Question

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

Answer 1

Hint: Weight of car, distance between front and back axles and distance of center of gravity from front axle is given. We know that at equilibrium position, angular momentum about the center of gravity is zero. Using this we can find the force acting on the back axle. Therefore, force acting on the back axle can be determined. Since there are two wheels on the front and back axles, the force acting on each axle will be the half of the force acting on each axle.
Formula used:
\[{{N}_{f}}+{{N}_{b}}=mg\]

Complete step by step answer:
Given that,
Weight of car, \[W=1800Kg\]
Distance between front and back axles, \[d=1.8m\]
Distance of center of gravity from front axle, \[{{L}_{1}}=1.05m\]
Distance of center of gravity from back axle, \[{{L}_{2}}=1.8-1.05=0.75m\]
Since, forces are balanced; force acting on front and back wheels is equal to the weight of the object. Then,
\[{{N}_{f}}+{{N}_{b}}=mg\]
Where,
\[{{N}_{f}}\]is the normal force acting on front wheel
\[{{N}_{b}}\]is the normal force acting on back wheel
Then,
\[{{N}_{f}}+{{N}_{b}}=mg=1800\times 9.8=17640\]
\[{{N}_{f}}+{{N}_{b}}=17640\]---------- 1
Where,
\[m\]is mass of the car
\[g\]is acceleration due to gravity
We have,
\[\tau =FL\]
Where,
\[F\]is the force acting on the body
\[L\]is the distance from center of gravity
Angular momentum about the center of gravity is zero. Then,
\[{{N}_{f}}{{L}_{1}}={{N}_{b}}{{L}_{2}}\]
Where,
\[{{\text{L}}_{\text{1}}}\text{ and }{{\text{L}}_{\text{2}}}\] are the distances from center of gravity from front and back axles respectively.
We have, \[{{\text{L}}_{\text{1}}}\text{=1}\text{.05m}\] and \[{{\text{L}}_{\text{2}}}=0.75m\]
Then,
\[{{N}_{f}}\times 1.05={{N}_{b}}\times 0.75\]
\[{{N}_{f}}=\dfrac{5}{7}{{N}_{b}}\] --------- 2
Substitute 2 in equation 1, we get,
\[\dfrac{5}{7}{{N}_{b}}+{{N}_{b}}=17640\]
\[{{N}_{f}}+10290=17640\]
\[{{N}_{f}}=17640-10290=7350N\]
Since there are two wheels in front and back axles,
Force on each front wheel \[=\dfrac{7350}{2}=3675N\]
Force on each back wheel \[=\dfrac{10290}{2}=5145N\]

Note:
If no external forces are acting on a system then, the angular momentum of the system is conserved. When no external forces are acting, the net torque will be zero. Hence, angular momentum is conserved in a system.