Sunday, January 17, 2010

Discussion Question

Question 1

Pascal's Principle

The hydraulic car lift relates to Pascal's Principle. It’s used in garages to lift a car off the ground for repairing. A small force by a small-area piston can be converted to a large force at a large area piston. It works relatively close to a lever, where a small force passes through a great distance to transport a heavy object a short distance. The work is the same when lifting the heavy object or when applying a small force. In the case of a lever a bar and a fulcrum convert the work, in the hydraulic lift the fluid performs the work.

Question – A hydraulic car lift has a pump piston with radius r1 = 0.0120 m. The
piston has a radius of r2 = 0.150 m. The total weight of the car and
plunger is F2 = 2500 m. If
the bottom ends of the piston and plunger are at the same height, what
input force is
required to stabilize the car and output plunger?


We need to use the area for circular objects, A=3.14r2 for both the piston
and plunger. Apply Pascal's Principle:

F1= F2(A1/A2) = F2(3.14r2/3.14r2) = (20 500N)[(0.01202)/(0.1502)] = 131 N

Question 2


A standard 1 kilogram weight is a cylinder 48.5 mm in height and 51.5 mm in diameter. What is the density of the material?


pi= 22/7,

h = 48.5
mm = 4.8
cm and d = 51.5 = 5.15
Volume = (1/4) x (pi) x [d²] x h = 101 cc = 101/ (100)³ m³
mass is 1 kg hence density = mass/volume = 1/ [101/ (100)³] kg/m³= 9894.16 kg/m³

Question 3

Archimede's Principle and Buoyancy Force
A large beaker of water is placed on a scale, and the scale reads 40 N. A block of wood with a weight of 10 N is then placed in the beaker. It floats with exactly half of its volume submerged. Assuming that none of the water spilled out of the beaker when the block was added, what does the scale read now?


draw the free body diagram of the block of wood,since the wood is submerged to half of its volume it means that the bouyance force is balancing its weight as per newtons third law since the water is excerting a force of 10N on the block theblock should excert 10N force on the water(that force will not be called bouyance force).

so the reading of the scale would be = weight of the beaker+force on the beaker =50N

Question 4


The deepest point in any ocean is in the Mariana Trench, which is about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 X 10⁸N/m².

Calculate the change in volume of 1 m3 of seawater carried from the surface to this deepest point.


B = -ΔP/ [(ΔV/Vi)] where:
B = bulk modulus, ΔP = change in pressure, ΔV = change in volume and Vi = initial volume.

When you use 0.21 X 10¹º for the bulk modulus of water (which apparently the online assignment website wants this for water and not seawater), you get:

ΔV = -(1.13 X 10⁸)(1)/(0.21 X 10¹º) = -0.0538 m²

Question 5

Why is a mercury barometer used as a scientific standard for measuring atmospheric pressure rather than an aneroid barometer on the basis of what advantages of an aneroid barometer in general use


1. Barometers are instruments to measure atmospheric pressure. There are two main types of barometers - mercury and aneroid. The mercury barometers measure the atmospheric pressure by means of height of mercury column pushed up a glass tube by the atmospheric. The aneroid barometer works on the basis of expansion and contraction of a metal chamber under the influence of atmospheric.
The mercury barometer is more accurate but less sensitive. That is though the mercury barometer indicates more accurately the absolute atmospheric pressure, it is not suitable for detecting very small changes in the pressure. Not only that the mercury barometer measures atmospheric pressure using using the primary effect of of atmospheric pressure that defines it. In comparison, the aneroid barometer measures atmospheric pressure based on its effect in compressing a metal chamber. Thus while the atmospheric pressure can be directly measured by a mercury barometer, in case of aneroid barometer it needs to be graduated by comparing the pressure shown by it against pressure known by other methods. Thus the aneroid barometer itself depends on mercury barometer for its correct graduation

Question 6
A tank 4m long, 3m wide, and 2m deep is filled to the brim with paraffin (density 800 kg/m^3). Calculate the pressure on the base. What is the thrust (force) on the base


The area of the base is 4m x 3m = 12m².
The volume of the paraffin = volume of the tank = length x breadth x depth = 4×3×2 = 24m³
Therefore, the mass of the paraffin = volume of paraffin×its density = 24 m³×800kg/m³ = 19200kg.
So the force on the base due to the weight of paraffin = 19200g N, where g is the acceleration due to gravity, and its value is assumed here to be 9.81 m/s².
Therefore, the pressure on the base due due to the weight of paraffin = 19200×g N = 19200×9.81 N/Base area = 19200×9.81 N/12 = 15696 N/m² = 15696 pascals. If you want to include the standard atmopsheric pressure to this , then the pressure is = (15696+101,325) Pa = 117021 pa.

Question 7

In finding the density or relative density of a liquid, why is a method using a density bottle more accurate than one involving the use of measuring cylinder?


In practical terms measuring the density of a liquid has been accomplished easily by use of a hydrometer. Early brewers needed to know the ratio of water with sugars vrs alcohol and the amount of conversion accomplished by yeast. To answer your question there is a greater chance of inaccuracy using measuring cylinders for density then the use of a hydrometer. For one thing you won't need a scale to measure weight.

Question 8

A 12N force & B 5N force both act at the same point. What are the greatest resultant, least resultant & if the two forces act at right angles to each other, the size and direction of their resultant.


1. We represent the forces,12N and 5N acting at apoint O, as adjascent sides of a parallelogram. Then the resultant force is represented by the diagonal of the parallelogram from the point of acting of forces both in magnitude and direction.If A and B are the forces acting at a point , then by the law of parallelogram of forces, the resultant force F is given by:
F = (a²+b²+2ab cos x)½ and is in the direction of the diagonal from the point of acting of forces.Here A= 12N, B=5N
let x be the angle between the forces.
F = [12²+5²+2(12)(5) cos x]½
N= (169+120cos x)½ Newtons. The resultant force is maximum when angle x= 0 and F = 12+5 =17 N and minimum when x= 180 and F = 12-5 = 7N.
When x= 90 dgree,
F = (12²+5²+120×cos90)^(1/2) = 13N along the diagonal from O which makes an angle of Tangent inverse (5/12) =22.6199 degree with 12N force and Tangent inverse(12/5) = 67.3801 degree with 5N force

Question 9

What is the volume of 2400 kg of petrol?


Petrol is a mixture of several hydrocarbons, generally called alkanes having general molecular formula, Cn H2n+2. Generally, pentane (C5H12) to octane(C8H18) mixtures refined is gasoline in liquid form and volatile used for fuel in vehicles. It has a coefficient of volume expansion of 0.00075/degree centigrade. Therefore, 2400 kg petrol (gasoline) at 20 degree centigrade, havind density of 737.22 kg/cubic meter has a volume of 2400 kg/(737.22kg/cubic meter) = 3.25547 cubic meter.

Question 10

A piece of wax weighs 18g in air. It is completely immersed in water by tying it to a sinker, and together they weigh 28g. Given that the sinker alone weoghs 30g when it is completelyimmersed in water, find the density of the wax?


The buoyant mass, which is the mass weighted from the completely immersed object, behaves the absolute mass. Thus buoyant mass of piece of wax and the sinker is equal to sum of the buoyant mass of each object.
m_b = m_b,wax + m_b,sinker
28g = m_b,wax + 30g
m_b,wax = -2g

The buoyant mass of an object and its mass in air are related by:
m_b = m_object • ( 1 - ρ_fluid / ρ_object)

m_b,wax = m_wax • ( 1 - ρ_water / ρ_wax)
ρ_wax = ρ_water / (1 - m_b,wax/m_wax)
= 1000kg/m³ / (1 - (-2g)/18g)
= 818.18kg/m³

Question 11

A car of mass 1100kg tows a caravan of mass 750kg. The total resistance to their motion has a constant value?
continue with the question:

of 2000N and a half of this acts on the caravan. Calculate the force exerted on the caravan by the tow bar

(a) when the acceleration of the car and caravan is 1.0 m/s²
(b) when the car and caravan move at constant speed of 10m/s¹


• Always do these with a free body force diagram.
b) Draw a free body force diagram of the caravan. A rectangle with a force arrow for T, tension of tow bar, coming out of one side in the direction of motion. On the opposite side draw a force arrow for resistance of motion and label it 1000N. At the top draw a double headed arrow to represent acceleration, point it in the direction of motion and label it 1 m/s^2.

Now, Resultant Force, F = m × a .... (1)

Resultant Force, F = T - 1000

You already got mass = 1100kg and acceleration = 1 m/s² so substitute F,m and a into equation (1). Therefore:

T - 1000 = 1100 ×1
T = 1100 + 1000
T = 2100N

b) If they move at constant speed, acceleration is 0 and there is no resultant force. Therefore, the force exerted by the tow bar on the caravan is being cancelled out by the resistance to motion which is 1000N on the caravan.

So tow bar must be exerting 1000N in the opposite direction to cancel it out the resistance to motion.

Question 12

Density question (physics)?
Imagine a rectangular solid having a height of 1.20 cm and a density of 8.0 103 kg/m3 floating in an unknown liquid. Given that the block is submerged to a depth of 1.16 cm, find the density of the liquid.

• apparent immersed weight = weight - weight of displaced fluid

density of object/ density of fluid = weight / (weight - appearent immersed weight)

im going to solve this assuming the solid is a square not a rectangle, even though that might not help you out completely

the weight of the solid = 8.0103 kg/m3× (1.2E-2 m)³ = 1.384E-5 kg

the appearent immersed weight can be found using the concept of buoyancy, where the force due to gravity is equal to the force pushing up by the liquid

simplifyin you can say

m= pV

where p = density and V equals the displaced volume
and m equals the appearent weight

Question 13

Pressure question (physics)?
A rectangular tank 2.0 m by 2.0 m by 3.5 m high contains gasoline, with a density of 0.68 103 kg/m3,to a depth of 2.5 m. What is the gauge pressure anywhere 1.0 m below the surface of the gasoline?


The size of the tank does not matter.

If the weight of a cubic meter is 0.681 then that means it weighs .681 1 meter below the surface. So the pressure is .681 kg/meter2

Question 14

Stretching a wire. A steel wire 1.00 m long with a diameter d = 1.00 mm, 10.0-kg mass hung from it.
(a) How much will the wire stretch?
(b) What stress on the wire?
(c) What is the strain?


(a) The cross-sectional area of the wire is given by

A= πd²/4 = π (1.00 X 10-³m)²/4 = 7.85 X 10-⁷ m²

We assume that the cross-sectional area of the wire does not change during the stretching
Process. The force stretching the wire is the weight of the 10.0-kg mass, that is,

F= mg = (10.0 kg)(9.80 m/s²) = 98.0 N

Young’s modulus for steel is found in table 10.1 as Y = 21 X 10¹° N/m
The elongation of the wire, found from modifying equation 10.6, is

∆L = FL./AY

= (98.0 N) (1.00 m)/(7.85 X 10-⁷ m²) (21.0 X 10¹° N/m²)

=0.594 X 10-³ m = 0.594 mm

(b) The stress acting on the wire is
F/A = 98.0 N/(7.85 X 10-⁷ m²) = 1.25 X 10⁸ N/m²

(c) The strain of the wire is
∆L/L. = 0.594 X 10-³ m/1.00m = 0.594 X 10-³

Question 15

Elasticity of shear. A sheet of cooper 0.750 m long, 1.00m high, and 0.50 thick is acted on by a tangential force of 50,000 N, as shown in figure 10.9. value of S for cooper is 4.20 X 10¹° N/m². find (a)the shearing stress, (b) shearing strain, (c) the linear displacement ∆x.


(a) The area that the tangential force is acting over is
A= bt = (0.750 m) (5.00 X 10-³ m)
= 3.75 X 10-³ m²

Where b is the length of the base and t is thethickness of the copper wire shown in figure 10.9. the shearing stress is

Ft/A = 50,000 N /3.75 X 10-³ m² = 1.33 X 10⁷ N/m²

(b) The shearing strain, found from equation 10.15, is

Ø = Ft/AS = 1.33 X 10⁷ N/m²/4.20 X 10¹° N/m²
= 3.17 X 10-⁴ rad

(c) The linear displacement ∆x, found from equation 10.10, is

∆x = hø = (1.00 m) (3.17 X 10-⁴ rad)

=3.17 X 10-⁴ m
= 0.317 mm


1. An iron rod 4.00m long 0.500cm in cross section strenches 1.00 mm when a mass of 225kg is hung from its lower end. Compute young’s modulus for the iron.

L.= 4.00 m
y= FL.

225x9.81 = 2207.25 N
∆L.= 1x10-³m

F= mg
F= 225x9.81
= 2207.25 N

Y= 2207.25 (4.00)
[0.5x(10-²)² m²] [1x10-³] m
= 1.7658 x10¹¹
= 176.58 x10¹¹x10-²
= 176 x10⁹ Pa
= 176 GPa

2. A load of 50 kg is applied to the lower end of a steel 80cm long and 0.6cm in diameter. How much will the road stretch? Y= 190GPa for steel.

m = 50kg x 9.81
d = 0.6cm ÷ 2
= 0.3
J² 3X10-³ m

A = πJ²
=π (3x10-³)²
=2.83 x 10-⁵

L. = 80cm
Y = 190 GPa

AL. = FL.
=(490.5) (0.8)
(190x10⁹) (2.83x10⁵)
= 392.4

3. A platform is suspended by four wires at its cover. The wires are 3.0m long and have a diameter of 2.0mm. Young’s modulus for the material is 180GPa. Howfar will the platform drop ( due to elongation of the wires) if a 50kg load is placed at the center of platform?

Y= 180 GPa
=180x10² Pa
D= 2mm ÷ 2
=1x10-³ m

Y= FL. AL. = FL.

L. = 3m F= 50x9.81
=490.5 N
A= πJ² = (3.14x10-⁶) m
AL. = 490.5x3m²
(180) (3.14x10-⁶)
=1471.5 Nm
=2.6x10⁶ m
=0.65 mm

4. Two parallel and apposite forces, each 4000N, are applied tangentially to the upper and lower faces of a cubical metal black 25cm on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus for the metal is 80GPa.

Tan Q =(25×10 ̄ 7/ 25× 10 ̄ 2) = 8.0 × 10 ̄ 7 rad.
Displament = 4000 / (2.8 × 10 ̄ ²)²
80 Gpa = 64000 / Ʃ
Ʃ = 64000 / 80 G
= 80 × 10 ̄ 7 (25 × 10 ̄ ² )
= 2 × 10 ̄ 7.

5. The bulk modulus of water is 2.1 GPa. Compute the volume control contraction of 100mL of water when subjected to a pressure of 1.5 MPa.

D = 2.1 Gpa
Vᴏ = 100 ml / 1000 =0.1 L
A = 1.5 mpa
AV = ( Aᴏ) ( Vᴏ) / B = ( 15 mpa ) ( 0.1 L) / 2.1 Gpa
= 15 000 /2.1 71.43 × 10 ³ = 0.0714 ml.

6. The compressibility of water is 5.0 x 10 . find the decreas in volume of 100mL of water when subjected to a pressure of 15 mPa.

Bulk strains = ∆Vᴏ / Vᴏ
AVᴏ = B × Nᴏ
= 15 × 10 6pa ( 15.0 × 10 ̄ 1 NM² )
= 7.5 × 10 ̄ ³
= 0.75 ml


1. Find the pressure due to the fluid at a depth of 76cm in still,
a) Water (Pω= 1.00g/cm )
b) Mercury (P= 13.6 g/cm )

P = ρgh
= ( 0.001 kg/m³) (9.81m/s²) ( 0.76m)
= 7.4556 kPa
= 7.5 kPa

( pm= 13.6g / cm³)
P = ρgh
= ( 0.0136 kg / m³) ( 9.81m/s²) (0.76m)
= 101.3962 kPa
= 101.4 kPa

2. A weighted piston confines a fluid density P in a closed container. The combined weight of piston and weight is 200N, and the cross- sectional area of the piston is A= 8.0 cm . Find the total pressure at point B if the fluid is mercury and h= 25cm. What would aan ordinary pressure gauge read mercury and h= 25cm. What would an ordinary pressure gauge read at B?

P = F / A + Pgh
= [200N/8 x 10 ̄4] + [(13600)(9.8m/s²)(0.25)]
= 250000+33354
=283,354 @ 283.4 kPa .
Pm =(13.6g/cm³)(9.81)(76)
= 0.136 kg/m (9.81)(76)
= 1.01 x 10 5 Nm
P = 283.4 kPa + Pm
= (283.4kPa) + (1.01 x 10 5 Nm)
= 384.4 kPa.

3. A vertical test tube has 2.0cm of oil (P=0.8g/cm ) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?

P 1 = ρgh
= 800 kg/m ( 9.8 m/s ̄ ² ) (2 x 10 ̄ ²)
= 156.96
P 2 = ρgh
= 1000 ( 9.8 m/s ²) ( 8 x 10 ̄ ²)
= 784
P1 + P2 = 156.96 + 784
= 0.94 kpa

4. The U-tube device conneted to the tank. What is the pressure in the tank it atmospheric pressure is 76cm of mercury? The density of mercury is 13.6 g/cm .

ρ = 76 cm – 13.6 g/cm³ (9.81 mg²)(5 x 10 ̄ ²)
= 76 cm – 13600 kg /cm³ ( 9.81 ms²)(5 x 10 ̄ ²)
= 101.325 kpa – 66708
= 95 kpa

5. The mass of a block of aluminium is 25.0g.
a) What is it’s volume?
b) What will be tension in a string that suspends the block when the block is totally submerged in water?
The density of aluminium is 2700 kg/m .

F = vg( e- pai)
= (9.26 x 10̄̄ ̄6 x 9.8 ) ( -1700 )
=9.0748 x 10 ̄ 5 (-1700)
= 0.154N
V = m/p
= 0.025/ 2700
= 9.259 x 10 ̄ 6
= 9.26 X 10 ̄ 6

6. A solid aluminium cylinder with P= 2700kg/mpadu has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB = Ví , Pí , a
PT = FB / Vi a
FB + Fr = mg
FB = mg – Fr
= ( 0.067 x 9.81 ) – ( 0.045 x 9.81 )
= 0.65727 – 0.44145
= 0.21582
V1 = M / P = 0.067 / 2700
= 2.481 x 10 ̄ 5
P1 = 0.21582
= ( 2.481 x 10 ̄ 5 ) ( 9.81 m / s²)
= 8.9 x 10 ² kg / m³