Pearson Edexcel IAL Mathematics M1.5 Statics of a particle Question Bank

Practise statics of particles with vectors, equilibrium equations and limiting friction on planes, strings, rings, pulleys and rough surfaces.

Exam points

resolve horizontal and vertical forces to find tensions, masses or applied forces in equilibrium
calculate a resultant pulley force from two equal tensions using components or a force triangle
compare required friction with limiting friction to decide whether a particle remains at rest

Covered syllabus points

M1.5.1 - Forces treated as vectors M1.5.2 - Equilibrium of a particle under M1.5.3 - Coefficient of friction

[Maximum number: 7]

A particle is acted upon by two forces F and G. The force F has magnitude 8 N and acts in a direction with a bearing of $240^{\circ}$ . The force G has magnitude 10 N and acts due South.

Given that R=F+G, find

(a)

the magnitude of R,

[ 3 ]

R=F+G

$\begin{array}{ll} R^{2}= 8^{2}+10^{2}-2 \times 8 \times 10 \cos 120^{\circ} \text { oe }(244) \text { OR: } R^{2}=8^{2}+10^{2}+2 \times 8 \times 10 \cos 60^{\circ} \text { OR: } R^{2}=(8 \sin 60)^{2}+\left(10+8 \cos 60^{\circ}\right)^{2} \text { OR: } R^{2}=(10 \sin 60)^{2}+\left(8+10 \cos 60^{\circ}\right)^{2} \end{array}$

M1A1

$R=\sqrt{244}=15.620499 \ldots \mathrm{~N}$

A1

(b)

the direction of R, giving your answer as a bearing to the nearest degree.

[ 4 ]

$\begin{aligned} \frac{\sin \alpha}{8}=\frac{\sin 120^{\circ}}{\sqrt{244}}\left(\text { allow } \sin 60^{\circ}\right) \text { OR } \frac{\sin \beta}{10}=\frac{\sin 120^{\circ}}{\sqrt{244}}\left(\text { allow } \sin 60^{\circ}\right) \text { OR } 8^{2}=(\sqrt{244})^{2}+10^{2}-2 \times \sqrt{244} \times 10 \cos \alpha \text { OR } 10^{2}=(\sqrt{244})^{2}+8^{2}-2 \times \sqrt{244} \times 8 \cos \beta \text { OR } \tan \alpha=\frac{8 \sin 60}{10+8 \cos 60^{\circ}} \text { or } \sin \alpha=\frac{8 \sin 60}{\sqrt{244}} \text { or } \cos \alpha=\frac{10+8 \cos 60^{\circ}}{\sqrt{244}} \text { (or reciprocal of tan) } \text { OR } \tan \beta=\frac{10 \sin 60}{8+10 \cos 60^{\circ}} \text { or } \sin \beta=\frac{10 \sin 60}{\sqrt{244}} \text { or } \cos \beta=\frac{8+10 \cos 60^{\circ}}{\sqrt{244}} \text { (or reciprocal of tan) } \end{aligned}$

M1A1

$\alpha=26 .(3 . .)^{\mathrm{o}} \quad$ OR $\beta=33.67 \ldots{ }^{\mathrm{o}}$ (accept 34)

A1

Bearing is $206^{\circ}$ (nearest degree)

A1

(7)

Notes for question 5

[Maximum number: 12]

Figure 3

A washing line ABCD is fixed at the points A and D.
There are two heavy items of clothing hanging on the washing line, one fixed at B and the other fixed at C.

The washing line is modelled as a light inextensible string, the item at B is modelled as a particle of mass 3 kg and the item at C is modelled as a particle of mass $M \mathrm{~kg}$ .

The section AB makes an angle $\alpha$ with the horizontal, where $\tan \alpha=\frac{3}{4}$ , the section BC is horizontal and the section CD makes an angle $\beta$ with the horizontal, where $\tan \beta=\frac{12}{5}$ , as shown in Figure 3.

The system is in equilibrium.

(a)

Use of $\sin \alpha=\frac{3}{5}$ or $\cos \alpha=\frac{4}{5}$ or $\alpha=37$ or better

B1

Vert equilibrium at B

M1

$T_{A B} \sin \alpha=3 g$

A1

$T_{A B}=49(\mathrm{~N})$

A1

(4)

(a)

Find the tension in AB.

[ 4 ]

Use of $\sin \alpha=\frac{3}{5}$ or $\cos \alpha=\frac{4}{5}$ or $\alpha=37$ or better

B1

Vert equilibrium at B

M1

$T_{A B} \sin \alpha=3 g$

A1

$T_{A B}=49(\mathrm{~N})$

A1

(4)

(b)

Find the tension in BC.

[ 3 ]

Horiz equilibrium at B

M1

$T_{A B} \cos \alpha=T_{B C}$

A1

$T_{B C}=39$ or $39.2 \quad(\mathrm{~N})$

A1

(3)

(c)

Find the value of M.

[ 5 ]

Use equilibrium to form at least one equation in $T_{CD}$ .

Examples:
$T_{CD}\frac{5}{13}=T_{BC}$ $T_{CD}\frac{12}{13}=Mg$

Form an equation in M only, for example
$\frac{12}{5}T_{BC}=Mg$ M=9.6

M1 A1 A1 dM1 A1

Marking guidance for this part:
Only penalise use of 9.81 once per whole question, and only penalise over- or under-accuracy once per whole question.

[Maximum number: 4]

Figure 5

A parcel of mass 2 kg is pulled up a rough inclined plane by the action of a constant force.

The force has magnitude 18 N and acts at an angle of 40^ to the plane.
The line of action of the force lies in a vertical plane containing a line of greatest slope of the inclined plane.

The plane is inclined at an angle of 30^ to the horizontal, as shown in Figure 5.
The coefficient of friction between the plane and the parcel is 0.3
The parcel is modelled as a particle P

The points A and B lie on a line of greatest slope of the plane, where A B=5 m and B is above A. Particle P passes through A with speed 2 ms^-1 in the direction AB.

The force of 18 N is removed at the instant P passes through B. As a result, P comes to rest at the point C.

Determine whether P will remain at rest at C. You must show all stages of your working clearly.

$R=2g\cos30^\circ=g\sqrt3$
B1

$\text{Friction}=0.3\times2g\cos30^\circ$
OR $0.3\times2g\sin30^\circ$ .
M1

Compares friction with the weight component parallel to the plane, e.g.
$2g\sin30^\circ-0.3(2g\cos30^\circ)=2a$
OR
$0.3(2g\cos30^\circ)-2g\sin30^\circ=2a$
dM1

(a)>0 OR (a)<0.
Concludes that P will not remain at rest.
A1

All question bank results loaded

Covered syllabus points

M1.5 - Statics of a particle question 1

Question (a)

Question (b)

M1.5 - Statics of a particle question 2

Question (a)

Question (b)

Question (c)

M1.5 - Statics of a particle question 3