In the given figure the weight of the section ABCD and PQRS is 50 N. Find the distance of the centre of gravity from the base SR:

CONCEPT:

Centre of gravity:

• The Centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated.
• The centre of gravity of a body is a point where the total gravitational torque on the body is zero.
• The centre of gravity of a body may lie inside or outside the body.
• The centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity-free space.
• If the body is so extended that gravitational acceleration varies from part to part of the body, then the centre of gravity and centre of mass will not coincide.
• Let a body consists of n number of particles and the gravitational acceleration is uniform for all the particles of the body.
• Then the position of the centre of gravity is given as,

\(⇒ r=\dfrac{w_1r_1+w_2r_2\ +\ ...\ +\ w_nr_n}{w_1+w_2\ +\ ...\ +\ w_n}\)

Where w1, w2,..., and wn = weight of the particles

CALCULATION:

Given w1 = w2 = 50 N, y1 = 11 cm and y2 = 5 cm

• In the given diagram the centre of gravity of the section ABCD and PQRS will be at point 1 and point 2 respectively.

• So the position of the centre of gravity from the base is given as,

\(⇒ y=\dfrac{w_1y_1+w_2y_2}{w_1+w_2}\)

\(⇒ y=\dfrac{(50\times11)+(50\times5)}{50+50}\)

\(⇒ y=\dfrac{550+250}{100}\)

⇒ y = 8 cm

• Hence, option 2 is correct.