# CIE A level Math -Mechanics :Topic : 4.1 Forces and equilibrium- equilibrium: Exam Style Questions Paper 4

### Question A small ring P is threaded on a fixed smooth horizontal rod AB. Three horizontal forces of magnitudes
4.5 N, 7.5 N and F N act on P (see diagram).
(i) Given that these three forces are in equilibrium, find the values of F and 1.

(ii) It is given instead that the values of F and 1 are 9.5 and 30 respectively, and the acceleration of
the ring is 1.5$$ms^{-2}$$. Find the mass of the ring.

(i) Resolve forces either horizontally or vertically

7.5cos60+ 4.5cos20=Fcosθ [= 7.97861]

7.5sin60– 4.5sin20=Fsinθ [= 4.95609]  $$f=\sqrt{(7.98^{2}+4.96^{2})}$$ $$\Theta =tan^{-1}(\frac{4.96}{7.98})$$

F= 9.39 and θ = 31.8

Alternative method for question (i)

$$\frac{F}{sin80}=\frac{4.5}{sin(120+\Theta )}=\frac{7.5}{sin(160-\Theta )}$$

[4.5sin(160–θ) = 7.5sin(120+θ)]

Use the
θ value found by valid trigonometry to find F
= 9.39 and θ = 31.8

(i) Alternative method for question (i)

Forces 4.5, 7.5, F opposite angles 60 –θ, θ + 20, 100

$$[f^{2}=4.5^{2}+7.5^{2}-2\times 4.5\times 7.5\times cos100]$$

$$\left [ \frac{9.39}{sin100} =\frac{4.5}{sin(60-\Theta)}=\frac{7.5}{sin(\Theta +20)}\right ]$$

F= 9.39 and θ = 31.8

(ii)$$9.5\cos 30^{\circ}-7.5\cos 60^{\circ}-4.5\cos 20^{\circ}=m\times 1.51$$
$$8.227-3.75-4.23=m\times 1.51$$
m=0.166kg

### Question The diagram shows three coplanar forces acting at the point O. The magnitudes of the forces are 6 N, 8 N and 10 N. The angle between the 6 N force and the 8 N force is 90Å. The forces are in equilibrium. Find the other angles between the forces.

Using Lami’s Theorem

$$\Rightarrow \frac{10}{\sin 90^{\circ}}=\frac{6}{\sin (90^{\circ}+\alpha )}=\frac{8}{\sin(90+\beta ) }$$
$$\Rightarrow \frac{10}{\sin 90^{\circ}}=\frac{6}{\cos\alpha }=\frac{8}{\cos \beta }$$
$$\Rightarrow 10=\frac{6}{\cos\alpha }$$ and $$10=\frac{8}{\cos \beta }$$
$$\Rightarrow \cos \alpha =\frac{6}{10}$$ and $$\cos \beta =\frac{8}{10}$$
$$\Rightarrow \alpha =53.13^{\circ}$$ and $$\Rightarrow \beta =36.86^{\circ}$$

Angle between 8N and 10N is $$\alpha =143.1$$
Angle between 6N and 10N is $$\beta =126.9$$

### Question Coplanar forces, of magnitudes F N, 3F N, G N and 50 N, act at a point P, as shown in the diagram.
(i) Given that F = 0, G = 75 and$$\alpha= 60^{\circ}$$, find the magnitude and direction of the resultant force.

(ii) Given instead that G = 0 and the forces are in equilibrium, find the values of F and$$\alpha$$.

(i)$$\Rightarrow \sum F_{x}=75+50 \cos 60^{\circ}$$
$$=75+50\times \frac{1}{2}$$
=100N
$$\Rightarrow \sum F_{x}=50\sin 60^{\circ}$$
=43.30N
$$\Rightarrow F_{R}=\sqrt{F_{x}^{2}+F_{y}^{2}}$$
$$\Rightarrow F_{R}=\sqrt{100^{2}+43.30^{2}}$$
$$\Rightarrow F_{R}=109^{\circ}$$
Angle of Resultant Vector,say $$\alpha$$
Then,$$\tan \alpha =\frac{F_{y}}{F_{x}}$$
$$\tan \alpha =\frac{43.3}{100}$$
$$\tan \alpha =0.433$$
$$\alpha =\tan^{-1}(0.433)$$
$$\alpha =23.41^{\circ}$$

(ii)Forces are in equilibrium Hence,forces along x-direction and y-direction will be zero.
Now,
\sum F_{x}=0
50\cos \alpha -F\cos 50^{\circ}=0
50\cos \alpha =F\cos 50^{\circ}
\cos \alpha=\frac{F\cos 50^{\circ}}{50}
50\sin \alpha-3F-F\sin50^{\circ}= 0
50\sin \alpha=3F +F\sin50^{\circ}
\sin \alpha=\frac{3F +F\sin50^{\circ}}{50}
$$tan\alpha =\frac{3F+Fsin50^{\circ}}{Fcos50^{\circ}}$$
\tan \alpha =\frac{3R+0.766F}{0.642F}
\tan \alpha =\frac{3.766F}{0.642F}
$$\alpha =\tan^{-1}80.3$$
$$\alpha =80.31^{\circ}$$

### (a) Question Four coplanar forces act at a point. The magnitudes of the forces are 10 N, F N, GN and 2F N. The directions of the forces are as shown in the diagram.

Given that the forces are in equilibrium, find the values of F and G.

Ans:

Attempt to resolve vertically or horizontally

G sin 60° + 2 F sin 40° − 10 = 0

F + G cos 60° − 2 F cos 40° = 0

Attempt to solve simultaneously for F or G

F = 4.53, G = 4.82

### (b) Question

Given instead that F = 3, find the value of G for which the resultant of the forces is perpendicular to the 10 N force.

Ans:

G sin 60° + 2 × 3 sin 40° − 10 = 0

G = 7.09 to 3 sf

### Question Coplanar forces of magnitudes 24 N, P N, 20 N and 36 N act at a point in the directions shown in the diagram. The system is in equilibrium.

Given that sin α =    find the values of P and θ.

Ans:

Attempt at resolving in any direction

P cos θ =(36 – 24) cos36.9
or
P cos θ = (36 – 24) × 0.8

P sin θ + 20 =  (24 + 36) sin36.9 = 14.4 + 21.6

or

P sin θ + 20 = 60 × 0.6 = 36

P cos θ = 9.6, P sin θ = 16 P = $$\sqrt{16^{2} + 9.6^{2}}$$

$$\theta = tan^{-1}\left ( \frac{16}{9.6} \right )$$

P =18.7
θ = 59 [ .0]

### Question Three coplanar forces of magnitudes 10 N, 25 N and 20 N act at a point O in the directions shown in
the diagram.
(a) Given that the component of the resultant force in the x-direction is zero, find a, and hence find
the magnitude of the resultant force.                                                                                                                      

(b) Given instead that a = 45, find the magnitude and direction of the resultant of the three forces.              

Ans

(a) 20cos30 = 25cos60 + 10cos α
[17.32 = 12.5 + 10cos α , → cos α = 0.4821]

α = 61.2

Resultant = 20sin 30 + 10sin 61.2 − 25sin60
[= 10 + 8.761 − 21.651]

Magnitude of resultant force = 2.89 N

(b) X =25cos 60 + 10cos45 − 20cos 30

= 12.5 + 7.07107 − 17.32051 = 2.25056

Y = 20sin 30 + 10sin 45 − 25sin 60

= 10 + 7.07107 − 21.65064 =−4.57957

$$R=\sqrt{X^{2}+Y^{2}}$$

$$a=\tan ^{-1}\frac{Y}{X}$$

Resultant = 5.10 N,
Direction = 63.8° below positive x-axis

### Question Coplanar forces of magnitudes 8 N, 12 N, 10 N and P N act at a point in the directions shown in the
diagram. The system is in equilibrium.
Find P and θ.                                                                                                                                                                                

Resolve forces either horizontally or vertically
P cos θ = 12 + 8 cos 30 – 10 cos 45 [= 11.857]
P sin θ = 10 sin 45 – 8 sin 30 [= 3.071]

$$P=\sqrt{(11.857^{2}+3.071^{2})}$$

$$\theta =\tan ^{-1}\left ( \frac{3.071}{11.857} \right )$$

P = 12.2 and θ = 14.5

### Question Three coplanar forces of magnitudes 100 N, 50 N and 50 N act at a point A, as shown in the diagram.
The value of cos $$\alpha$$ is $$\frac{4}{5}$$
Find the magnitude of the resultant of the three forces and state its direction.                                                   

Resultant = 100 – 2× 50cos α
20 N
Direction is to the left (or equivalent)

### Question One end of a light inextensible string is attached to the highest point A of a solid fixed sphere with centre O and radius 0.6 m. The other end of the string is attached to a particle P of mass 0.2 kg which rests in contact with the smooth surface of the sphere. The angle AOP = 60° (see diagram). The sphere exerts a contact force of magnitude R N on P and the tension in the string is T N.

(i) By resolving vertically, show that R + (√3)T = 4. 

P is now set in motion, and moves with angular speed ω rad s−1 in a horizontal circle on the surface of the sphere.

(ii) Find an equation involving R, T and ω. 

(iii) Hence

(a) calculate R when ω = 2, 

(b) find the greatest possible value of T and the corresponding speed of P. 

Ans: ### Question Coplanar forces, of magnitudes FN, 3N, 6N and 4N, 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 3N force, find F.
(b) Given instead that the four forces are in equilibrium, find the values of F and $$\alpha$$.

Ans:

(a) $$[4 sin 30+Fsin60-6=0]$$
Correct equatioin
F = 4.62
(b) Resolve forces either vertically or horizontally
$$F sin \alpha +4sin 30 – 6 = 0$$
and
$$Fcos \alpha + 3 – 4cos 30=0$$
$$[F^2=4^2+0.464^2]$$
or
$$[F=\frac{4}{sin83.4}=\frac{0.464}{cos83.4}]$$
$$[\alpha=tan^{-1}(\frac{4}{0.464}]$$
or
$$[\alpha =sin^{-1}(\frac{4}{4.03})=cos^{-1}(\frac{0.464}{4.03})]$$
F = 4.03 and $$\alpha=83.4$$

### Question Given that$$tan\alpha =\frac{12}{5} and tan\Theta =\frac{4}{3}$$,show that the coplanar forces shown in the diagram are in equilibrium.

(X =) $$78× 5/13 – 50×3/5=78 cos 67.4 – 50 cos 53.1(Y =) 78× 12/13+ 50 × 4/5 – 112= 78 sin 67.4+ 50 sin 53.1 –112$$

[X = 30 – 30 = 0

Y = 72 + 40 – 112 = 0]

X = 0 andY =0

$$\frac{112}{sin59.5}$$=$$\frac{50}{sin157.5}$$=$$\frac{78}{sin143.1} \frac{112}{\frac{56}{65}}$$=$$\frac{50}{\frac{5}{13}}$$=$$\frac{78}{\frac{3}{5}}$$=130

### Question Four coplanar forces of magnitudes F N, 5 N, 25 N and 15 N are acting at a point P in the directions shown in the diagram. Given that the forces are in equilibrium, find the values of F and$$\alpha$$

F cos (\Alpha\) = 15 cos 20 – 5 (= 9.095…)

F sin (\Alpha\) = 15 sin 20 + 25 (= 30.13….)

F=$$\sqrt{(15cos20-5)^{2}+(15sin20+25)^{2}}$$

∝=$$tan^{-1}\left [ (15sin20+25/(15cos20-5) \right ]$$

α = 73.2 and F = 31.5

### Question Coplanar forces, of magnitudes 15 N, 25 N and 30 N, act at a point B on the line ABC in the directions
shown in the diagram.
(i) Find the magnitude and direction of the resultant force.

(ii) The force of magnitude 15 N is now replaced by a force of magnitude F N acting in the same
direction. The new resultant force has zero component in the direction BC. Find the value of F, and find also the magnitude and direction of the new resultant force.

(i) 25 cos 30
– 15 cos 40 (= 10.1599…)

25 sin 30+ 15 sin 40 – 30 (= –7.8581…)

Magnitude = $$\sqrt{(10.15..^{2}+7.858..^{2})}=12.8N= 12.8 N$$

Angle 37.7° below the horizontal in the direction BA

(ii) F cos 40= 25 cos 30

F = 28.3

New resultant force

= 28.26…sin 40+ 25 sin 30–30= 0.667 N upwards

### Question The three coplanar forces shown in the diagram are in equilibrium. Find the values of 1 and P.

EITHER:
2P sin θ = P sin 60

θ = 25.7
2P cos θ + P cos 60 = 10
P = 4.34

θ = 25.7
Use a second Lami equation
P = 4.34

OR2:
Use sine or cosine rule with triangle of forces
using forces P, 2P and 10 and with angles 60,
θ and 120 – θ between

θ = 25.7
Use a second relationship from the triangle of
forces

P = 4.34

### Question Three coplanar forces of magnitudes F N, 2F N and 15 N act at a point P, as shown in the diagram. Given that the forces are in equilibrium, find the values of F and$$\alpha.$$

2F + Fcos60 = 15cosα

Fsin60 = 15sinα

F = 5.67 and α = 19.1

### Question Coplanar forces of magnitudes 50 N, 40 N and 30 N act at a point O in the directions shown in the diagram, where tan $$\alpha = \frac{7}{24}$$ .

(i) Find the magnitude and direction of the resultant of the three forces. 

(ii) The force of magnitude 50 N is replaced by a force of magnitude P N acting in the same direction. The resultant of the three forces now acts in the positive x-direction. Find the value of P.  ### Question A small bead Q can move freely along a smooth horizontal straight wire AB of length 3 m. Three horizontal forces of magnitudes F N, 10 N and 20 N act on the bead in the directions shown in the diagram. The magnitude of the resultant of the three forces is R N in the direction shown in the diagram.

(i) Find the values of F and R. 

(ii) Initially the bead is at rest at A. It reaches B with a speed of 11.7 m s−1. Find the mass of the bead. 

Ans:

(i) Fcos70 + 20 – 10 cos 30
= Rcos15

10sin30 – F sin70 = R sin15

F = 1.90 N and R = 12.4 N

Alternative method for (i)

[X = 0.342 F + 11.34
Y = 0.94 F – 5]

(0.342 F + 11.34)2 + (0.94 F – 5)2
= R2

tan15
= (5 – 0.94F) / (0.342F + 11.34)

F = 1.90 N and R = 12.4 N

(ii) 11.72 = 0 + 2a × 3
a = 22.815

R cos15 = m × 22.815

Mass of bead = 0.526 kg

### Question Three horizontal forces of magnitudes F N, 63 N and 25 N act at O, the origin of the x-axis and y-axis. The forces are in equilibrium. The force of magnitude F N makes an angle θ anticlockwise with the positive x-axis. The force of magnitude 63 N acts along the negative y-axis. The force of magnitude 25 N acts at tan−1 0.75 clockwise from the negative x-axis (see diagram). Find the value of F and the value of tan 1. 

Fx = F cosθ = 25 × 0.8 = 20,
Fy = F sinθ = 63 – 25 × 0.6 = 48

F = 52 N or tanθ = 2.4

tanθ = 2.4 or F = 52 N

### Question Four coplanar forces act at a point. The magnitudes of the forces are 5 N, 4 N, 3 N and 7 N, and the directions in which the forces act are shown in the diagram. Find the magnitude and direction of the resultant of the four forces

.

X = 5 – 7cos 60o – 3cos 30o

(= – 1.098)

Y = 7sin 60o– 3sin 30o – 4 (= 0.5622)

Resultant is 1.23 N and
Direction is 152.9o

anticlockwise from

+ve x-axis oe

### Question A particle P of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point X. A horizontal force of magnitude F N is applied to the particle, which is in equilibrium when the string is at an angle α to the vertical, where tan α = $$\frac{8}{15}$$ (see diagram). Find the tension in the string and the value of F. 

[Tcosα = mg]
Tension is 3.4 N
[F = Tsinα]
F = 1.6

### Question A particle P of mass 0.5 kg lies on a smooth horizontal plane. Horizontal forces of magnitudes F N,
2.5 N and 2.6 N act on P. The directions of the forces are as shown in the diagram, where tan $$\alpha =\frac{12}{5}$$ and tan $$\beta =\frac{7}{24}$$.

(i) Given that P is in equilibrium, find the values of F and tan θ. 

(ii) The force of magnitude F N is removed. Find the magnitude and direction of the acceleration with which P starts to move. 

Ans:  ### Question Three coplanar forces of magnitudes 8 N, 12 N and 2 N act at a point. The resultant of the forces has magnitude R N. The directions of the three forces and the resultant are shown in the diagram. Find R and θ.

X = 12cos25o – 8cos10o (= 2.9972….)

Y = 12sin25o + 8sin10o – 2 (= 4.4606….)

R = 5.37

θ = 33.9

### Question A small ring of mass 0.2 kg is threaded on a fixed vertical rod. The end A of a light inextensible string
is attached to the ring. The other end C of the string is attached to a fixed point of the rod above A. A horizontal force of magnitude 8 N is applied to the point B of the string, where AB = 1.5 m and BC = 2 m. The system is in equilibrium with the string taut and AB at right angles to BC (see diagram).
(i) Find the tension in the part AB of the string and the tension in the part BC of the string.  The equilibrium is limiting with the ring on the point of sliding up the rod.

(ii) Find the coefficient of friction between the ring and the rod.

(i)

$$T_{C}$$ × (2/2.5) –$$T_{A}$$× (1.5/2.5) = 0
$$T_{C}$$ × (1.5/2.5) +$$T_{A}$$ × (2/2.5) = 8
[0.6 TC + 0.8$$(4T_{C}/3)$$ = 8 → (5/3) TC = 8 or
0.6(0.75TA) + $$0.8T_{A}$$= 8 → $$1.25T_{A}$$ = 8 ]

Tension in AB is 6.4 N; tension in BC is 4.8 N

(ii)

F + 0.2 g =$$T_{A}$$ × (1.5/2.5)  N = TA × (2/2.5)
[ μ = (3.84 – 2 )/5.12]

horizontal Coefficient is 0.359

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