Informative line

# Concept of Equilibrium

• If a body has no acceleration in any direction, then the body is said to be in equilibrium.
• In other words, if a body is in equilibrium, then sum of forces in any direction is zero.
• On this basis, unknown forces can be calculated.
• For instance : on a block, the forces F1 & Fare acting, as shown in figure. The block will be in equilibrium, if F1 – F= 0

#### A body is in equilibrium, calculate force F1, as shown in figure.

A 5 N

B 15 N

C 2 N

D 3 N

×

The body is in equilibrium, thus, the sum of all the forces applying on it, is zero along any direction.

Forces along $$x$$- axis,

10 + 5 = 15 N

Force along $$(-x)$$ axis,

F1

Thus, 15 - F1 = 0

F1 = 15 N

### A body is in equilibrium, calculate force F1, as shown in figure.

A

5 N

.

B

15 N

C

2 N

D

3 N

Option B is Correct

# Independence of Components of Forces along two Mutually Perpendicular Axes

• The components of forces along two mutually perpendicular directions are independent of each other.
• If a force acts along $$x$$– axis, then its component along y – axis is zero.
• If a force acts along y – axis, then its component along $$x$$– axis is zero.
• If the forces act along perpendicular directions, then their components along their respective perpendicular directions will be zero.

#### If the value of force F1 is Changed, then what happens?

A Value of F2 changes

B Force in x– direction changes

C Force in y- direction changes

D Force in (-y) – direction changes

×

If the forces act along perpendicular directions, then their components along their respective perpendicular directions will be zero.

The force Fis acting in y–direction and F1 and  F2 are perpendicular to each other, thus, force only in y–direction will change.

### If the value of force F1 is Changed, then what happens?

A

Value of F2 changes

.

B

Force in x– direction changes

C

Force in y- direction changes

D

Force in (-y) – direction changes

Option C is Correct

# Components of Force Along $$x$$ and $$y$$– axis when Force is Oblique to $$x$$ or $$y$$–axis

• Consider a force vector, as shown in figure.

• To calculate force along $$x$$ and $$y$$–axis, due to the impact of this force, the force is resolved in $$x$$ and $$y$$–axis.
• Suppose the force vector $$\vec{F}$$ makes angle $$\theta$$ with positive $$x$$–axis.

The components of force can be resolved as follows :

If the force vector $$\vec{F}$$ makes angle $$\theta\,$$with positive $$y$$–axis, then the components of $$\vec{F}$$ will be as follows :

#### What will be the $$x$$ and $$y$$– component of the force shown in figure?

A $$F_x = 10 \; cos \,\theta \\ F_y = 10\; sin \,\theta$$

B $$F_x = 20 \; cos\,\theta\\ F_y = 30 \; sin \,\theta$$

C $$F_x = 10 \; sin \,\theta \\ F_y = 10 \; cos \,\theta$$

D $$F_x = 10\; cos \,\theta\\ F_y = 10 \; cos \,\theta$$

×

Since, the force makes $$\theta\,$$angle with positive $$x$$–axis

Therefore, the components of force will be

$$F_x = 10 \;cos \,\theta \\ F_y = 10\; sin \,\theta$$

### What will be the $$x$$ and $$y$$– component of the force shown in figure?

A

$$F_x = 10 \; cos \,\theta \\ F_y = 10\; sin \,\theta$$

.

B

$$F_x = 20 \; cos\,\theta\\ F_y = 30 \; sin \,\theta$$

C

$$F_x = 10 \; sin \,\theta \\ F_y = 10 \; cos \,\theta$$

D

$$F_x = 10\; cos \,\theta\\ F_y = 10 \; cos \,\theta$$

Option A is Correct

# Condition of Equilibrium

• If a body has no acceleration in any direction, then the body is said to be in equilibrium.
• In other words, if a body is in equilibrium, then sum of forces in any direction is zero.
• On this basis, unknown forces can be calculated.
• For instance : on a block, the forces F1 & Fare acting, as show in figure. The block will be in equilibrium, if F1 – F= 0

• For example :

Two persons are applying forces on a block.

The force F1 is applied by person 1 and the force F2 is applied by person 2 on the block.

The block will not displace from its position, if the forces become equal.i.e., the block will be in equilibrium if F1 and F2 become equal.

## Condition of Equilibrium

The sum of all forces along a direction is zero.

Or in other words, the forces from opposite sides should be equal.

#### Calculate tension in the string shown in figure, if the body is in equilibrium.( g = 10 m/s2 )

A 10 N

B Zero

C 20 N

D 30 N

×

The body is in equilibrium, thus, the forces from opposite sides should be equal.

Thus,

Force applied = Tension + gravitational pull

F = T + mg

10 = T + 1 × 10

10 = T + 10

T = 0

Hence, the tension in the string is zero.

### Calculate tension in the string shown in figure, if the body is in equilibrium.( g = 10 m/s2 )

A

10 N

.

B

Zero

C

20 N

D

30 N

Option B is Correct

# Calculation of Force under Condition of Equilibrium

• Consider a force vector, as shown in figure.

• To calculate force along $$x$$ and $$y$$–axis, due to the impact of this force, the force is resolved in $$x$$ and $$y$$–axis.
• Suppose the force vector $$\vec{F}$$ makes angle $$\theta$$ with positive $$x$$–axis.

The components of force can be resolved as follows :

If the force vector $$\vec{F}$$ makes angle $$\theta$$ with positive $$y$$–axis, then the components of $$\vec{F}$$ will be as follows :

## Condition of Equilibrium

The sum of all forces along a direction is zero.

Or in other words, the forces from opposite sides should be equal.

#### The system shown in figure is in equilibrium. Calculate tension in the string, if $$\theta = 30 °$$ and mass (m) of block is $$10 \; kg$$ . Given:- $$g =10\;m/s^2$$

A $$100 \,N$$

B $$150 \,N$$

C $$160 \,N$$

D $$200 \,N$$

×

Mass of block, $$m = 10 \; kg$$

Acceleration due to gravity, $$g = 10 \; m/s^2$$

Tension in string = T

Resolving tension into its components :

component along y-axis is $$T\,sin\; 30°$$

Total force acting along y-axis is $$2T\,sin\; 30°$$

Resolving in $$x$$-direction :

$$\sum\; F_x = 0$$

$$T \; cos \; 30° = T \; cos \; 30°$$

So block is in equilibrium in $$x$$-direction

The system is in equilibrium.Thus, the force acting on opposite sides of block must be equal in y -direction .

Thus ,

$$T \; sin\; 30 ° + T \; sin\; 30 ° = mg$$

$$2T \; sin \;30 ° = 10 \times10$$

$$T \left(\dfrac{1}{2}\right) = \dfrac{100}{2}$$

$$T = 100\; N$$

### The system shown in figure is in equilibrium. Calculate tension in the string, if $$\theta = 30 °$$ and mass (m) of block is $$10 \; kg$$ . Given:- $$g =10\;m/s^2$$

A

$$100 \,N$$

.

B

$$150 \,N$$

C

$$160 \,N$$

D

$$200 \,N$$

Option A is Correct

# Tension in string

• Consider a block A which is in equilibrium, as shown in figure.

• The mass of block is m and the block is attached to the surface of earth by a string.
• An upward force is applied on the block.
• Since the body is in equilibrium, therefore, the forces from opposite sides of block should be equal.
• Thus,

Force applied = Tension + gravitational pull

F = T + mg

Here, T is tension in the string.

If the force applied is equal to the gravitational pull, then tension in the string becomes zero.

#### A bag of cement of weight W hangs in equilibrium from three strings, as shown in figure.  Two of the strings make angle $$\theta_1$$ and $$\theta_2$$ with the horizontal. Assuming the system is in equilibrium, the tension (T1) in the left hand side string is:

A $$\dfrac{W\; cos\;\theta_2}{sin(\theta_1 +\theta_2)}$$

B $$\dfrac{W\; sin \; \theta_2}{sin(\theta_1 +\theta_2)}$$

C $$\dfrac{W\; tan\; \theta_1}{cos(\theta_1 +\theta_2)}$$

D $$\dfrac{W\; cos \; \theta_2}{cos(\theta_1 +\theta_2)}$$

×

Resolving the components of tension T1 and T2:

Component of $$T_1$$ along y–axis is $$T_1 \; sin \; \theta _1$$

Component of $$T_1$$ along $$x$$–axis is $$T_1 \; cos \; \theta _1$$

Component of $$T_2$$ along y–axis is $$T_2 \; sin \; \theta _2$$

Component of $$T_2$$ along $$x$$–axis is $$T_2 \; cos \; \theta _2$$

Tension T3 is canceled out

The body is in equilibrium, thus, the sum of all the forces applying on it, is zero along any direction.

$$Force\;along\;x– direction :$$

$$T_2 \; cos \; \theta_2 - T_1\; cos \; \theta_1 = 0$$

$$T_2 \; cos \;\theta_2 = T_1 \; cos \; \theta_1$$

$$T_2 = \dfrac{T_1 \; cos \,\theta_1}{cos \,\theta_2}\;\;\;\ —( 1)$$

$$Force\;along\; y–direction\,:$$

$$T_1\; sin\;\theta_1 + T_2 \; sin\; \theta_2 – W = 0$$ { Since T3 is internal force }

$$T_1 \; sin \; \theta_1 + \left(\dfrac{T_1 \; cos \; \theta_1}{cos\; \theta _2} \right)sin\; \theta_2 =W$$

$$T_1 (sin\; \theta_1) (cos\;\theta_2) + T_1 (cos\; \theta_1)(sin\;\theta_2) = W \; cos\; \theta_2$$

$$T_1 \left[(sin\; \theta_1) (cos\;\theta_2) + (cos\; \theta_1)(sin\;\theta_2)\right] = W \; cos\; \theta_2$$

$$T_1 [sin (\theta_1 +\theta_2)] = W \; cos \; \theta_2$$

$$T_1 \; = \dfrac{W \; cos \;\theta_2}{sin\; (\theta_1 +\theta_2)}$$

### A bag of cement of weight W hangs in equilibrium from three strings, as shown in figure.  Two of the strings make angle $$\theta_1$$ and $$\theta_2$$ with the horizontal. Assuming the system is in equilibrium, the tension (T1) in the left hand side string is:

A

$$\dfrac{W\; cos\;\theta_2}{sin(\theta_1 +\theta_2)}$$

.

B

$$\dfrac{W\; sin \; \theta_2}{sin(\theta_1 +\theta_2)}$$

C

$$\dfrac{W\; tan\; \theta_1}{cos(\theta_1 +\theta_2)}$$

D

$$\dfrac{W\; cos \; \theta_2}{cos(\theta_1 +\theta_2)}$$

Option A is Correct