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Q7.9 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.

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The FBD of the car is shown below : 

                               Rotational motion,    20122

We will use conditions of equilibrium here :

                                           R_f\ +\ R_b\ =\ mg

                                                                   =\ 1800\times 9.8\ =\ 17640\ N                ....................................(i)

          For rotational equilibrium :

                                           R_f\left ( 1.05 \right )\ =\ R_b(1.8-1.05)

or                                        1.05R_f\ =\ 0.75R_b

                                            R_b\ =\ 1.4R_f                                                                         ..............................(ii)

From (i) and (ii) we get : 

                                          R_f\ =\ \frac{17640}{2.4}\ =\ 7350\ N

and                                    R_b\ =\ 17640-7350\ =\ 10290\ N

Thus force exerted by the front wheel is = 3675 N

and force exerted by the back wheel = 5145 N. 

Posted by

Devendra Khairwa

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