In the previous section we saw conservation of momentum. In this section we will see equilibrium of a particle
We will write the steps:
1. ‘Equilibrium of a particle’ refers to that state of the particle in which net force on it is zero
2. Consider the following situation:
• A person points to a particle and says that, ‘it is in equilibrium’.
• If his statement is verified, we can say this: Net force acting on that particle is zero
3. This can mean any one of the following two:
(i) No force is acting on the particle at all
(ii) Forces are acting, but they cancel each other out
4. Further, it may be noted that, ‘zero net force’ need not imply ‘no motion’
• The particle can be in motion even when there is no net force.
• In such cases, the particle will be in ‘uniform motion’. We have seen such cases in previous sections.
■ So, a particle in ‘uniform motion’ can also be said to be in equilibrium.
5. If there are only two forces $\mathbf\small{\vec{F_1}\: and \: \vec{F_2}}$ acting on a particle which is in equilibrium, we can write:
■ The resultant of the two forces is a null vector
• That is., the vector sum of the two vectors is zero
• That is., $\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
• We have already seen how to calculate vector sum in a previous section
6. Similarly, if a particle is in equilibrium under the action of 3 forces, we can write:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
7. The above equation for 3 forces can be rearranged in 3 different ways:
(i) $\mathbf\small{\vec{F_1}+\vec{F_2}=-\vec{F_3}}$
(ii) $\mathbf\small{\vec{F_1}+\vec{F_3}=-\vec{F_2}}$
(iii) $\mathbf\small{\vec{F_2}+\vec{F_3}=-\vec{F_1}}$
This means that, we can group 'any two' together. The vector sum of those two, should be equal and opposite to the third
8. This can be explained using an example
• Consider fig.5.27(a) below:
• Three forces are acting on a particle. If the particle is in equilibrium, then:
We can represent the forces by the sides of a triangle ABC. This is shown in fig.b
[If the particle is not in equilibrium, then we will not be able to form the triangle ABC]
• Also note that, strict order should be maintained while forming the triangle
♦ That is., the head of a vector should coincide with the tail of the next vector.
♦ Then only we will be able to form the triangle.
♦ See triangle method of vector addition
9. Note that, in fig.c, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_2}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_3}}$
♦ This is same as the analytical result in 7(i) above
• In fig.d, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_2}}$
♦ This is same as the analytical result in 7(ii) above
• In fig.e, the resultant of $\mathbf\small{\vec{F_2}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_1}}$
♦ This is same as the analytical result in 7(iii) above
We will write the steps:
1. Consider the following scenario. See fig.5.28(a) below:
• We see a particle in equilibrium
♦ Three forces are acting on it
♦ We know the details (magnitude and direction) about 2 forces
♦ We want the third force.
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{3x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{3y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{3x}}}$ and $\mathbf\small{\vec{F_{3y}}}$ will give the required $\mathbf\small{\vec{F_{3}}}$
Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.28(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side can be easily drawn
This is shown in red color in fig.c
• This third side is called the 'closing side'
♦ The length of the closing side will give the magnitude of the unknown force
♦ The direction of the closing side will give the direction of the unknown force.
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
3. In the above example, three forces were acting. The third force was unknown. We successfully calculated it. Let us see another case related to the same example:
• Two forces $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$ are acting on the particle. See fig.5.29(a) below:
• The particle is not in equilibrium
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium
• So, for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}-\vec{R_{x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$
Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.29(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction
• This is shown in orange color in fig.c
♦ The length of the orange vector will give the magnitude of the resultant
♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
■ Comparing figs.5.28 and 5.29, we find that: $\mathbf\small{\vec{F_{3}}=-\vec{R}}$
We will write the steps:
1. ‘Equilibrium of a particle’ refers to that state of the particle in which net force on it is zero
2. Consider the following situation:
• A person points to a particle and says that, ‘it is in equilibrium’.
• If his statement is verified, we can say this: Net force acting on that particle is zero
3. This can mean any one of the following two:
(i) No force is acting on the particle at all
(ii) Forces are acting, but they cancel each other out
4. Further, it may be noted that, ‘zero net force’ need not imply ‘no motion’
• The particle can be in motion even when there is no net force.
• In such cases, the particle will be in ‘uniform motion’. We have seen such cases in previous sections.
■ So, a particle in ‘uniform motion’ can also be said to be in equilibrium.
5. If there are only two forces $\mathbf\small{\vec{F_1}\: and \: \vec{F_2}}$ acting on a particle which is in equilibrium, we can write:
■ The resultant of the two forces is a null vector
• That is., the vector sum of the two vectors is zero
• That is., $\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
• We have already seen how to calculate vector sum in a previous section
6. Similarly, if a particle is in equilibrium under the action of 3 forces, we can write:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
7. The above equation for 3 forces can be rearranged in 3 different ways:
(i) $\mathbf\small{\vec{F_1}+\vec{F_2}=-\vec{F_3}}$
(ii) $\mathbf\small{\vec{F_1}+\vec{F_3}=-\vec{F_2}}$
(iii) $\mathbf\small{\vec{F_2}+\vec{F_3}=-\vec{F_1}}$
This means that, we can group 'any two' together. The vector sum of those two, should be equal and opposite to the third
8. This can be explained using an example
• Consider fig.5.27(a) below:
 |
| Fig.5.27 |
We can represent the forces by the sides of a triangle ABC. This is shown in fig.b
[If the particle is not in equilibrium, then we will not be able to form the triangle ABC]
• Also note that, strict order should be maintained while forming the triangle
♦ That is., the head of a vector should coincide with the tail of the next vector.
♦ Then only we will be able to form the triangle.
♦ See triangle method of vector addition
9. Note that, in fig.c, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_2}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_3}}$
♦ This is same as the analytical result in 7(i) above
• In fig.d, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_2}}$
♦ This is same as the analytical result in 7(ii) above
• In fig.e, the resultant of $\mathbf\small{\vec{F_2}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_1}}$
♦ This is same as the analytical result in 7(iii) above
So how do we put the above information to practical use?
1. Consider the following scenario. See fig.5.28(a) below:
 |
| Fig.5.28 |
♦ Three forces are acting on it
♦ We know the details (magnitude and direction) about 2 forces
♦ We want the third force.
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{3x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{3y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{3x}}}$ and $\mathbf\small{\vec{F_{3y}}}$ will give the required $\mathbf\small{\vec{F_{3}}}$
Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.28(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side can be easily drawn
This is shown in red color in fig.c
• This third side is called the 'closing side'
♦ The length of the closing side will give the magnitude of the unknown force
♦ The direction of the closing side will give the direction of the unknown force.
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
3. In the above example, three forces were acting. The third force was unknown. We successfully calculated it. Let us see another case related to the same example:
• Two forces $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$ are acting on the particle. See fig.5.29(a) below:
 |
| Fig.5.29 |
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium
• So, for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}-\vec{R_{x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$
Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.29(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction
• This is shown in orange color in fig.c
♦ The length of the orange vector will give the magnitude of the resultant
♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
■ Comparing figs.5.28 and 5.29, we find that: $\mathbf\small{\vec{F_{3}}=-\vec{R}}$
Another example:
1. Consider the following scenario. See fig.5.30(a) below:
• We see a particle in equilibrium
♦ Four forces are acting on it
♦ We know the details (magnitude and direction) about 3 forces
♦ We want the fourth force
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+\vec{F_4}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}+\vec{F_{4x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}+\vec{F_{4y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{4x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{4y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{4x}}}$ and $\mathbf\small{\vec{F_{4y}}}$ will give the required $\mathbf\small{\vec{F_{4}}}$
Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.30(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side can be easily drawn
This is shown in red color in fig.c
• This fourth side is called the 'closing side'
♦ The length of the closing side will give the magnitude of the unknown force
♦ The direction of the closing side will give the direction of the unknown force.
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
3. In the above example, four forces were acting. The fourth force was unknown. We successfully calculated it. Let us see another case related to the same example:
• Three forces $\mathbf\small{\vec{F_{1}}}$, $\mathbf\small{\vec{F_{2}}}$ and $\mathbf\small{\vec{F_{3}}}$ are acting on the particle. See fig.5.31(a) below:
• The particle is not in equilibrium
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$, $\mathbf\small{\vec{F_{2}}}$ and $\mathbf\small{\vec{F_{3}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium
• So for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}-\vec{R_{x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$
Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.31(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction
• This is shown in orange color in fig.c
♦ The length of the orange vector will give the magnitude of the resultant
♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
■ Comparing figs.5.30 and 5.31, we find that: $\mathbf\small{\vec{F_{4}}=-\vec{R}}$
1. Consider the following scenario. See fig.5.30(a) below:
 |
| Fig.5.30 |
♦ Four forces are acting on it
♦ We know the details (magnitude and direction) about 3 forces
♦ We want the fourth force
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+\vec{F_4}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}+\vec{F_{4x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}+\vec{F_{4y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{4x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{4y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{4x}}}$ and $\mathbf\small{\vec{F_{4y}}}$ will give the required $\mathbf\small{\vec{F_{4}}}$
Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.30(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side can be easily drawn
This is shown in red color in fig.c
• This fourth side is called the 'closing side'
♦ The length of the closing side will give the magnitude of the unknown force
♦ The direction of the closing side will give the direction of the unknown force.
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
3. In the above example, four forces were acting. The fourth force was unknown. We successfully calculated it. Let us see another case related to the same example:
• Three forces $\mathbf\small{\vec{F_{1}}}$, $\mathbf\small{\vec{F_{2}}}$ and $\mathbf\small{\vec{F_{3}}}$ are acting on the particle. See fig.5.31(a) below:
 |
| Fig.5.31 |
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$, $\mathbf\small{\vec{F_{2}}}$ and $\mathbf\small{\vec{F_{3}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium
• So for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}-\vec{R_{x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$
Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.31(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here)
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction
• This is shown in orange color in fig.c
♦ The length of the orange vector will give the magnitude of the resultant
♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
■ Comparing figs.5.30 and 5.31, we find that: $\mathbf\small{\vec{F_{4}}=-\vec{R}}$
• We saw the case when 3 forces are acting on a particle
• We saw the case when 4 forces are acting on a particle
■ Now we can write the general case:
A. To find the unknown force when n forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_n}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The closing side will give the unknown force
B. To find the resultant when (n-1) forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_{(n-1)}}-\vec{R}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The 'reverse of the closing side' will give the resultant force
■ This graphical method is known as: Polygon law of vector addition
It states that, if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.
Three forces are acting on a body as shown in fig.5.32(a) below. If the body is in equilibrium, find the magnitude of $\mathbf\small{\vec{F_2}\: \: \text{and}\,\,\vec{F_3}}$
Solution:
1. Since the body is in equilibrium, we have:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
2. Substituting the values, we get:
(i) $\mathbf\small{0-(|\vec{F_2}|\cos 60)\hat{i}+(|\vec{F_3}|\cos 30)\hat{i}=0}$
$\mathbf\small{\Rightarrow |\vec{F_2}|=\sqrt{3}\left [|\vec{F_3}|\right ]}$
(ii) $\mathbf\small{-100+(|\vec{F_2}|\sin 60)\hat{j}+(|\vec{F_3}|\sin 30)\hat{j}=0}$
$\mathbf\small{\Rightarrow \sqrt{3}\left [|\vec{F_2}|\right ]+|\vec{F_3}|=200}$
3. Solving 2(i) and 2(ii), we get:
$\mathbf\small{|\vec{F_2}|=50\sqrt{3}\: N \,\,\text{and}\,\,|\vec{F_3}|=50\: N}$
Solved example 5.15
Three forces acting on a body are shown in the fig.5.32(b) above. What is the minimum additional force to be applied so that, the resultant of the four forces will be along the vertical direction only?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
1. Vector sum of the horizontal components of the 3 forces already present:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}}$
= $\mathbf\small{-(4 \times \cos 60)\hat{i}+(2 \times \cos 60)\hat{i}+(1 \times \cos 60)\hat{i}}$
= $\mathbf\small{-0.5 \hat{i}}$
2. So the available resultant in the horizontal direction is $\mathbf\small{-0.5 \hat{i}}$
That is., in the horizontal direction, a force of magnitude 0.5 N is acting towards the left
3. If we apply an additional force of magnitude 0.5 N in the horizontal direction, towards the right, then there will be no net force in the horizontal direction
• The resultant of the 4 forces will then be in the vertical direction only
Solved example 5.16
Two forces of equal magnitude are acting on an object. The angle between them is 60o. If the magnitude of their resultant is 40√3 N, What is the magnitude of each force?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
1. Let the magnitude of each force be 'p' N
• Also let one of the forces be towards the positive side of the x axis
• This is shown in fig.5.32(c) above
2. We have:
$\mathbf\small{\vec{R_x}=p\: \hat{i}+(p \times \cos 60)\hat{i}=\frac{3p}{2}\: \hat{i}}$
$\mathbf\small{\vec{R_y}=0+-(p \times \sin 60)\hat{j}=-\frac{p\sqrt{3}}{2}\: \hat{j}}$
3. We have:
• Magnitude of the resultant = $\mathbf\small{|\vec{R}|=\sqrt{|\vec{R_x}|^2+|\vec{R_y}|^2}}$
• Substituting the values, we get:
$\mathbf\small{|\vec{R}|=\sqrt{\left (\frac{3p}{2} \right )^2+\left (\frac{p\sqrt{3}}{2} \right )^2}=\sqrt{3p^2}=p\sqrt{3}}$
4. But given that, magnitude of the resultant is 40√3 N
• So we can write: 40√3 = p√3
• Thus we get: Magnitude of each force = p = 40 N
• We saw the case when 4 forces are acting on a particle
■ Now we can write the general case:
A. To find the unknown force when n forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_n}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The closing side will give the unknown force
B. To find the resultant when (n-1) forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_{(n-1)}}-\vec{R}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The 'reverse of the closing side' will give the resultant force
■ This graphical method is known as: Polygon law of vector addition
It states that, if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.
Solved example 5.14
 |
| Fig.5.32 |
1. Since the body is in equilibrium, we have:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
2. Substituting the values, we get:
(i) $\mathbf\small{0-(|\vec{F_2}|\cos 60)\hat{i}+(|\vec{F_3}|\cos 30)\hat{i}=0}$
$\mathbf\small{\Rightarrow |\vec{F_2}|=\sqrt{3}\left [|\vec{F_3}|\right ]}$
(ii) $\mathbf\small{-100+(|\vec{F_2}|\sin 60)\hat{j}+(|\vec{F_3}|\sin 30)\hat{j}=0}$
$\mathbf\small{\Rightarrow \sqrt{3}\left [|\vec{F_2}|\right ]+|\vec{F_3}|=200}$
3. Solving 2(i) and 2(ii), we get:
$\mathbf\small{|\vec{F_2}|=50\sqrt{3}\: N \,\,\text{and}\,\,|\vec{F_3}|=50\: N}$
Solved example 5.15
Three forces acting on a body are shown in the fig.5.32(b) above. What is the minimum additional force to be applied so that, the resultant of the four forces will be along the vertical direction only?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
1. Vector sum of the horizontal components of the 3 forces already present:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}}$
= $\mathbf\small{-(4 \times \cos 60)\hat{i}+(2 \times \cos 60)\hat{i}+(1 \times \cos 60)\hat{i}}$
= $\mathbf\small{-0.5 \hat{i}}$
2. So the available resultant in the horizontal direction is $\mathbf\small{-0.5 \hat{i}}$
That is., in the horizontal direction, a force of magnitude 0.5 N is acting towards the left
3. If we apply an additional force of magnitude 0.5 N in the horizontal direction, towards the right, then there will be no net force in the horizontal direction
• The resultant of the 4 forces will then be in the vertical direction only
Solved example 5.16
Two forces of equal magnitude are acting on an object. The angle between them is 60o. If the magnitude of their resultant is 40√3 N, What is the magnitude of each force?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
1. Let the magnitude of each force be 'p' N
• Also let one of the forces be towards the positive side of the x axis
• This is shown in fig.5.32(c) above
2. We have:
$\mathbf\small{\vec{R_x}=p\: \hat{i}+(p \times \cos 60)\hat{i}=\frac{3p}{2}\: \hat{i}}$
$\mathbf\small{\vec{R_y}=0+-(p \times \sin 60)\hat{j}=-\frac{p\sqrt{3}}{2}\: \hat{j}}$
3. We have:
• Magnitude of the resultant = $\mathbf\small{|\vec{R}|=\sqrt{|\vec{R_x}|^2+|\vec{R_y}|^2}}$
• Substituting the values, we get:
$\mathbf\small{|\vec{R}|=\sqrt{\left (\frac{3p}{2} \right )^2+\left (\frac{p\sqrt{3}}{2} \right )^2}=\sqrt{3p^2}=p\sqrt{3}}$
4. But given that, magnitude of the resultant is 40√3 N
• So we can write: 40√3 = p√3
• Thus we get: Magnitude of each force = p = 40 N
In the next section we will see some common forces in mechanics
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