Verify Second Condition of Equilibrium.

Hints:

1. Take a meter rod and find its center of gravity G by balancing it on a sharp wedge.

2. Suspend three weights WI on right side ofthe meter rod and W2 and W3 on its left side.

3. Balance the meter rod on wedge again and note the positions of each weight.

4. Apply 2nd condition i.e. T miclocksise =T dackstice Center of gravity of meter rod = G=…………………………M

NO.of.Obs.	Weight suspended			Position of				Distance
	W1	W2	W3	W1	W2	W3	AG		BG	CG
				A	B	C
	N	N	N	m	m	m	m		m	m

No.of  Obs.	Anticlockwise Torque		Clockwise Torque	⅀t =0
	T2=wBG	T2=WCG	T=-w1AG	⅀t =t1+t2+t3=0
	N.m	N.m	N.m	N.m
1 2 3

Precautions:

1. The meter rod should be suspended edge wise.

2. The height of each end of meter rod must be same from ground.

3. Reading on spring balance should be noted when meter rod become stationary.

Viva Voce:

Q.1. What is center of gravity?

Ans. The point where the whole weight of the body acts.

Q.2. Define torque.                                                         

Ans. It is tuning effect of the force and equal to the product of force and moment arm.

Q.3. What is moment arm?

Ans. The perpendicular distance between the line of action of the force and axisof rotation.

Q.4. Define 1st condition for equilibrium.

Ans. Sum of external forces action the body must be zero.

Q.5. Define 2nd condition for equilibrium.

Ans. Sum of all the external torques acting on the body is equal zero