WB20 In the diagram, AOC = 90° and BOC = q °.
A particle at O is in equilibrium under the action of three coplanar forces.
The three forces have magnitudes 8 N, 12 N and X N and act along OA, OB and
OC respectively. Calculate
(a) the value, to one decimal place, of θ
R

12 cos  90  8
8
 cos  90  
12
B
   90  48.2 1dp
12 N
   138.2 1dp

O
XN
C
(b) the value, to 2 decimal places, of X.
R

X  12 sin   90
8N
 8.94 N (2dp)
A
Now try Ex 4A, Q1,4,14 then 7
Q1
R
 Q  5 cos 30   4.33 N (3sf)
R  P  5sin 30  2.5 N
PN
30
5 3
2
QN
5N
QN
Q4
PN
R
 Q sin 60  6 sin 45
6 sin 45
 2 6 N  4.90 N (3sf)
Q
sin 60
120 60
45
R
6N
 P  Q cos 60  6 cos 45



6  3 2 N  6.69 N (3sf)
R
Q14
8N
PN

P  8 sin 45  10 cos 30
 P  10 cos 30  8 sin 45

45
R
QN
30

 5 3 - 4 2 N  3.00 N (3sf)
10 N
 8 cos 45  Q  10 sin 30
 Q  8 cos 45  10 sin 30


 4 2 - 5 N  0.657 N (3sf)
Q7
PN
7N
35
R
 P sin   7 sin 35  8 cos 40 1
R  P cos  8 sin 40  7 cos 35 2
1  P sin   8 cos 40  7 sin 35
2 P cos  7 cos 35  8 sin 40

40
8N
40 7 sin35
 tan   87 cos
cos 35 8 sin 40
   74.4 (3sf)
Sub in (1)  P sin 
 8 cos 40  7 sin 35
 P  2.19 N (3sf)
WB21 A particle has mass 2 kg.
It is attached at B to the ends of two light inextensible strings AB and BC.
When the particle hangs in equilibrium, AB makes an angle of 30° with the vertical.
The magnitude of the tension in BC is twice the magnitude of the tension in AB.
(a) Find, in degrees to one decimal place, the size of the angle that BC makes with
the vertical.
A
2T sin 
T sin 30
C
R

 21 T  2T sin 
2T
T
T sin 30  2T sin 
 sin  
30°
   14.5 (1dp)

B
1
4
(b) Hence find, to 3 significant figures,
the magnitude of the tension in AB.
2g
R

T cos 30  2T cos   2 g
 T cos 30  2 cos   2g
2g
 6.99 N (3sf)
T 
cos 30  2 cos 
Now try Ex 4B, Q1,2,6,5,4
45
45
Q1
T
R
T
 2T sin 45  5 g
T 
5 kg
5g
2 sin 45
 35 N (2sf)
5g
R
Q2
30
 T  sin1030  20 N
T
m kg
10 N
R
 T cos 30  mg
m
mg
Q6
R
30
m kg
T
mg
20 cos 30
g
 1.8 N (2sf)
 2  T sin 30  T sin 60
 T  sin30 2sin60  1.46 N (3sf)
T
60
 T sin 30  10
2N
R
 T cos 30  T cos 60  mg
 m  T cos 30 gT cos 60  0.055 kg (2sf)
30
60
Q5
R
 T cos 30  T cos 60  2
T 
T
T
2N
m kg
R
60
T1
45
T2
6 kg
6g
 5.46 N (3sf)
 T sin 60  T sin 30  mg
 m  T sin 60 gT sin30  0.76 kg (2sf)
mg
Q4
2
cos 30 cos 60
R
R
 T1 cos 60  T2 cos 45
45
 T1  cos
2T2
cos 60 T2 
 T1 sin 60  T2 sin 45  6g
 2T2 sin 60  T2 sin 45  6 g
 T2

 T2 

2 sin 60  sin 45  6 g
6g
2 sin 60 sin 45
 T1  43 N (2sf)
 30 N (2sf)
WB23 A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the
action of a force of magnitude X newtons acting up a line of greatest slope of the
plane, as shown in the diagram above. The plane is inclined at 20° to the horizontal.
The coefficient of friction between P and the plane is 0.4. The particle is in limiting
equilibrium and is on the point of moving up the plane. Calculate
(a) the normal reaction of the plane on P,
R

R  52 g cos 20  23 N (2sf)
R
P
(b) the value of X.
R
XN
 X  52 g sin 20  R
 18 N (2sf)
R
5
2
20º
The force of magnitude X newtons is now removed.
(c) Show that P remains in equilibrium on the plane.
Fmax  R  0.4  g cos 20  9.2 N (2sf)
5
2
F  g sin 20  8.4 (2sf)
5
2
 F  Fmax
20º
g
R
P
XN
F
5
2
g
 remains in equilibrium
Now try Ex 4B, Q7,8,12,13
5N
R
Q7
R

FN
R
6
tan   34
5
4

 sin  
PN
5
R  5 sin   6
 R  6  5  54  2 N
R
1N
R
2


 F  5 35  3 N
4
5
R

13
5 cos  F
 cos   35
3
Q8

12
tan   125
12
 sin   13
 cos   135

P cos  1
 P  1 135  2.6 N
 R  P sin   2
12
 R  135  13
 2  4.4 N
Q12
R
T
A
R
 R  2 g cos
For B:
R
 T  5g
For A:
R

For A:
T
2 kg
5 kg
FN

B
2g
5g
5

4
tan   34
3
 sin   35
 cos   54
 85 g  16 N (2sf)
F  2 g sin   T
 F  T  2 g sin 
 5 g  56 g
 195 g
 37 N (2sf)
Q13
R
T
Q
FN
For 2kg mass:
T
5 kg
For Q:
2 kg
R
F
2g
5

4
 sin   35
 cos  
5 g sin T
cos

4
5
3 g 2 g
4
5
 54 g  12 N (2sf)
tan   34
3
 T  2g
 F cos   T  5 g sin 
5g

R
For Q:
R

R  5 g cos   F sin 
 4 g  34 g
 194 g  47 N (2sf)
WB22 Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal
pole. The coefficient of friction between each ring and the pole is m.
The rings are attached to the ends of a light inextensible string.
A smooth ring C, of mass 3m, is threaded on the string and hangs in equilibrium below
the pole.
The rings A and B are in limiting equilibrium on the pole, with BAC = ABC = θ,
where tan   34 , as shown in the diagram above.
(a) Show that the tension in the string is 52 mg
R

2T sin   3mg
3mg
T 
2 sin 
3mg 5
 T  6  2 mg
R
A (2m)
5
B (2m)
R
T
T
(b) Find the value of μ.
For B:
R
R
 R  T sin   2mg  72 mg

R  T cos 
 R  2mg
2mg 4
 7

7
2 mg
2mg
C (3m)
3 mg
tan   34  sin   35  T sin   32 mg
 cos   54  T cos   2mg
WB24 A parcel of mass 5 kg lies on a rough plane inclined at an angle a to the
horizontal, where tan   34
The parcel is held in equilibrium by the
R
action of a horizontal force of magnitude
20 N, as shown in the diagram above.
20 N
The force acts in a vertical plane through
R
a line of greatest slope of the plane.
The parcel is on the point of sliding down
5g
the plane. Find the coefficient of friction
between the parcel and the plane.
R

R 
R  5 g cos   20 sin 
R  20 cos   5 g sin 
1
2
R
5 g cos   20 sin 
 sin   35
 cos   54
2    5 g sin   20 cos 
1    5 g sin   20 cos 
tan   34

3 g  16
 0.26 (2sf)
4 g  12
WB25 A box of mass 1.5 kg is placed on a plane which is inclined at an angle of 30°
to the horizontal. The coefficient of friction between the box and plane is 13
The box is kept in equilibrium by a light string which lies in a vertical plane containing
a line of greatest slope of the plane. The string makes an angle of 20° with the plane,
as shown in the diagram above. The box is in limiting equilibrium and is about to
move up the plane. The tension in the string is T newtons.
The box is modelled as a particle.
Find the value of T.
R
 R  T sin 20  32 g cos 30 1
R  T cos 20  32 g sin 30  R 2
1  R  g cos 30  T sin 20 3 
2 R  3T cos 20  92 g sin 30 4
TN
R
20
3
2
  31
Equating (3) and (4):
 3T cos 20  92 g sin 30  32 g cos 30  T sin 20
R
30
3
2
g
 T sin 20  3 cos 20  92 g sin 30  32 g cos 30
g sin 30  32 g cos 30
 11N (2sf)
T 
sin 20  3 cos 20
9
2
Now try Ex 4C, Q8-12
Q8
R
R
Q
20 N
1.5 g
20 cos 30  1.5 g sin 30  10 N (2sf)
 F acts down the slope
1.5 kg
30

R
 R  1.5 g cos 30  20 sin 30  23 N (2sf)
For equilibrium, F  R   
F
R
F
R
30 1.5 g sin30
 0.44 (2sf)
 120.5cos
g cos30 20 sin30
   0.44 (2sf)
Q9
R
Q
XN
3 kg
3
10
R
40
3g
R
 R  3 g cos 40  X sin 40 1
R  X cos 40  3 g sin 40  103 R 2
2 R  103 X cos 40  10 g sin 40 3 
1  3  3 g cos 40  X sin 40  103 X cos 40  10 g sin 40
 103 X cos 40  X sin 40  3 g cos 40  10 g sin 40
X
3 g cos 40 10 g sin 40
10 cos 40  sin 40
3
 45 N (2sf)
Sub in (1)  R  51N (2sf)
Q10
R
22 kg
1
8
R
22 g
35
Q11a)
R
T
R
kg
R
40
1
2
 Fmax  228 g cos 35
T  Fmax  22 g sin 35
R
20
1
5

R  22 g cos 35
 T  22 g sin 35  228 g cos 35  100 N (2sf)
T
1
2
R

g
3  4
 R  T sin 20  21 g cos 40
R  T cos 20  21 g sin 40  51 R
1  R  21 g cos 40  T sin 20
2  R  5T cos 20  52 g sin 40
1
2
3 
4
 21 g cos 40  T sin 20  5T cos 20  52 g sin 40
 5T cos 20  T sin 20  21 g cos 40  52 g sin 40
T 
1 g cos 40  5 g sin 40
2
2
5 cos 20  sin 20
 3.9 N (2sf)
Q11b)
R
R
T
20
1
2
kg
1
2
40
1
5
R
g
3  4
 R  T sin 20  21 g cos 40
R  T cos 20  51 R  21 g sin 40
1  R  21 g cos 40  T sin 20
2  R  52 g sin 40  5T cos 20
1
2
3 
4
 21 g cos 40  T sin 20  52 g sin 40  5T cos 20
 5T cos 20  T sin 20  52 g sin 40  21 g cos 40
T 
Q12
10 N
R
20
1 kg
R
40
g
5 g sin 40  1 g cos 40
2
2
5 cos 20 sin 20
 2.8 N (2sf)
 R  10 sin 20  g cos 40 1
R  10 cos 20  g sin 40  R 2
1  R  g cos 40  10 sin 20
40
 0.76 (2sf)
2    10 cos 20R g sin40  10g coscos402010g sin
sin 20
R