Transcript Practice questions

1. The following diagram shows a system of forces acting on a
particle in a plane. A third force is added so that the particle rests
in equilibrium. Find the magnitude of this force and the angle
that make with the horizontal.
8N
12 N
30
Components diagram
8 sin 30
12
8 cos 30
F cos
F sin
Horizontal component:
8 cos 30 + F cos  12 = 0
 F cos = 12 8 cos 30
Vertical component:  :
[1]
8 sin 30  F sin = 0
 F sin = 8 sin 30
[2]
Divide [2] by [1]:
F sin 
8 sin 30

F cos  12  8 cos 30
 tan  
4
 0.7887
5.0718
  = 38.3
Sub  into [2]:
F sin 38.3 = 4
 F = 6.45 N
Added force is 6.45 N, acting at an angle of 38.3 below the
horizontal.
2. A particle is acted on by a force of 15 N which acts on a
bearing of 020º, and another force of 4 N which acts on a
bearing of 230º. Find the magnitude of a third force which
will keep the system in equilibrium, stating the angle of its
line of action as a bearing.
Let the third force be P N acting at θ to the horizontal, as shown.
The force diagram is:
North
20º
230º
4N
θ
PN
Components diagram
15sin70
4 cos 40
Pcos θ
15cos70
4 sin 40
Psin θ
15 cos 70 – P cos θ – 4 cos 40 = 0
(1)
(2)
From (1):
15 sin 70 – P sin θ – 4 sin 40 = 0
P cos θ = 15 cos 70 – 4 cos 40
From (2):
P sin θ = 15 sin 70 – 4 sin 40
(4)
Horizontally:
Vertically:
(4) divided by (3)
tan   5.5777

(3)
  79.8
Substitute into (3) 
P = 11.7 N
P acts on a bearing of 190º and has size 11.7 N
3. The diagram shows a particle suspended from a horizontal beam
by two unequal, light and inextensible strings. Given that the
tension in the left string is 8 N and it makes an angle of 40º to the
beam, and the other string makes an angle of 60º
to the beam, find the tension in the other string and the mass of the
particle.
40
60
40
Force diagram
60
T
8N
60
40
Mg
Components diagram
8sin 40
T sin60
T cos60
8cos 40
Mg
Resolving horizontally:T cos60  8cos40  0
(1)
Resolving vertically: 8 sin 40 + T sin 60 – Mg = 0
(2)
T cos 60  8 cos 40  0
Resolving horizontally:
Resolving vertically:
8 sin 40 + T sin 60 – Mg = 0
From (1):
T
8cos 40
 12.3 N
cos 60
The tension in the other string is 12.3 N
Substitute in (2):
M  8 sin 40 T sin 60
g
 1.61 kg
The mass of the particle is 1.61 kg.
(1)
(2)
4. A car of mass 1500 kg is broken down on a rough plane inclined
at an angle of θ° to the horizontal, where sin-1 = (7/25). It is being
pulled up the plane by means of a towrope, which is acting at 9º
above the line of greatest slope of the plane.
The resistance between the plane and the car has a magnitude of
1000 N. The car is at rest in equilibrium and is about to move up the
plane. Find the tension in the towrope and the magnitude of the
force of the plane on the car.
Force diagram
T
R
1000N
9º

1500g
Components diagram
Rn T sin9
T cos9
1000
1500g sin
1500g cos
Res parallel:
T cos9 – 1000 – 1500g (7/25) = 0
T = 5180 N
Res perpendicular: Rn + 5100 sin9 – 1500g (24/25)
Rn= 13300 N
5. A light inextensible string has its ends attached to two fixed
points A and B. The point A is vertically above B. A smooth ring R
of mass m kg is threaded on the string and is pulled by a force of
magnitude 3N acting upwards at 400 to the horizontal. The
section AR of the string makes an angle of 300 with the
downward vertical and the section BR is horizontal (see diagram).
The ring is in equilibrium with the string taut.
(i)
Find the tension in the string.
A
2N
30
(ii)
Find the value of m.
B
R 45
mg kg
Components diagram
T sin 60 2 sin45
T cos60
2 cos45
T
mg
Resolve Horizontally:
T + T cos 60 = 2 cos 45
 T = 0.942809 ..
Resolve vertically:
T sin 60 + 2 sin 45 = mg
 m = 0.228 kg
(a) T = 0.943 N
(b) m = 0.228 kg