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. - PowerPoint PPT Presentation

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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.

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1. The following diagram shows a system of forces acting on a ... Pcos ?. Horizontally: (1) Vertically: (2) From (1): From (2): (3) (4) (4) divided by (3) ... – PowerPoint PPT presentation

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Title: 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.


1
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.
Components diagram
Horizontal component?
8 cos 30 F cos? ? 12 0
? F cos? 12? 8 cos 30 1
Vertical component ?
8 sin 30 ? F sin? 0
? F sin? 8 sin 30 2
2
Divide 2 by 1
? ? 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.
3
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 ? to the
horizontal, as shown. The force diagram is
P N
4
Components diagram
15 cos 70 P cos ? 4 cos 40 0
Horizontally
(1)
(2)
15 sin 70 P sin ? 4 sin 40 0
Vertically
P cos ? 15 cos 70 4 cos 40
From (1)
(3)
P sin ? 15 sin 70 4 sin 40
From (2)
(4)
(4) divided by (3)
P 11.7 N
Substitute into (3)
P acts on a bearing of 190º and has size
5
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.
6
Force diagram
Components diagram
T sin60
T cos60
Mg
(1)
Resolving horizontally
Resolving vertically
8 sin 40 T sin 60 Mg 0
(2)
7
Resolving horizontally
(1)
Resolving vertically
(2)
8 sin 40 T sin 60 Mg 0
From (1)
The tension in the other string is 12.3 N
Substitute in (2)
The mass of the particle is 1.61 kg.
8
4. A car of mass 1500 kg is broken down on a
rough plane inclined at an angle of ? 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

1500kg
1500g
9
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
10
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. (ii) Find
the value of m.
11
Components diagram
2 sin45
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
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