass
resolve forces and set the net moment to zero about a point that removes unknown reactions
at limiting equilibrium, apply friction or move the weight line through the contact edge

Covered syllabus points

3.2.1—Moments 3.2.2—Centre of mass 3.2.3—Centre of mass 3.2.4—Centre of mass 3.2.5—Rigid body equilibrium 3.2.6—Toppling and sliding

[Maximum number: 7]

Figure for Question 1 — CAIE A-Level Further Math AS

A uniform lamina is in the form of an isosceles triangle ABC in which A C=2 a and angle $A B C=90^{\circ}$ . The point D on AB is such that the ratio D B: A B=1: k. The point E on CB is such that DE is parallel to AC. The triangle DBE is removed from the lamina (see diagram).

(a)

Find, in terms of k, the distance of the centre of mass of the remaining lamina ADEC from the midpoint of AC.

When the lamina ADEC is freely suspended from the vertex A, the edge AC makes an angle $\theta$ with the downward vertical, where $\tan \theta=\frac{5}{18}$ .

[ 4 ]

Mass is proportional to area

Figure for Answer to Question (a) — CAIE A-Level Further Math AS

At least two areas correct, at least one distance correct

$\bar{x}\left(-\frac{1}{2} \times 2\left(\frac{a}{k}\right)^{2}+a^{2}\right)=\frac{1}{3} a \times a^{2}-\left(a-\frac{2}{3}\left(\frac{a}{k}\right)\right) \times\left(\frac{a}{k}\right)^{2}$
Moments equation, dimensionally correct, correct number
of terms.

$\bar{x}=\frac{a\left(k^{2}+k-2\right)}{3 k(k+1)}$
Allow unsimplified single fraction $\frac{a\left(k^{3}-3 k+2\right)}{3 k\left(k^{2}-1\right)}$ .

4

(b)

Find the value of k.

Figure for Question (b) — CAIE A-Level Further Math AS

Two smooth vertical walls meet at right angles. The smooth sphere A, with mass m, is at rest on a smooth horizontal surface and is at a distance d from each wall. An identical smooth sphere B is moving on the horizontal surface with speed u at an angle $\theta$ with the line of centres when the spheres collide (see diagram). After the collision, the spheres take the same time to reach a wall. The coefficient of restitution between the spheres is $\frac{1}{2}$ .

[ 3 ]

$\tan \theta=\frac{\bar{x}}{a} \quad\left[=\frac{5}{18}\right]$
With their $\bar{x}$ from part (a).

So $18\left(k^{2}+k-2\right)=5\left(3 k^{2}+3 k\right)$ $k^{2}+k-12=0$
Obtain a polynomial in k only, e.g. $k^{3}-13 k+12=0$ ,
may be implied.

(k+4)(k-3)=0,

k=3 only
CWO

3

[Maximum number: 8]

Figure for Question 2 — CAIE A-Level Further Math AS

A uniform square lamina of side 2 a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle $\theta$ with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is $\frac{1}{2}$ .

Given that the vertex B is about to slip up the wall, find the value of $\tan \theta$ .

Let N be normal reaction at B and F the frictional force acting downwards
$\uparrow T \cos \theta=F+W$ $\rightarrow T \sin \theta=N$
Moments about $B: T a=W \sin \theta \times a+W \cos \theta \times a$
A moments equation with all relevant forces.

$F=\frac{1}{2} N$ used
$(\cos \theta+\sin \theta)\left(\cos \theta-\frac{1}{2} \sin \theta\right)=1$ ое
Combine to obtain equation in $\theta$ .
Equation in trigonometric functions only.

$\begin{aligned} \frac{1}{2} \cos \theta \sin \theta=\frac{3}{2}(\sin \theta)^{2} \sin \theta(\cos \theta-3 \sin \theta)=0 \end{aligned}$
Solve trigonometric equation.

$\tan \theta=\frac{1}{3}$
8

[Maximum number: 8]

Figure for Question 3 — CAIE A-Level Further Math AS

An object is formed from a solid hemisphere, of radius 2 a, and a solid cylinder, of radius a and height d. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line OC forms a diameter of the base of the cylinder, where C is the centre of the plane face of the hemisphere and O is common to both circumferences (see diagram). Relative to axes through O, parallel and perpendicular to OC as shown, the centre of mass of the object is ( $\bar{x}, \bar{y}$ ).

(a)

Show that $\bar{x}=\frac{32 a^{2}+3 a d}{16 a+3 d}$ and find an expression, in terms of a and d, for $\bar{y}$ .

The object is placed on a rough plane which is inclined to the horizontal at an angle $\theta$ where $\sin \theta=\frac{1}{6}$ . The object is in equilibrium with CO horizontal, where CO lies in a vertical plane through a line of greatest slope.

[ 5 ]

[Mass is proportional to volume]

Distance of

Distance of

Volume centre of mass centre of mass from vertical axis from OC

$\left(\frac{2}{3} \pi(2 a)^{3}+\pi a^{2} d\right) \bar{x}=\frac{16}{3} \pi a^{3} \times 2 a+\pi a^{2} d \times a$
Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.

Simplify to $\bar{x}=\frac{32 a^{2}+3 a d}{16 a+3 d}$
AG. At least one line of intermediate working.

$\left(\frac{2}{3} \pi(2 a)^{3}+\pi a^{2} d\right) \bar{y}=\frac{16}{3} \pi a^{3} \times\left(-\frac{3}{8} \times 2 a\right)+\pi a^{2} d \times \frac{1}{2} d$
Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.

$\bar{y}=\frac{3\left(d^{2}-8 a^{2}\right)}{2(16 a+3 d)}$
AEF

5

PUBLISHED

(b)

Find d in terms of a.

[ 3 ]

$\sin \theta=\frac{2 a-\bar{x}}{2 a}$ $\begin{aligned} 2 a \times \frac{1}{6}=2 a-\frac{32 a^{2}+3 a d}{16 a+3 d} \frac{5}{3}(16 a+3 d)=(32 a+3 d) \end{aligned}$
Remove fractions

$d=\frac{8}{3} a$
3

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (b)