In the previous section we talked about the forces in equilibrium. Here we'll discuss about the concept of center of gravity, moment.

What is

-Since any object of concern is not a point mass, but composed of many point masses(atoms), the vector sum of the weights of all the point masses may be taken as acting at a single point. So, basically

$$\tau_{couple}=F_{\perp}\times d$$

$$\tau_{force}=F_{\perp}\times d$$

As we have seen in the previous section that a body is in equilibrium if there is no resultant force(i.e.$\sum_{i=1}^{n}F_{i}=0$). And so similarly a body is in equilibrium also if no resultant torque is acting on it. (i.e.$\sum_{i=1}^{n}\tau_{i}=0$).

$$\sum \tau_{clockwise}=\sum \tau_{anti-clockwise}$$

What is

**centre of gravity?**-Since any object of concern is not a point mass, but composed of many point masses(atoms), the vector sum of the weights of all the point masses may be taken as acting at a single point. So, basically

**centre of gravity**of a body is the point at which the weight of the body**appears**to act. The image below shows an example of the principle of centre of gravity to keep balance on the bird.wikipedia.com |

**Couples:**A couple is a pair of forces(same in magnitude but opposite in direction) which tends to produce rotations only.**Torque of couple**: It is the product of one of the forces and the perpendicular distance between their lines of action.$$\tau_{couple}=F_{\perp}\times d$$

**Torque of force(moment):**The moment or torque of a force is the product of the force and the perpendicular distance between the line of action of the force and the pivot.photocourtesy:learneasy.info |

$$\tau_{force}=F_{\perp}\times d$$

As we have seen in the previous section that a body is in equilibrium if there is no resultant force(i.e.$\sum_{i=1}^{n}F_{i}=0$). And so similarly a body is in equilibrium also if no resultant torque is acting on it. (i.e.$\sum_{i=1}^{n}\tau_{i}=0$).

**Principle of moments: "**For a body in equilibrium under the action of coplanar forces, the no resultant torque condition is due to the fact that clock-wise torque is equal to the anti-clockwise torque about the same point.**"**$$\sum \tau_{clockwise}=\sum \tau_{anti-clockwise}$$

**Ex:**Consider a beam(blue color) of length $L$ and weight $W$. And let $F_{1},F_{2}$ be acting at a distance of $\frac{1}{8}L$ and distance of $\frac{1}{4}L$ from its end as shown in the diagram below. Find the $F_{1},F_{2}$ in terms of $W$.

**Sol$^{n}$: $F_{1}=(2/5)W$ and****$F_{2}=(3/5)W$**
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