Notes on Forces, Equilibrium, Density and Pressure

1. Examples - Forces,density and pressure 1
Density
Example
Duralumin has been used for many years in
the aviation industry. It consists of:
 4.4% copper,
 1.5% magnesium,
 0.6% manganese,
 93.5% aluminium.
The densities of the metals are given below.
What is the density of duralumin?
Answer
Assume that the volume is 1 m3.
Use mass = density × volume:
4.4% copper = 0.044 ×8930 = 393 kg.
1.5% magnesium = 0.015 × 1738 = 26.1 kg
0.6% manganese = 0.006 × 7210 = 42.3 kg
93.5% aluminium = 0.935 × 2712 = 2536 kg
Total mass in 1 m3
= 393 + 26.1 + 42.3 + 2536 = 2997.4
Density = 3000 kg m-3
As well as being very low density, duralumin
is also very hard and durable.
Example
Titanium metal has a density of 4.5 g cm–3. A
cube of titanium of mass 48 g contains
6.0 × 1023 atoms.
(i) Calculate the volume of the cube.
(ii) Estimate
1. the volume occupied by each atom in the
cube,
2. the separation of the atoms in the cube.
(9702/22 N09 Q2)
Answer
(i) volume = (48/4.5) = 10.7 cm3
(ii) 1. volume = 10.7 / (6.0 × 1023)
= 1.8 × 10-23 cm3
2. separation = 3√(1.8 × 10-23)
= 2.6 × 10-8 cm
Pressure in fluids
Example
Estimate the height of the atmosphere if
atmospheric density at the Earth’s surface is
1.29 kg m-3.
(Atmospheric pressure= 101 kPa)
Answer
7 900 m.
This figure is too small because it assumes
the density of the air is constant. In fact
density decreases with height.
Upthrust
Examples
1. A 20 kg spherical hollow steel buoy of
volume 0.06 m3 is tethered to the bottom of a
fast flowing river by a cable so that the cable
makes an angle of 40° with the base of the
river. Calculate the tension (T) in this cable.
Answer
Resolving vertically and taking g = 9.8 m s-2
Forces on the buoy:
Upthrust = mg + T[cos40]
= 20g + T[cos40]
But Upthrust = 0.06 x 1000 x g
Therefore: 20g + T[cos40]
= 0.06 x 1000 x g
so T cos 40 = 60g – 20g
= 40g
T = 511 N
2. A hot air balloon with a volume of 200 m3
hangs in the air. If the density of the hot air is
0.8 kg m-3 and that of the cool air outside the
balloon is 1.2 kg m-3 what is the biggest load
it can support if the fabric of the balloon and
the basket have a total mass of 60 kg.
(Take g = 9.8 ms-2)
Answer
Weight of balloon and basket
= 60 g = 588 N
Upthrust = Weight of air displaced
= 200x1.2x g
= 2352 N
Weight of hot air in the balloon
= 200 x 0.8 x 9.8
= 1568 N
Total weight of balloon and hot air
= 1568 + 588
= 2156 N
Metal density/ kg m-3
Aluminium 2712
Copper 8930
Magnesium 1738
Manganese 7210

2. Examples - Forces,density and pressure 2
Therefore additional load that can be
supported by the balloon
= 2352 – 2156 = 196 N
Turning effects of forces
Example
A uniform metre rule supports 4.5 N weight at
its 100 mm mark. The rule is balanced
horizontally on a horizontal knife edge at its
340 mm mark. Sketch the arrangement and
calculate the weight of the rule. (PiC)
Answer
6.75 N
Triangle of forces
Example
The three forces shown are in equilibrium
Using the triangle of forces, find P and Q.
P
Q
60° 40°
20 N
Answer
P = 15.6 N Q = 10.2 N
Forces in equilibrium
Example- Parallel forces
Suspend a 3 N weight from the metre rule as
shown.
Set x1 = 0.2 m and x2 = 0.6 m. Ignore the
weight of the metre rule.
Find R1 and R2.
Answer
(i) The resultant force on the body must be
zero in any direction
3 N = R1 + R2
(ii) The resultant torque about any point must
be zero
0.2 R1 = 0.6 R2
R1 = 3 R2
and so 4 R2 = 3 N
R2 = 0.75 N
and R1= 2.25 N
Example
A van and trailer cross the bridge above, the
axel loads and the position of the vehicles are
shown. The single span bridge is supported
at points 21m apart.
(a) Calculate the vertical forces at each of the
supports caused by the van and trailer on the
bridge.
(b) The support forces are higher than you
calculated, explain why.
Answer
(a)Taking moments about the LHS and
working in kN.
(7 𝑥 4 𝑘𝑁) + (10.5 𝑥 10 𝑘𝑁)
+ (15.75 𝑥 8 𝑘𝑁)
= 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑥 21
28 + 105 + 126 = 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑥 21
So vertical force = 259/21 = 12.3 kN
A similar calculation can be performed by
taking moments around the RHS of the
bridge.
(b) In equilibrium upward force = Downward
force
22 𝑘𝑁 = 12.3 𝑘𝑁 + 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑅𝐻𝑆 𝑜𝑓 𝑏𝑟𝑖𝑑𝑔𝑒.
Force = 9.7 kN
(b) The weight of the bridge has been ignored
so upward forces will be larger.
Answer 12.3 kN and 9.7 kN

3. Examples - Forces,density and pressure 3
Example- Non parallel forces
A sign of weight 50 N is hung from the end of
a uniform bar of length L and weight 20 N.
The bar is hinged to a wall and held
horizontal by a wire attached to the wall at
30° to the horizontal. Find
(a) the tension in the wire,
(b) the reaction at the hinge. (Dunc)
T
30°
20 N 50 N
Answer
R
T sin30
T
θ 30°
20 N 50 N
(a) There is no rotational acceleration; we
take moments about the hinge to exclude the
reaction force at the hinge.
(20 𝑥 𝐿/2) + (50 𝑥 𝐿) = 𝑇𝑠𝑖𝑛30 𝑥 𝐿
T= 120 N
(b) There is no translational acceleration;
resolve the tension and reaction into
horizontal and vertical components at each
end of the rod. There is no resultant force
vertically and horizontally. Divide.
Vertically:
𝑅 𝑠𝑖𝑛 𝜃 + 𝑇𝑠𝑖𝑛30 = 20 + 50
R sin θ + 120 sin 30 = 70
R sin θ = 10 (i)
Horizontally: 𝑅 𝑐𝑜𝑠 𝜃 = 𝑇 𝑐𝑜𝑠 30
R cos θ = 120 cos 30 = 103.9 (ii)
Dividing
𝑅 𝑠𝑖𝑛 𝜃/ 𝑅 𝑐𝑜𝑠 𝜃 = 10/ 103.9
tan θ = 0.0962
θ = 5.5̊
Substitute for θ:
𝑅 𝑠𝑖𝑛 5.5 = 10
R= 104 N
Answer: The tension in the wire is 120 N, and
the reaction at the hinge is 104 N at 5.5° to
the beam.
Example- Non parallel forces
X
C
R
S
4m
3 m F
200 N
A uniform ladder 5 m long of a weight 200 N
rests with its upper end against a smooth wall
and its lower end 3 m from the wall, on rough
ground. Calculate
(a) the frictional force F between the ladder
and the ground,
(b) the normal reaction S at the ground, and
at the wall R,
(c) the magnitude and direction of the force X
exerted at the bottom of the ladder.
Answer
There is no resultant vertical force,
so 𝑆 = 𝑊,
S = 200 N
Taking moments about the top of the ladder,
to exclude R:
( 𝐹 𝑥 4) 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 + (𝑊𝑥1.5) 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 =
(𝑆 𝑥 3) 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
(F x 4) + (200x1.5) = (200x 3)
F = 75 N
Taking moments about the bottom of the
ladder
(𝑅 𝑥 4) 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 = (𝑊 𝑥 1.5) 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
(R x 4) = (200 x 1.5)
R = 75 N

4. Examples - Forces,density and pressure 4
The line of action of the weight meets R at
point C. At this point, R and W have zero
moment (no turning effect).
X must pass through C to have zero moment,
so the ladder can be in equilibrium. (X is also
equal to the resultant of S and F).
The triangle of forces for the ladder is shown.
Using Pythagoras’s theorem,
200 N X
θ
R = 75 N
X2 = 2002 + 752
X = 213.6 N
tan θ = 200/75
θ = 69.4°
Answer: The force X exerted at the bottom of
the ladder = 214 N at 70° to the horizontal