- 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 F
_{1}& F_{2 }are acting, as shown in figure. The block will be in equilibrium, if F_{1}– F_{2 }= 0

- 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.

A Value of F2 changes

B Force in x– direction changes

C Force in y- direction changes

D Force in (-y) – direction changes

- 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 :

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 \(F_x = 10\; cos \,\theta\\ F_y = 10 \; cos \,\theta\)

- 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 F
_{1}& F_{2 }are acting, as show in figure. The block will be in equilibrium, if F_{1}– F_{2 }= 0

- For example :

Two persons are applying forces on a block.

The force F_{1} is applied by person 1 and the force F_{2} 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 F_{1} and F_{2} become equal.

The sum of all forces along a direction is zero.

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

A 10 N

B Zero

C 20 N

D 30 N

- 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 :

The sum of all forces along a direction is zero.

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

A \(100 \,N\)

B \(150 \,N\)

C \(160 \,N\)

D \(200 \,N\)

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

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

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

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