NCERT solution Class 11 system of particles and rotational motion

NCERT Solutions

Class 11 Physics

System of Particles and Rotational Motion

NCERT Questions

Q. 6.1
Q. 6.2
Q. 5.3
Q. 6.4
Q. 6.5
Q. 6.6
Q. 6.7
Q. 6.8
Q. 6.9
Q. 6.10
Q. 6.11
Q. 6.12
Q. 6.13
Q. 6.14
Q. 6.15
Q. 6.16
Q. 6.17

Q. 6.9

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Given:

Mass of the car, m = 1800 kg

Distance between the front and back axles, d = 1.8 m

Distance between the C.G. (centre of gravity) and the back axle = 1.05 m

The various forces acting on the car are shown in the following figure.

R_f and R_bare the forces exerted by the level ground on the front and back wheels respectively.

At translational equilibrium:

(R_f + R_b)= mg

= (1800 × 9.8)

= 17640 N ... (i)

For rotational equilibrium, on taking the torque about the C.G., we have:

R_f(1.05) = R_b(1.8 – 1.05)

(R_f x 1.05) = (R_b x 0.75)

(R_f / R_b) = (0.75)/ (1.05)

(R_f / R_b) = (5/7)

(R_b / R_f) = (7/5)

R_b = (1.4) R_f (ii)

Solving equations (i) and (ii), we get:

(1.4) R_f+ R_f = 17640

R_f = (17640/2.4)

=7350N

Therefore, R_b = (17640 – 7350)

= 10290 N

Therefore, the force exerted on each front wheel, (7350)/ (2) = 3675N

The force exerted on each back wheel, (10290)/ (2) = 5145N

Video solution:

System of Particles and Rotational Motion Class 11 Physics NCERT Chapter 6 NCERT Solutions

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