TUTORIAL

ELASTICITY


1.An iron rod 4.00m long and 0.500cm²in cross section stretches 1.00mm when amass of 225 Kg is hung from its lower end.compute young’s modulus for the iron.

σ = F/A = (225 kg) (9.8m/s²) / (0.500 X 10⁻⁴ m²)
=4.41×10⁷ Pa
ε = ∆s/S∘
= 1.00 x 10 ⁻³m / 4.00 m
=2.5×10⁻⁴
Y= stress/strain
= σ/ε
=4.41x10⁷ / 2.5x10⁻⁴
=1.764×10¹¹
=176 Gpa


2. A load of 50 kg is applied to the lower end of steel rod 80 cm long and 0.60 cm in diameter, How much will the rod strectch?

m=50 kg×9.81 j²=3×10⁻³m ℓ∘ = 80cm
d=0.6cm÷2 A=πj² = 80×10⁻²
=0.3 =πj²(3×10⁻³)² = 0.8
=2.83×10⁻⁵m y=190kPa
Aℓo=Fℓo / Y∆
= (4.90 x 0.8)/(190x 10⁹)(2.83x 10⁵)
=392.4 / 5.377 x 10⁸
=72.97×10⁻⁷
=73μm


3. A platform is suspended by four wires at its corner. The wires are 3.0m long and lave diameter of 2.0mm. Young’s modulus for material is 180GPa.How far will the platform drop (due to elongation of the wires)if a50kg load is placed at the center of the platform?

Y=180GPa
=180×10⁹ Pa
D = 2mm÷2
=1×10×⁻³
Y = Fℓo / YA
ℓo=3m A=πj²
F=50×9.81 = (3.14×10⁻⁵)m
=490.5N
Aℓo=490.5N (3m²) / (180)(3.14 x 10 ˉ⁶)
=1471.5Nm/5.652x 10⁻⁴
=2.6×10⁶m
=2.6/4
=0.65mm

4. Two parallel and opposite forces,each 4000N,aree aplied tangentialy to the upper and lower faces of a cubical metal block 25cm on a side.Find the angel of shear and the displament of the upper surface relative to the lower surface.The shear modulus for the metl is 80GPa.

tanθ=2 x 10ˉ⁷ / 25 x 10⁻²
= 8.0×10⁻⁷ rad
Displament=4000 / (2.5 x 10ˉ²)
80GPa = 64000/∑
∑=64000/80G
=8×10⁻⁷(25×10⁻²)
=2×10⁻⁷

5.The bolk moduls of water is 2.1 Gpa. Compute the volume comtraction of 100ml of water when subjected to a pressure of 1.5 Mpa.

B= 2.1GPa
Vo= 100mL / 1000
=0.1 mL
∆p = 1.5Mpa
∆v = (∆p)(Vo) / B
=(1.5 Mpa)(0.1mL) / 2.1 GPa)
=150000 / 2.1GPa
= -71.43×10³
= -0.0714mL

6. The compressibility of water is 5.0×10⁻¹⁰m²/N,find the decrease in volume of 100ml of water subjected to a pressure of 15mPa.

Bulk strains = ∆Vo / Vo
AVo=B×VoA
=15×10⁶Pa(5.0×10⁻¹Nm⁻²)
=7.5×10⁻³
=0.75mL

TUTORIAL

HYDROSTATIC


1. Find the pressure due to the fluid at a depth of 76cm in still
Water (pw=1.00g/cm³)

P=pgh
=(0.001kg/m³)(9.8m/s²)(0.76m)
=7.448kpa = 7.5 kPa
(b)mecury.(p=13.6 g/cm³)
P=pgh
=(0.0136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵N/m²=101.3×10³ N/m²

2. A weight piston confines afluid density p in a closed container. The combined weight of piston and weight is 200N, and the cross-sentional area of piston is A=8.0cm².Find the total pressure at point B if the mercury and h=25cm.What would an ordinary pressure gouge read at B?

P = F/A + Pgh (2000/8 x 10⁻⁴)+13600(9.8m/s²)(25×10⁻²)
=2500000+33354
= 2.8×10⁵Pa
P=2.8×10⁵Pa+Pm
=2.8×10⁵Pa+1.01×10³Nm
=3.8×10⁵Pa
Pm=(13.6g/cm³)(9.81)90.76)
=0. 136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵Nm

3. A vertical test tube has 2.0 cm of oil (p=0.8g/cm³)flooting 8.0cm of water. What is the pressure at botton of the tube due to the fluid in it ?

P₁=Pgh
=800kg/m (9.81m/s²)(2×10⁻²)
=156.96
P₂=Pgh
=1000kg/m (9.81m/s²)(8×10⁻²)
=784.8
P₁+P₂ = 156.96+784.8
=0.94 kPa

4. The u-tube devuce connected to the tank. Wahat is the pressure in the tank if atmosphera pressure is 76cm of mercury? The density of mercury is 13.6g/cm³.

P= 76cm – 13.6g/cm³(9.81ms²)(5×10⁻²)
= 76cm – 13600kg/cm³(9.81ms²)(5×10⁻²)
= 101.325 kPa -66708
=94.65×10³
=95 kPa

5. The mass of a block of aluminiumis 25.0g. (a) what is its volume?(b0what will be the tension in a string that suspends the block when the block is totally submerged in water?the density of aluminium is 2700kg/m²?

V=m/p ft=vg(e-pal)
=0.025/2700 =(9.26×10⁻⁶×9.81)(1000-2700)
= 9.26 =0.154N

6. A solid aluminium cylinder with p= 2700kg/m³ has a measured of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB=V₁P₁а
PT = FB / V1a
FB+Fr=mg
FB=mg-Fr
=(0.067×9.81)-(0.045×9.81)
=0.65727-0.4415
=0.21582
V1 = m/p = 0.067 / 2700
=2.481×10⁻⁵
P₁=0.21582
=(2.481×10⁻⁵)(9.81m/s²)
=8.9×10²kg/m³

5th WEEK:THERMAL PROPERTIES OF MATTER

QUESTION 1

1. A sheet of copper has an area of 500cm³ at 0˚C. Find the area of this sheet at 80˚C.

ANSWER 1

1. The coefficient of area expansion is to good approximation 2α. Thus,
∆A = 2αA ∆t = 2(1.67 x 10ˉ⁵)(500)(80˚ - 0˚)
= 1.34cm²
The new area is 500 + 1.34 = 501.3cm²

QUESTION 2

2. Determine the change is volume of a block of iron 5 cm x 10 cm x 6 cm, when the temperature changes from 15˚C to 47˚C.

ANSWER 2

2. Vo = 300cm³ ; β=3α. Then,
∆V = 3α Vo∆T = 3(1.2 x 10ˉ⁵˚C⁻ ¹) (300cm³) (32˚C)
= 0.35 cm³

QUESTION 3

3. The coefficient of linear expansion of glass is found. If a bottle holds 50.000cm³ at 15˚C, what is its capacity at 25˚C?

ANSWER 3

3. The glass expands uniformly in all dimensions, so the capacity expands in proportion to the glass. The volume coefficient of expands to the linear coefficient by β = 3α. Then 3α. Then,
V = Vo (1+3α ∆T )
= (50.000cm³)[1+3(8.3 x 10ˉ⁶ ˚C⁻ ¹)(10˚C)]
V = 50.012 cm³.

QUESTION 4

4. Calculate the increase in volume of 100 cm³ of mercury when its temperature changes from 10 to 35˚C.

ANSWER 4

4. ∆V = β Vo∆T. The volume coefficient of expansion of mercury can be gotton.
β = 1.8 x 10ˉ⁴˚C⁻ ¹. Then,
∆V = (1.8 x 10ˉ⁴˚C⁻ ¹)(100cm³)(25˚C)
= 0.45 cm³.


QUESTION 5

5. Compute the increase in length of 50m of copper wire when its temperature changes from 12 to 32˚C?

ANSWER 5

5. ∆L = α Vo ∆T
= (1.67 x 10ˉ⁵˚C⁻ ¹)(50m)(20˚C)
= 16.7mm.

4th WEEK: TEMPERATURE AND HEAT

QUESTION 1

1) Hydrogen may be liquefied at-253˚C under the pressure of 20 atm. What is temperature on the Fahrenheit scale?

ANSWER

tf = 1.8tc + 32
=1.8(-253)+32
= - 391˚

QUESTION 2

2) A rod 3m long is found to have expanded 0.091cm in length for a temperature rise of 60˚C. what is α for the material of the rod?

ANSWER

α = 1/L∘ (∆L)/(∆T)
= (0.091x 10⁻²m)/((3m)(60k))
=5.1 x 10⁻⁶K⁻ ¹


QUESTION 3

3) Starting at 20˚C, how much heat is required to heat 0.3kg of aluminum to its melting point and then to convert it all liquid?

ANSWER

Q = mc ∆t + mL. the specific heat of aluminum c is 0.22 kcal/kg. ˚C, heat of fusion L is 76.8 kcal/kg, and the melting point is 660˚C

Q = 0.3 (0.22) (660˚-20˚) + 0.3 (76.8)
= 42.24 + 23.04
=65.3kcal

QUESTION 4

4) A certain 6-g bullet melts at 300˚C and has specific heat capacity of 0.20cal/g. ˚C and heat of fusion of 15cal/g. how much heat to melt the bullet if it is originally at 0˚C?

ANSWER

The bullet must first be heated to 300˚C and then melted.
Heat needed = (6g) (0.20cal/g.˚C) (300-0)˚C + (6g) (15cal/g)
= 450cal, or 1880J

QUESTION 5

5) Refer to prob (4) .what is the slowest speed at wich the bullet can travel if it is to just melt when suddenly stopped?

ANSWER

1880J is to be supplied as
K= (mv²)/2
1880J =(1/2)(0.006kg)v²min
So that
Vmin= 790m/s.

3th WEEK:ELASTICITY & HYDROSTATIC

QUESTION 1

1. The height of the mercury column in barometer is 760 mm. Find the pressure of the atmosphere in pascals.

ANSWER 1

1. p = pgh
= ( 13.6 X 10 ³ kg/ m ³ ) ( 9.80 m/s ² )( 0.760 m)
= 1.013 X 10 ⁵ Pa.

QUESTION 2

2. A metal ball weighs 0.096N. When suspended in water it has an apparent weight of 0.071N. Find the density of the metal.


ANSWER 2 ;

2. The desired density is given by p = m/V. But since the volume V of the ball is also the volume of displaced water, the buoyant force is given by B = p water g V. Thus,
p = (mg)p water / B
= (0.096 N)(1 X 10 ³ kg/m ³ ) / ( 0.096-0.071) N
= 3840 kg/m ³ .

QUESTION 3

3. Air has a density of 1.29kg/m ³ under standard conditions. What is the mass of air in a room with dimensions
10m x 8m x 3m?


ANSWER 3


3. m = p V
V = (10 m)(8 m)(3 m) = 240 m ³
m = (1.29 kg/m ³ )(240 m ³ )
= 310 kg.

QUESTION 4

4. Approximately how large a force is required to stretch a 2.0 cm diameter steel rod by 0.01 percent?
Y= 195000 M Pa


ANSWER 4

4. F = AY( ∆ L/L )
= ∏(0.01) ² (195 X 10 ⁹ ) (10 ⁻⁴)
= 6100 N


QUESTION 5

5. A copper wire 2.0 m long and 2 mm in diameter is stretched 1 mm. What tension is needed? Young's modulus for copper is 117600 M Pa


ANSWER 5

5. F/A = Y ( ∆ L/ L)
F = Y ( ∆ L/ L) A
= (117.6 X 10 ⁹ ) 0.001 / 2 ( ∏ ( 0.001 ) ² )
= 184.7 N.

2nd WEEK:ELASTICITY

QUESTION 1

1. A load of 100 lb is applied to the lower end of a steel rod 3 ft long and 0.20 in a diameter, How much will the rod stretch?

ANSWER 1

1. Y = F/A ÷ ∆ L/L
= (3 ft)(100 lb) ÷ ∏ (0.1in ²) (3.3 X 10 ⁷lb /in ²)

= 2.9 X 10 ⁻⁴ ft or ∆ L = 3.5 x 10 ⁻³ in



QUESTION 2


2. A solid metal cylinder has r = 0.60cm and L = 8.0cm. Its mass is 77g. Find the density of the metal.


ANSWER 2


2. The cylindrical volume is ∏r L= ∏(0.60) ²(8.0) = 9.05cm ³ and density is p = m / V
= 71 / 9.05 = 7.85g / cm ³= 7850 kg / m ³


QUESTION 3

3. Find the density and specific gravity of ethyl alcohol if 63.3 g occupies 80.0 mL.


ANSWER 3

3. p = density = mass / volume = 63.3 g / 80.0 mL
= 0.791 g / mL = 791 kg / m ³
specific gravity = density of alcohol / density of water
= 791 kg / m ³ ÷ 1000 kg / m ³
= 0.791

QUESTION 4

4. Give a 2.0 m length of steel wire with 1.0 mm diameter, about how much will the wire stretch under a 5.0 kg load?
Y = 195000 M Pa.


ANSWER 4

4. Since tensile modulus is stress / strain, we have ∆ L/L = (F/A) /Y = (5(9.8) )/( ∏ (5 X 10 ˉ⁴) ²(195 x 10 ⁹) )
= 3.2 x 10 ˉ⁴ and ∆L = 6.4 x 10 ˉ⁴ m


QUESTION 5

5. What increase in pressure is required to decrease the volume of 200L of water by 0.004 percent? Find ∆ V.
( B = 2100 M Pa)

ANSWER 5

5. ∆V = 0.00004( 200L) = 0.008L
= ∆ p = B ( - ∆ V / V )
= (2100 M Pa) (0.008 / 200 L) = 0.084 M Pa = 84 k Pa.

1st WEEK:HYDROSTATICS

QUESTION 1

1. By how much does the pressure under the seawater for every 10 m increase in depth?

ANSWER 1:
1. p2 – p1 = ρsea g Δh
= (1025 kg/m3)(9.8 m/s2)(10 m)
= 100.45 N/m2

• In this question, the pressure p1 in the air above the water is already at 1 atmosphere= 1 atm = 101.3 kPa. So for the 10 m depth increase, the pressure in the water increases by very close to 1 atmosphere. Therefore, p2 = p1 + 100.45 kPa = 202 kPa.Usually, p2– p1 would be called the “gauge-pressure”(measured relative to 1 atm). While p2 by itself is called the “absolute pressure”.

pabs = patm + pgauge
• The SI unit of pressure is 1 pascal = 1 Pa = 1 N/m2 (a very small amount of pressure).
Also, 1 atm = 14.7 lbs/in2 = 760 mm-Hg. (height of mercury column supported by atmospheric pressure), so 1 mm-Hg = 133 N/m2.

QUESTION 2


2. A solid has a radius of 1.5 cm and a mass of 0.038kg. what is the specific gravity of the
sphere?

use 1.5cm= 1.5 x 10 ˉ² m for the radius

ANSWER 2;

2. V= 4/3 ∏R³ = 4/3 ∏ (1.5 x 10ˉ² )³ = 1.413 x 10ˉ⁵ m³
d = m/v = 0.038/1.413 x 10 ˉ⁵ = 2690 kg/m ³
specific gravity = 2690kg/m ³ ÷ 1000kg/m ³

QUESTION 3

3. An 80kg metal cyclinder 2 m long and with each end of area 25 cm ² stands vertically
on one end. what pressure does the cylinder exert on the floor?

ANSWER 3;

3. p = normal force/ area
= (80kg)(9.81 m/s ² ) / 25 x 10 ˉ ⁴ m ²

= 3.14 x 10 ⁵N/m ²

= 314 kPa


QUESTION 4

4. Find the pressure at a depth of 10m in water when the atmospheric pressure is that
corresponding to a mercury column of height 760mm. The densities of water and
mercury are 10 ³kg/m ³ and 13.6 x 10 ³kg/m ³.


ANSWER 4;

4. p = patm + pgy = phghgh + pgy
= (13.6 x 10 ³)(9.8)(0.760) + (10 ³)(9.8)(10)
= 1.99 x 10 ⁵ Pa = 199 kPa


QUESTION 5

5. If the blood vessels in a human being acted as simple pipes(which they do not), what
would be the difference in blood pressure between the blood in a 1.80-m- tall man's
feet and his head when he is standing? Assume the specific gravity of blood to be
1.06.

ANSWER 5.

5. Δp = pgh
= 1060(9.8)(1.8)
= 18.7 kPa