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: 1]

A block of mass 8 kg slides down a rough plane inclined at $30^{\circ}$ to the horizontal, starting from rest. The coefficient of friction between the block and the plane is $\mu$ . The block accelerates uniformly down the plane at $2.4 \mathrm{~m} \mathrm{~s}^{-2}$ .

Draw a diagram showing the forces acting on the block.

[Maximum number: 7]

Figure for Question 2 — CAIE A-Level Mathematics A2

Two forces of magnitudes 20 N and $F \mathrm{~N}$ act at a point P in the directions shown in the diagram.

(a)

Given that the resultant force has no component in the y-direction, calculate the value of F.

[ 2 ]

$F=20 \sin 60$
Attempt to resolve in y-direction;
2 terms;
must be $20 \cos 60$ or $20 \sin 60$
and must be linked to F (can be implied by the correct answer seen only).

$=17.3 \mathrm{~N}$
AWRT 17.3 (17.320508…) or $10 \sqrt{3}$ .

(b)

Given instead that F=10, find the magnitude and direction of the resultant force.

[ 5 ]

For resolving in any direction

*M1

Correct number of terms;
allow sin/cos mix;
allow sign errors.

$\begin{aligned} (\text { Horizontal component }=X=R \cos \theta)= \pm(20 \cos 60)[ \pm 10] (\text { Vertical component }=Y=R \sin \theta)= \pm(20 \sin 60-10) {[= \pm 7.3205 \ldots]} \end{aligned}$
For both correct.

$\begin{aligned} \text { Magnitude }=\sqrt{(20 \sin 60-10)^{2}+(20 \cos 60)^{2}} {[=12.393136 \ldots]} \end{aligned}$
OE - correct number of terms.

Angle $=\tan ^{-1}\left(\frac{20 \sin 60-10}{20 \cos 60}\right)[=36.206023 \ldots]$
OE (e.g. reciprocal) - correct number of terms.

Magnitude $=12.4 \mathrm{~N}$
and Direction $=36.2^{\circ}$ above (positive) x-axis
OE for direction e.g. $36.2^{\circ}$ anticlockwise from (positive) x-direction, $36.2^{\circ}$ above the horizontal.

Possibly seen on a diagram.

(Radians: 0.63191... to 3sf or better)

[Maximum number: 6]

Figure for Question 3 — CAIE A-Level Mathematics A2

Four coplanar forces of magnitude $P \mathrm{~N}, 10 \mathrm{~N}, 16 \mathrm{~N}$ and 2 N act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium.

Find the values of $\theta$ and P.

Resolving either direction to get an equation

*M1

Correct number of relevant terms;
allow sign errors;
allow
$\sin / \cos$ mix.

$10 \cos 25=2 \cos 40+16 \sin \theta$ $\left[\begin{array}{c} 9.06307787=1.532088886+16 \sin \theta 7.530988984=16 \sin \theta \sin \theta=0.4706868115 \end{array}\right]$ $\begin{aligned} P=10 \sin 25+16 \cos \theta+2 \sin 40 {\left[\begin{array}{c} P=4.226182617+16 \cos \theta+1.285575219 P=5.511757837+16 \cos \theta \end{array}\right]} \end{aligned}$
This may be with their $\theta$ .

Attempt to solve for $\theta=\sin ^{-1}\left(\frac{10 \cos 25-2 \cos 40}{16}\right)$
From equation(s) with correct number of relevant terms.
Must be a numerical expression for $\theta$ .

Attempt to solve for $P=10 \sin 25+16 \cos ($ their $\theta)+2 \sin 40$
From equation(s) with correct number of relevant terms.
Using their $\theta$ .
Must be a numerical expression for P.

$\theta=28.1$ AND P=19.6
28.07888819 and 19.6285636.
AWRT 28.1 and AWRT 19.6 from correct work.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3