27:0:0ELASTIC PROPERTIES OF SOLID:
In solid, the molecules are very closely packed together and are held in relatively fixed position by strong intermuscular forces. As a result of this, it is difficult for an external force that is applied on the solid to displace the molecules of the solid due to the resistance of the intermuscular forces between the molecules.
ELASTICITY:
Elasticity is the ability of a substance to regain its original shape and size after being deformed by an external force or it is the ability of a substance to regains its original shape and size after the force that caused the deformation is removed.
ELASTIC MATERIALS:
Elastic materials are those materials that regains their original shape and size after the force that caused the deformation is removed.
DISTORTION OF MATERIALS:
Materials can be distorted by a force by stretching the materials or compressing the materials. The amount by which a material is stretched or compressed depends on the amount of force that is applied on the material. If large force or more force is applied, the material will stretch out / expands more or will compress / shorten more. If the force is small, the material will expand little or compress / shorten small or will not expand or compress.
HOOKE’SLAW:
Hooke’s law states that the extension, e, of a material is directly proportional to the force, F, that is applied on the material, provided the elastic limit is not exceeded.
FORMULA OF HOOKE’S LAW:
The statement of Hooke’s law can be translated into formula that is used for calculation thus:
Mathematically:
Force F is directly proportional to extension e
Force F ∞ extension, e
Force F = K * extension e . F = Ke
Recall that force F = mass * acceleration. F = m* a. ( for a mass moving along the horizontal surface)
F = mass * acceleration due to gravity. F = m * g ( when the object is lifted up from the ground. A changes to g)
K is constant of proportionality called elastic constant or force constant or stiffness of the .
UNIT OF ELASTIC CONSTANT:
To find the unit of K, we have to make K the subject of the formula thus:
Force constant K = Force F ( measured in newton ) / extension e ( measured in meter )
. K = Newton / meter or N / m or Nm^-1
Therefore, the unit of force constant is Newton per meter or Nm^-1.
DEFINITION OF ELASTIC CONSTANT K:
Elastic constant of a material is the force that is required to produce an extension of one meter. Or
You can define elastic constant of a material as the force per meter.
Elastic constant of a material is a measure of how stiff or tuff the material is. It can also help to suggest the amount of force that is required to stretch or compress a material.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*If a force of 0.75 N stretched a spring by 25cm, find the stiffness of the spring.
SOLUTION:
Data given in the question:
Force F = 0.75N, extension e = 25cm = 25 / 100 = 0.25m. Stiffness K = ?
Formula: Force F = stiffness K * extension e
Substitution: 0.75 N = K * 0.25 m
Make K the subject of the formula: K = 0.75 / 0.25 = 3Nm^-1
*A material of length 200mm is stretched by a force of 5.2N to 235mm. Calculate the elastic constant of the material.
SOLUTION:
Data given in the question:
Initial length of material L1 = 200mm = 200 /1000 = 0.2m, length when stretched L2 = 235mm = 235 / 1000 = 0.235m, force F = 5.2N.
Formula: Force F = elastic constant K * extension e. F = K * e
We have to find the extension e. Extension e = L2 - L1. e = 0.235 – 0.200 = 0.035m
Substitution: 5.2 N = K * 0.035
Make k the subject of the formula: K = 5.2N / 0.035 m. K = 148.57 Nm^-1.
The calculated value of K shows that the material is very stiff.
*A force of 4.0N stretches an elastic material by 30cm. What additional force will stretch the material 35mm? Assuming the elastic limit is not exceeded.
SOLUTION:
Data given in the question:
Force F1 = 4.0N, extension e1 produced by 4.0N = 30cm = 30 /100 = 0.3meter. Additional force F2 = 4 + F, extension e2 = 35mm = 35 / 100 = 0.35meter
Note:
In the question, we have two forces and two extensions. Use the F1 and e1 to calculate the elastic constant of the material.
Since it is the same material that is used in the second statement, you have to use the calculated value of the elastic constant K and the value of e2 to calculate F2. You then find the difference between F1 and F2 to know the additional force that will produce extension of 35cm.
Formula: Force F = elastic constant * extension e
Substitution: 4.0 N = K * 0.3 m
Make K the subject of the formula: K = 4.0N / 0.3m. K = 13.33 Nm^-1
Now, you use the this value of K and the value of e2 to calculate F2 thus:
F = K * e
Substitution: F2 = 13.33 Nm^-1 * 0.35 m
F2 = 4.67 N
To find the additional force, F2 - F1 = 4.67 – 4.0 = 0.67N. additional force = 0.67N.
Take note:
The problem can also be solved by combining the two formula to form one like this:
For the first part of the sentence. F1 = K * e1,
Make K the subject of formula. K = F1 / e1
Also, for the second part of the sentence, F2 = K * e2
Also make K the subject of formula: K = F2 / e2.
Since the same material is used in the sentences, the constant K is the same. The you have equate the Ks thus:
K = K.
Substitute for the Ks: F1 / e1 = F2 / e2
Substitute for the letters: 4.0 / 0.3 = F2 / 0.35
Make F2 the subject of formula: 4.0 * 0.35 = F2 * 0.3 . F2 = 4.0 * 0.35 / 0.3 ., F2 = 4.67N
Additional force F2 - F1 = 4.67 – 4.0 = 0.67N.
EXPERIMENT TO FERIFY HOOKE’S LAW:
AIM:
To verify Hooke’s law.
APPARATUS:
Stand and clamp, meter rule, spring, pan, 6 equal masses , Plasticine and optical pin.
SETUP DIAGRAM:
PROCEDURES:
Weigh and record the mass of the scale pan as W1
Attach one end of the spring to a scale pan and with plasticine, fix a pointer at the end of the srping and suspend the other end on the clamp.
Place a vertical scale ( meter rule ) by the side of the setup, over which the reading can be taken.
Read and record the initial reading of the pointer when no mass is added as Lo
Add a mass or weight ( W1 ) to the scale pan. Read and record the new position ( reading ) of the pointer as L1.
Find the extension e1 as e1 = L1 – Lo.
Add another weight to the previous one in the scale pan and call them W2.
Read and record the new position of the pointer as L2.
Find the extension e2 as e2 = L2 – Lo.
Repeat the experiment for 4 other masses.
Continue to add more weight to the scale pan, read and record the pointer reading and calculate the extension in each cases.
Carry out the experiment back ward by offloading the masses one after the other.
In cases, read and record the pointer position and find the contraction of the spring as the weight is being removed. Let the contraction of the spring be c.
Prepare the table of values as shown below:
S/N. Weight W (N). Pointer position for. Pointer position for. Average value. Extension e ( m )
increasing weight. decreasing weight. Of L ( cm ) 1
1 W1 L1.1 L2.1. ( L1.1 + L1.2 ) / 2. e1
2 W2. L2.1. L2.2. ( L2.1 + L2.2 ) / 2 e2
GRAPH OF HOOKE’S LAW:
The graph of Hooke’s law is obtained by plotting the force, N ( weigh ) on the vertical axis against extension e on the horizontal axis. This graph is call force extension graph. The graph is as shown below:
Graph here:
The graph is a straight line graph. The graph shows that the extension of the elastic material is directly proportional to the force acting on the material, provided the elastic limit of the material is not exceeded.
SLOPE OF GRAPH OF HOOKE’S LAW:
The slope of the graph of Hooke’s law represent the elastic constant of the material. Let us see it.
Slope = ∆ force or weight ( N ) / ∆ extension
Slope = change in force / change in extension
Slope = F2 - F1 / e2 - e1 . slope = F / e.
Recall from Hooke’s law that F = K * e. Then, K = F / e. If we compare the two formulae, you will see that slope represent the constant K. Slope = Elastic constant K
Therefore, slope of the graph of Hooke’s law represent the elastic constant of the material.
If the experiment is continued further, that is more weights are added, the elastic limit of the material would be exceeded. At this point, the extended of the material would no more be proportional to the force applied. Small increase in weight would produce large extension. The material would no more be able to return to it original length when the force is removed.
At this point, if the values of the forces ( weights ) are plotted against extension, the graph would not be a straight line graph rather the shape of the graph would be as shown below.
Graph here:
Explanation:
From point O to point P, Hooke’s law is obeyed. The extension is proportional to the force applied.
Point P is the proportionality limit of the material. the material obey Hooke’s law.
Point E is the elastic limit of the material. Up to this point, the elasticity of the material still holds. The object will return to its original size when the force is removed. Beyond the elastic limit, the material will losses its elasticity.
OE is called the elastic region. That is the material is still elastic in its behaviour. Beyond E, the wire extends rapidly when more force is applied. In this region, the material does not return to its original length.
Point Y is called the yielding point. At this point, the material loss all its elasticity permanently and become plastic. This change from elastic to plastic by the material is noticed a sudden and rapid extension when a when a small force is applied.
Point B is called the breaking point. Maximum extension is reach at this point. Any further addition of force to extend the material beyond this point will cause the material to brake or snap.
STRESS OR TENSILE STRESS:
Stress is the ratio of force that act on a material to the area of the material. It is the force per unit area of a material.
FORMULA OF STRESS:
From the definition of stress,
Stress = Force F / area A . Stress = F / A
UNIT OF STRESS:
The unit is derived by involving the unit of each quantities that are present in the formula thus:
Stress = Force F ( measured in Newton )/ area A ( measured in meter )
Unit of stress = Newton / meter² unit of stress = Newton per meter² = N/m² = Nm^-2
APPLICATION OF THE FORMULA OF STRESS:
EXAMPLES:
STRAIN OR TENSILE STRAIN:
Strain is the ratio of the extension of a material when a force act on it to the original length of the material. Strain has no unit because the meter which is the unit of extension cancels the meter which unit of length. I.e meter cancels meter.
FORMULA OF STRAIN:
From the definition,
Strain = extension e / original length l
Strain = e /L
APPLICATION OF THE FORMULA OF STRAIN TO SOLVE PROBLEMS:
EXAMPLES:
*Calculate the strain on a material of length 25 cm if a force of 5 N caused the length of the material to become e 37.2m.
SOLUTION:
Data given in the question:
Force F = 5 N , original length L1 = 25 cm = 25 /100 = 0.25m, street he’d length L2 = 37.2mcm = 37.2 / 100 = 0.372 m.
Extension e = L2 - L1 = 0.372 – 0.25 = 0.122 m
Formula: Strain = extension e / original length L
Substitution: Strain = 0.122 / 0.25 = 0.488
*The strain on a spring is 0.023. If the original length of the material is 3.2m, calculate the extension and the length of the material when it is stretched.
SOLUTION:
Data given in the question:
Strain = 0.023, original length L1 = 3.2 m, extension e = ? Stretched length L2 =?
Formula: Strain = extension e / original length L1
Substitution: 0.023 = e /3.2
Make e the subject of the formula: extension e = 1.023 * 3.2 = 0.074m
Extension e = L2 - L1. 0.074 = L2 – 3.2 . L2 = 0.074 + 3.2. . L2 = 3.274m
YOUNG MODULUS:
Hooke’s law can also be stated that the stress on a material is directly proportional to the strain on the material.
Mathematically:
Stress is directly proportional to strain
Stress = K * strain
Make k the subject of the formula: K = stress / strain. K is constant of proportionality. It is called Young Modulus.
Therefore, young modulus Y = stress / strain
Substitute for stress and strain: young modulus Y =( F/A ) / ( e / L )
Young modulus Y = F/A ÷ e/L. Y = F/ L*B ÷ e/L
APPLICATION OF THE FORMULA OF YOUNG MODULUS TO SOLVE PROBLEMS:
EXAMPLES:
WORK DONE IN SPRING AND ELASTIC MATERIALS:
Work is done on an elastic material of original length L when a force (F) is applied on the material to extend it by e (cm).
When the force is applied, the force increase from zero to maximum value F. Then the average force is (0 + F)/2 = F/2 = ½F
Going by the formula for calculating work done, which is
Work done = force * distance.
For the elastic material, force = Average Force F = ( 0 + F ) / 2 = F / 2 = ½F, distance = extension e.
Therefore,
Work done = F/2 * e = ½F * e
Recall from Hooke’s law that, F = K * e. You will have to substitute for force F in the above formula thus:
Work done = ½F * e = ½ * K * e * e . Work done = ½*K*e² = ½Ke²
Take note that the formula of work done can be modified base on the DAT that is given in a question.
From F = K*e, You can make e the subject of the formula and substitute in the formula of work done. Then, e = F/K.
Therefore, work done = ½F*e = ½*F*F/K work done = ½F²/K.
APPLICATION OF FORMULA OF WORK DONE TO SOLVE PROBLEMS:
EXAMPLES:
*A spring is stretched 25cm by a constant force of 21N. What is the work done on the spring?
SOLUTION:
Data given in the question:
Force F = 21N , extension e= 25cm = 25/100 = 0.25m, work done =?
Formula: work done = ½F*e
Substitution: work done = ½*21N*0.25m. work done = 2.63 Joules
*If the work done in stretching a material is 450J, calculate the extension of the material if applied was 12N. What is the stiffness of the material?
SOLUTION:
Data given in the question:
Work done = 450Joules, force F = 12N, extension e = ?, stiffness K = ?
Formula: work done = ½F*e
Substitution: 450 = ½*12*e
Make e the subject of the formula: e = 450*2/12
E = 75m
To calculate K, you use the formula, F = K*e
Substitution: 450 = K * 75
Make K the subject of the formula: K = 450/75. K = 6N/m
*Given that the force constant of a spring is 0.25Nm^-1, calculate the work done if a force of 120N act on the spring.
SOLUTION:
Data given in the question:
Force constant K = 0.25Nm^-1, force F = 12N, work done= ?
Formula: work done = ½Fe or
work done= ½Ke² or
work done = ½F²/K
Substitution: work done = ½F²/K = ½*12²/0.25 = 12²/2*0.25
work done = 288 Joules
ENERGY STORED IN AN ELASTIC MATERIALS:
When an elastic material is compressed or stretched, work is done on the material. The energy that is stored in a compressed elastic material or stretched elastic material is equal to the work done in compressing or stretching the material. The type of energy that is stored in the material is called elastic potential energy. Elastic potential energy of an elastic material can be defined as the ability of a compressed or stretched elastic material to do work.
FORMULA FOR CALCULATING ENERGY STORED IN A COMPRESSED OR STRETCHED ELASTIC MATERIALS:
Since the energy that is stored in an elastic material is equal to the work done in compressing or stretching the material, therefore,
Work done = energy stored
Applying the formula: work done = energy stored = ½Fe or work done = energy stored = ½Ke²
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*A spring of length 20cm is compressed by 0.02m. What is the energy stored in the spring if the force constant is . 350Nm^-1?
SOLUTION:
Data given in the question:
Length of spring = 20cm = 20/100 = 0.2m, compression e = 0.02m, force constant K = 350Nm^-1
Formula: energy stored = work done = ½F*e = ½K*e²
Substitution: energy stored = work done = ½*350*0.2²
energy stored = work done = 7.0 Joules.
*A stone of mass 35 g was let gone by a catapult which was stretched through 5cm. If the force constant of the catapult was 250Nm^-¹, calculate the velocity with which the stone leave the catapult.
SOLUTION:
Data given in the question:
Mass of stone = 35g = 35/1000 = 0.035kg, extension e = 5cm = 5/100 = 0.05m, force constant K = 250Nm^-¹
Take note that the question want us to calculate the velocity of the stone.
When a catapult is pulled, work is done. The work that is done in pulling the catapult becomes the potential energy stored in the catapult and the stone. When the catapult is released, the potential energy becomes the kinetic energy which moves the stone.
Therefore,
Work done in pulling catapult =elastic potential energy of catapult and stone = kinetic energy of the moving stone.
Formula: Work done = kinetic energy of stone
½K*e² = ½M*V²
Substitution: ½*250*0.05² = ½*0.035*V²
½ cancels ½: 250*0.05² = 0.035*V²
250*0.0025 = 0.035*V²
Make V the subject of the formula: 250*0.0025/0.035 = V²
V² = 17.215
V = √17.215 = 4.149 ms^-¹
Exercises:
In solid, the molecules are very closely packed together and are held in relatively fixed position by strong intermuscular forces. As a result of this, it is difficult for an external force that is applied on the solid to displace the molecules of the solid due to the resistance of the intermuscular forces between the molecules.
ELASTICITY:
Elasticity is the ability of a substance to regain its original shape and size after being deformed by an external force or it is the ability of a substance to regains its original shape and size after the force that caused the deformation is removed.
ELASTIC MATERIALS:
Elastic materials are those materials that regains their original shape and size after the force that caused the deformation is removed.
DISTORTION OF MATERIALS:
Materials can be distorted by a force by stretching the materials or compressing the materials. The amount by which a material is stretched or compressed depends on the amount of force that is applied on the material. If large force or more force is applied, the material will stretch out / expands more or will compress / shorten more. If the force is small, the material will expand little or compress / shorten small or will not expand or compress.
HOOKE’SLAW:
Hooke’s law states that the extension, e, of a material is directly proportional to the force, F, that is applied on the material, provided the elastic limit is not exceeded.
FORMULA OF HOOKE’S LAW:
The statement of Hooke’s law can be translated into formula that is used for calculation thus:
Mathematically:
Force F is directly proportional to extension e
Force F ∞ extension, e
Force F = K * extension e . F = Ke
Recall that force F = mass * acceleration. F = m* a. ( for a mass moving along the horizontal surface)
F = mass * acceleration due to gravity. F = m * g ( when the object is lifted up from the ground. A changes to g)
K is constant of proportionality called elastic constant or force constant or stiffness of the .
UNIT OF ELASTIC CONSTANT:
To find the unit of K, we have to make K the subject of the formula thus:
Force constant K = Force F ( measured in newton ) / extension e ( measured in meter )
. K = Newton / meter or N / m or Nm^-1
Therefore, the unit of force constant is Newton per meter or Nm^-1.
DEFINITION OF ELASTIC CONSTANT K:
Elastic constant of a material is the force that is required to produce an extension of one meter. Or
You can define elastic constant of a material as the force per meter.
Elastic constant of a material is a measure of how stiff or tuff the material is. It can also help to suggest the amount of force that is required to stretch or compress a material.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*If a force of 0.75 N stretched a spring by 25cm, find the stiffness of the spring.
SOLUTION:
Data given in the question:
Force F = 0.75N, extension e = 25cm = 25 / 100 = 0.25m. Stiffness K = ?
Formula: Force F = stiffness K * extension e
Substitution: 0.75 N = K * 0.25 m
Make K the subject of the formula: K = 0.75 / 0.25 = 3Nm^-1
*A material of length 200mm is stretched by a force of 5.2N to 235mm. Calculate the elastic constant of the material.
SOLUTION:
Data given in the question:
Initial length of material L1 = 200mm = 200 /1000 = 0.2m, length when stretched L2 = 235mm = 235 / 1000 = 0.235m, force F = 5.2N.
Formula: Force F = elastic constant K * extension e. F = K * e
We have to find the extension e. Extension e = L2 - L1. e = 0.235 – 0.200 = 0.035m
Substitution: 5.2 N = K * 0.035
Make k the subject of the formula: K = 5.2N / 0.035 m. K = 148.57 Nm^-1.
The calculated value of K shows that the material is very stiff.
*A force of 4.0N stretches an elastic material by 30cm. What additional force will stretch the material 35mm? Assuming the elastic limit is not exceeded.
SOLUTION:
Data given in the question:
Force F1 = 4.0N, extension e1 produced by 4.0N = 30cm = 30 /100 = 0.3meter. Additional force F2 = 4 + F, extension e2 = 35mm = 35 / 100 = 0.35meter
Note:
In the question, we have two forces and two extensions. Use the F1 and e1 to calculate the elastic constant of the material.
Since it is the same material that is used in the second statement, you have to use the calculated value of the elastic constant K and the value of e2 to calculate F2. You then find the difference between F1 and F2 to know the additional force that will produce extension of 35cm.
Formula: Force F = elastic constant * extension e
Substitution: 4.0 N = K * 0.3 m
Make K the subject of the formula: K = 4.0N / 0.3m. K = 13.33 Nm^-1
Now, you use the this value of K and the value of e2 to calculate F2 thus:
F = K * e
Substitution: F2 = 13.33 Nm^-1 * 0.35 m
F2 = 4.67 N
To find the additional force, F2 - F1 = 4.67 – 4.0 = 0.67N. additional force = 0.67N.
Take note:
The problem can also be solved by combining the two formula to form one like this:
For the first part of the sentence. F1 = K * e1,
Make K the subject of formula. K = F1 / e1
Also, for the second part of the sentence, F2 = K * e2
Also make K the subject of formula: K = F2 / e2.
Since the same material is used in the sentences, the constant K is the same. The you have equate the Ks thus:
K = K.
Substitute for the Ks: F1 / e1 = F2 / e2
Substitute for the letters: 4.0 / 0.3 = F2 / 0.35
Make F2 the subject of formula: 4.0 * 0.35 = F2 * 0.3 . F2 = 4.0 * 0.35 / 0.3 ., F2 = 4.67N
Additional force F2 - F1 = 4.67 – 4.0 = 0.67N.
EXPERIMENT TO FERIFY HOOKE’S LAW:
AIM:
To verify Hooke’s law.
APPARATUS:
Stand and clamp, meter rule, spring, pan, 6 equal masses , Plasticine and optical pin.
SETUP DIAGRAM:
PROCEDURES:
Weigh and record the mass of the scale pan as W1
Attach one end of the spring to a scale pan and with plasticine, fix a pointer at the end of the srping and suspend the other end on the clamp.
Place a vertical scale ( meter rule ) by the side of the setup, over which the reading can be taken.
Read and record the initial reading of the pointer when no mass is added as Lo
Add a mass or weight ( W1 ) to the scale pan. Read and record the new position ( reading ) of the pointer as L1.
Find the extension e1 as e1 = L1 – Lo.
Add another weight to the previous one in the scale pan and call them W2.
Read and record the new position of the pointer as L2.
Find the extension e2 as e2 = L2 – Lo.
Repeat the experiment for 4 other masses.
Continue to add more weight to the scale pan, read and record the pointer reading and calculate the extension in each cases.
Carry out the experiment back ward by offloading the masses one after the other.
In cases, read and record the pointer position and find the contraction of the spring as the weight is being removed. Let the contraction of the spring be c.
Prepare the table of values as shown below:
S/N. Weight W (N). Pointer position for. Pointer position for. Average value. Extension e ( m )
increasing weight. decreasing weight. Of L ( cm ) 1
1 W1 L1.1 L2.1. ( L1.1 + L1.2 ) / 2. e1
2 W2. L2.1. L2.2. ( L2.1 + L2.2 ) / 2 e2
GRAPH OF HOOKE’S LAW:
The graph of Hooke’s law is obtained by plotting the force, N ( weigh ) on the vertical axis against extension e on the horizontal axis. This graph is call force extension graph. The graph is as shown below:
Graph here:
The graph is a straight line graph. The graph shows that the extension of the elastic material is directly proportional to the force acting on the material, provided the elastic limit of the material is not exceeded.
SLOPE OF GRAPH OF HOOKE’S LAW:
The slope of the graph of Hooke’s law represent the elastic constant of the material. Let us see it.
Slope = ∆ force or weight ( N ) / ∆ extension
Slope = change in force / change in extension
Slope = F2 - F1 / e2 - e1 . slope = F / e.
Recall from Hooke’s law that F = K * e. Then, K = F / e. If we compare the two formulae, you will see that slope represent the constant K. Slope = Elastic constant K
Therefore, slope of the graph of Hooke’s law represent the elastic constant of the material.
If the experiment is continued further, that is more weights are added, the elastic limit of the material would be exceeded. At this point, the extended of the material would no more be proportional to the force applied. Small increase in weight would produce large extension. The material would no more be able to return to it original length when the force is removed.
At this point, if the values of the forces ( weights ) are plotted against extension, the graph would not be a straight line graph rather the shape of the graph would be as shown below.
Graph here:
Explanation:
From point O to point P, Hooke’s law is obeyed. The extension is proportional to the force applied.
Point P is the proportionality limit of the material. the material obey Hooke’s law.
Point E is the elastic limit of the material. Up to this point, the elasticity of the material still holds. The object will return to its original size when the force is removed. Beyond the elastic limit, the material will losses its elasticity.
OE is called the elastic region. That is the material is still elastic in its behaviour. Beyond E, the wire extends rapidly when more force is applied. In this region, the material does not return to its original length.
Point Y is called the yielding point. At this point, the material loss all its elasticity permanently and become plastic. This change from elastic to plastic by the material is noticed a sudden and rapid extension when a when a small force is applied.
Point B is called the breaking point. Maximum extension is reach at this point. Any further addition of force to extend the material beyond this point will cause the material to brake or snap.
STRESS OR TENSILE STRESS:
Stress is the ratio of force that act on a material to the area of the material. It is the force per unit area of a material.
FORMULA OF STRESS:
From the definition of stress,
Stress = Force F / area A . Stress = F / A
UNIT OF STRESS:
The unit is derived by involving the unit of each quantities that are present in the formula thus:
Stress = Force F ( measured in Newton )/ area A ( measured in meter )
Unit of stress = Newton / meter² unit of stress = Newton per meter² = N/m² = Nm^-2
APPLICATION OF THE FORMULA OF STRESS:
EXAMPLES:
STRAIN OR TENSILE STRAIN:
Strain is the ratio of the extension of a material when a force act on it to the original length of the material. Strain has no unit because the meter which is the unit of extension cancels the meter which unit of length. I.e meter cancels meter.
FORMULA OF STRAIN:
From the definition,
Strain = extension e / original length l
Strain = e /L
APPLICATION OF THE FORMULA OF STRAIN TO SOLVE PROBLEMS:
EXAMPLES:
*Calculate the strain on a material of length 25 cm if a force of 5 N caused the length of the material to become e 37.2m.
SOLUTION:
Data given in the question:
Force F = 5 N , original length L1 = 25 cm = 25 /100 = 0.25m, street he’d length L2 = 37.2mcm = 37.2 / 100 = 0.372 m.
Extension e = L2 - L1 = 0.372 – 0.25 = 0.122 m
Formula: Strain = extension e / original length L
Substitution: Strain = 0.122 / 0.25 = 0.488
*The strain on a spring is 0.023. If the original length of the material is 3.2m, calculate the extension and the length of the material when it is stretched.
SOLUTION:
Data given in the question:
Strain = 0.023, original length L1 = 3.2 m, extension e = ? Stretched length L2 =?
Formula: Strain = extension e / original length L1
Substitution: 0.023 = e /3.2
Make e the subject of the formula: extension e = 1.023 * 3.2 = 0.074m
Extension e = L2 - L1. 0.074 = L2 – 3.2 . L2 = 0.074 + 3.2. . L2 = 3.274m
YOUNG MODULUS:
Hooke’s law can also be stated that the stress on a material is directly proportional to the strain on the material.
Mathematically:
Stress is directly proportional to strain
Stress = K * strain
Make k the subject of the formula: K = stress / strain. K is constant of proportionality. It is called Young Modulus.
Therefore, young modulus Y = stress / strain
Substitute for stress and strain: young modulus Y =( F/A ) / ( e / L )
Young modulus Y = F/A ÷ e/L. Y = F/ L*B ÷ e/L
APPLICATION OF THE FORMULA OF YOUNG MODULUS TO SOLVE PROBLEMS:
EXAMPLES:
WORK DONE IN SPRING AND ELASTIC MATERIALS:
Work is done on an elastic material of original length L when a force (F) is applied on the material to extend it by e (cm).
When the force is applied, the force increase from zero to maximum value F. Then the average force is (0 + F)/2 = F/2 = ½F
Going by the formula for calculating work done, which is
Work done = force * distance.
For the elastic material, force = Average Force F = ( 0 + F ) / 2 = F / 2 = ½F, distance = extension e.
Therefore,
Work done = F/2 * e = ½F * e
Recall from Hooke’s law that, F = K * e. You will have to substitute for force F in the above formula thus:
Work done = ½F * e = ½ * K * e * e . Work done = ½*K*e² = ½Ke²
Take note that the formula of work done can be modified base on the DAT that is given in a question.
From F = K*e, You can make e the subject of the formula and substitute in the formula of work done. Then, e = F/K.
Therefore, work done = ½F*e = ½*F*F/K work done = ½F²/K.
APPLICATION OF FORMULA OF WORK DONE TO SOLVE PROBLEMS:
EXAMPLES:
*A spring is stretched 25cm by a constant force of 21N. What is the work done on the spring?
SOLUTION:
Data given in the question:
Force F = 21N , extension e= 25cm = 25/100 = 0.25m, work done =?
Formula: work done = ½F*e
Substitution: work done = ½*21N*0.25m. work done = 2.63 Joules
*If the work done in stretching a material is 450J, calculate the extension of the material if applied was 12N. What is the stiffness of the material?
SOLUTION:
Data given in the question:
Work done = 450Joules, force F = 12N, extension e = ?, stiffness K = ?
Formula: work done = ½F*e
Substitution: 450 = ½*12*e
Make e the subject of the formula: e = 450*2/12
E = 75m
To calculate K, you use the formula, F = K*e
Substitution: 450 = K * 75
Make K the subject of the formula: K = 450/75. K = 6N/m
*Given that the force constant of a spring is 0.25Nm^-1, calculate the work done if a force of 120N act on the spring.
SOLUTION:
Data given in the question:
Force constant K = 0.25Nm^-1, force F = 12N, work done= ?
Formula: work done = ½Fe or
work done= ½Ke² or
work done = ½F²/K
Substitution: work done = ½F²/K = ½*12²/0.25 = 12²/2*0.25
work done = 288 Joules
ENERGY STORED IN AN ELASTIC MATERIALS:
When an elastic material is compressed or stretched, work is done on the material. The energy that is stored in a compressed elastic material or stretched elastic material is equal to the work done in compressing or stretching the material. The type of energy that is stored in the material is called elastic potential energy. Elastic potential energy of an elastic material can be defined as the ability of a compressed or stretched elastic material to do work.
FORMULA FOR CALCULATING ENERGY STORED IN A COMPRESSED OR STRETCHED ELASTIC MATERIALS:
Since the energy that is stored in an elastic material is equal to the work done in compressing or stretching the material, therefore,
Work done = energy stored
Applying the formula: work done = energy stored = ½Fe or work done = energy stored = ½Ke²
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*A spring of length 20cm is compressed by 0.02m. What is the energy stored in the spring if the force constant is . 350Nm^-1?
SOLUTION:
Data given in the question:
Length of spring = 20cm = 20/100 = 0.2m, compression e = 0.02m, force constant K = 350Nm^-1
Formula: energy stored = work done = ½F*e = ½K*e²
Substitution: energy stored = work done = ½*350*0.2²
energy stored = work done = 7.0 Joules.
*A stone of mass 35 g was let gone by a catapult which was stretched through 5cm. If the force constant of the catapult was 250Nm^-¹, calculate the velocity with which the stone leave the catapult.
SOLUTION:
Data given in the question:
Mass of stone = 35g = 35/1000 = 0.035kg, extension e = 5cm = 5/100 = 0.05m, force constant K = 250Nm^-¹
Take note that the question want us to calculate the velocity of the stone.
When a catapult is pulled, work is done. The work that is done in pulling the catapult becomes the potential energy stored in the catapult and the stone. When the catapult is released, the potential energy becomes the kinetic energy which moves the stone.
Therefore,
Work done in pulling catapult =elastic potential energy of catapult and stone = kinetic energy of the moving stone.
Formula: Work done = kinetic energy of stone
½K*e² = ½M*V²
Substitution: ½*250*0.05² = ½*0.035*V²
½ cancels ½: 250*0.05² = 0.035*V²
250*0.0025 = 0.035*V²
Make V the subject of the formula: 250*0.0025/0.035 = V²
V² = 17.215
V = √17.215 = 4.149 ms^-¹
Exercises:
- A spring of length 25 cm is stretched to 35 cm by a load of 50N. What will be it length 2hen stretched by 100 N assuming the elastic limit is not exceeded?
- An elastic cord can be stretched to its elastic limit by a load of 2 N. If a 35 cm of the cord is extended 0.6 cm by a force of 0.5 N, what will be the length of the cord when the stretching for is 2.5 N?
- A spring balance designed for a maximum load of 400 N was used to weigh a load of 500 N. After the 500 N load had been removed, the pointer would not return to the zero Marion the scale. Why?
- A spring of force constant 1500 Nm^-¹ is acted upon by a force of 75 N. Calculate the potential energy stored in the spring.
- A catapult is used to project a stone. What is the energy conservation that took place.
- State Hooke’s law. Explain how you can verify it in the laboratory. A force 40 N stretched a wire through 3.0 cm. What force will stretch it through 5.0 cm and through what length will a force of 100 N stretch it? What assumption have you made?
- A stone of mass 5 g is projected with a catapult. if the catapult is stretched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released.
- Sketch the graph of the relation between the extension of a spiral spring and the load attached to it when it is gradually loaded up to the elastic limit. If the spring has a stiffness of 950 Nm^-¹, what work will be done in extending the string by 60 cm?
- How would you demonstrate that within some elastic limit, iron is perfectly elastic? With the aid of a graph, describe what is observed if the elastic limit is exceeded. What is elastic limit?
- Define each of the following: (I) strain (ii) stress (iii) young modulus (iv) what is the relationship between them? A spring of length l is stretched through a length e by a force F. If its cross sectional area is A and its young modulus is E, deduce the relationship between F, E, e and l.
- Define young modulus. What is the physical quantities one has to measure in order to determine the young modulus of a wire.
- State Hooke’s law of elasticity. An elastic material loaded with a piece of iron extends by 15.3cm in air. When the iron is fully submerged in water, the spring extends by 8.4 cm. Calculate the relative density of the metal ( Assume Hooke’s law is obeyed }.
Weldon
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