1.
A small box is pushed along a floor. The floor is modelled as a rough horizontal plane and the
box is modelled as a particle. The coefficient of friction between the box and the floor is
1
.
2
The box is pushed by a force of magnitude 100 N which acts at an angle of 30° with the floor,
as shown in the diagram above.
Given that the box moves with constant speed, find the mass of the box.
(Total 7 marks)
2.
A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of
magnitude P newtons. The force acts in the vertical plane containing the line of greatest slope of
the inclined plane which passes through the particle. The plane is inclined to the horizontal at
an angle α, where tan α =
3
, as shown in the diagram above.
4
The coefficient of friction between the particle and the plane is
1
.
3
Given that the particle is on the point of sliding up the plane, find
(a)
Dubai ELS
the magnitude of the normal reaction between the particle and the plane,
1
(5)
Dubai ELS
2
(b)
the value of P.
(5)
(Total 10 marks)
3.
A particle of mass m kg is attached at C to two light inextensible strings AC and BC. The other
ends of the strings are attached to fixed points A and B on a horizontal ceiling. The particle
hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 60° respectively, as
shown in the diagram above.
Given that the tension in AC is 20 N, find
(a)
the tension in BC,
(4)
(b)
the value of m.
(4)
(Total 8 marks)
Dubai ELS
3
4.
A pole AB has length 3 m and weight W newtons. The pole is held in a horizontal position in
equilibrium by two vertical ropes attached to the pole at the points A and C where AC = 1.8 m,
as shown in the diagram above. A load of weight 20 N is attached to the rod at B. The pole is
modelled as a uniform rod, the ropes as light inextensible strings and the load as a particle.
(a)
100 
5
Show that the tension in the rope attached to the pole at C is  W +
 N.
3 
6
(4)
(b)
Find, in terms of W, the tension in the rope attached to the pole at A.
(3)
Given that the tension in the rope attached to the pole at C is eight times the tension in the rope
attached to the pole at A,
(c)
find the value of W.
(3)
(Total 10 marks)
5.
A particle is acted upon by two forces F1 and F2, given by
Dubai ELS
4
F1 = (i – 3j) N,
F2 = (pi + 2pj) N, where p is a positive constant.
(a)
Find the angle between F2 and j.
(2)
Dubai ELS
5
The resultant of F1 and F2 is R. Given that R is parallel to i,
(b)
find the value of p.
(4)
(Total 6 marks)
6.
P
150 °
O
Q
Two forces P and Q act on a particle at a point O. The force P has magnitude 15 N and the force
Q has magnitude X newtons. The angle between P and Q is 150°, as shown in the diagram
above. The resultant of P and Q is R.
Given that the angle between R and Q is 50°, find
(a)
the magnitude of R,
(4)
(b)
the value of X.
(5)
(Total 9 marks)
Dubai ELS
6
7.
45 N
50°
30°
A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal. The package is
held in equilibrium by a force of magnitude 45 N acting at an angle of 50° to the plane, as
shown in the diagram above. The force is acting in a vertical plane through a line of greatest
slope of the plane. The package is in equilibrium on the point of moving up the plane. The
package is modelled as a particle. Find
(a)
the magnitude of the normal reaction of the plane on the package,
(5)
(b)
the coefficient of friction between the plane and the package.
(6)
(Total 11 marks)
8.
49 N
P
30 °
A particle P of mass 6 kg lies on the surface of a smooth plane. The plane is inclined at an angle
of 30° to the horizontal. The particle is held in equilibrium by a force of magnitude 49 N, acting
at an angle θ to the plane, as shown in the diagram above. The force acts in a vertical plane
through a line of greatest slope of the plane.
Dubai ELS
7
(a)
Show that cosθ =
3
5
(3)
(b)
Find the normal reaction between P and the plane.
(4)
Dubai ELS
8
The direction of the force of magnitude 49 N is now changed. It is now applied horizontally to P
so that P moves up the plane. The force again acts in a vertical plane through a line of greatest
slope of the plane.
(c)
Find the initial acceleration of P.
(4)
(Total 11 marks)
9.
PN
°
20
A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using a
rope. The rope makes an angle of 20° with the ground, as shown in the diagram above. The
coefficient of friction between the box and the ground is 0.4. The box is modelled as a particle
and the rope as a light, inextensible string. The tension in the rope is P newtons.
(a)
Find the value of P.
(8)
The tension in the rope is now increased to 150 N.
(b)
Find the acceleration of the box.
(6)
(Total 14 marks)
10.
(→) 100cos30 = F
F = 0.5 R seen
Dubai ELS
M1 A1
A1 (B1)
9
(↓) mg + 100cos60 = R
M1 A1
m = 13 kg or 12.6 kg
DM1 A1
[7]
11.
(a)
F = 13 R
B1
( ↑ ) R cos α − F sin α = 0.4 g
R = 23 g = 6.53 or 6.5
Dubai ELS
M1 A1
M1 A1
5
10
(b)
(→)P − F cos α − R sin α = 0
P=
26
45
M1 A2
g = 5.66 or 5.7
M1 A1
5
[10]
12.
(a)
R(→)
20cos30° = T cos60°
M1 A2 (1,0)
T = 20 3 , 34.6, 34.64,…
(b)
R(↑)
mg = 20sin 30° + T sin 60°
m=
A1
4
M1 A2 (1,0)
40
(≈ 4.1),4.08
8
A1
4
[8]
13.
(a)
M ( A)
W×1.5+20×3=Y×1.8
5
100
Y= W+
*
6
3
Dubai ELS
M1 A2 (1, 0)
cso
A1
4
11
(b)
X + Y = W + 20
↑
or equivalent
1
40
X= W–
6
3
M1 A1
A1
3
Alternative
M(C)
X × 1.8 + 20 × 1.2 =W × 0.3
M1 A1
1
40
X= W–
6
3
A1
5
100  1
40 
W+
= 8 W –

6
3
6
3 

(c)
W = 280
M1 A1 ft
A1
3
[10]
14.
(a)
tan θ =
p
2p
⇒ θ = 26.6°
M1 A1
R = (i – 3j) + (pi + 2pj) = (1 + p)i + (–3 + 2p)j
(b)
R is parallel to i
⇒ (–3 + 2p) = 0
⇒ p=
2
M1 A1
DM1
3
2
A1
4
[6]
15.
(a)
15
R
30°
50°
X
(↑) 15 sin 30° = R sin 50°
R ≈ 9.79 (n)
Dubai ELS
M1A1
DM1A1
4
12
(b)
(→) X – 15 cos 30° = R cos 50°
X ≈ 19.3 (n)
Dubai ELS
ft their R
M1A2 ft
DM1A1
5
13
Alternatives using sine rule in (a) or (b); cosine rule in (b)
15
R
50°
30°
X
(a)
R
15
=
sin 30° sin 50°
M1A1
r ≈ 9.79 (N)
(b)
DM1A1
X
15
R
=
=
sin 100° sin 50° sin 30°
4
M1A2ft on R
X ≈ 19.3 (n)
DM1A1
5
X2 = R2 + 152 – 2 × 15 × R cos 100°
OR: cosine rule; any of R2 = X2 + 152 – 2 × 15 × X cos 30°
M1A2ft on R
152 = R2 + X2 – 2 × X × R cos 50°
X ≈ 19.3 (n)
DM1A1
5
[9]
16.
(a)
R
45 N
50°
R
4g
30°
R = 45cos40° + 4g cos 30°
R ≈ 68
(b)
Dubai ELS
Use of F = µR
M1A2(1,0)
accept 68.4
DM1A1
5
M1
14
F + 4g sin 30 = 45 cos 50°
Leading to µ ≈ 0.14
M1A2(1,0)
accept 0.136
DM1A1
6
[11]
Dubai ELS
15
17.
(a)
(b)
R (// plane): 49 cos θ = 6g sin 30
M1A1
⇒ cos θ = 3/5 *
A1
R (perp to plane): R = 6g cos 30 + 49 sin θ
M1A1
R ≈ 90.1 or 90 N
(c)
3
DM1 A1
R (// to plane): 49 cos 30 – 6g sin 30 = 6a
4
M1A2,1,0
⇒ a ≈ 2.17 or 2.2 m s
–2
A1
4
[11]
18.
(a)
R
P
20º
30g
Use of F = µR
P cos 20° = µR
M1A1
¡ R + Psin20° = 30g
M1 A1
P cos 20° = µ(30g – P sin 20°)
M1
0.4 × 30 g
cos 20° + 0.4 sin 20°
M1
P=
≈ 110 (N)
(b)
B1
i
accept 109
R + 150 sin 20° = 30 g
A1
8
M1A1
(R ≈ 242.7)
N2L
a≈
150 cos 20° – µR = 30a
M1A1
150 cos 20° − 0.4 × 242.7
30
= 1.5 (m s–2)
M1
accept 1.46
A1
6
[14]
Dubai ELS
16