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.

As an experienced tutor registered on UrbanPro, I can guide you through solving this physics problem step by step. First, let's break down the problem and identify the key concepts involved. Given: Mass of the car (m) = 1800 kg Distance between front and back axles (L) = 1.8 m Center of gravity behind the front axle (d) = 1.05 m To find: Force exerted by the level ground on each front wheel and each back wheel. Now, the key concept here is the distribution of weight and the concept of torque. The weight of the car is acting downwards from its center of gravity, and it exerts a torque about the front and back axles. This torque causes the ground to exert a reaction force on the wheels. First, let's find the torque exerted by the weight of the car about each axle. The torque (τ) is given by the formula: τ=F⋅rτ=F⋅r Where: F is the force (weight of the car) r is the perpendicular distance from the point of application of force to the pivot (axle in this case) For the front axle: τfront=m⋅g⋅dτfront=m⋅g⋅d For the back axle: τback=m⋅g⋅(L−d)τback=m⋅g⋅(L−d) Where: m is the mass of the car (1800 kg) g is the acceleration due to gravity (9.8 m/s²) Now, let's calculate these torques: For the front axle: τfront=1800 kg×9.8 m/s2×1.05 mτfront=1800kg×9.8m/s2×1.05m τfront=17640 N⋅mτfront=17640N⋅m For the back axle: τback=1800 kg×9.8 m/s2×(1.8−1.05) mτback=1800kg×9.8m/s2×(1.8−1.05)m τback=1800 kg×9.8 m/s2×0.75 mτback=1800kg×9.8m/s2×0.75m τback=13230 N⋅mτback=13230N⋅m Now, since the car is in equilibrium, the sum of torques about each axle must be zero. Sumoftorques=τfront−τbackSumoftorques=τfront−τback Sumoftorques=17640 N⋅m−13230 N⋅mSumoftorques=17640N⋅m−13230N⋅m Sumoftorques=4410 N⋅mSumoftorques=4410N⋅m This torque is balanced by the reaction forces at each axle. Now, to find the force exerted by the level ground on each front wheel and each back wheel, we can use the fact that the torque exerted by a force about a pivot is equal to the force times the perpendicular distance from the pivot. So, the force exerted by the ground on each wheel is: For each front wheel: Ffront=SumoftorquesDistancebetweenfrontwheelsFfront=DistancebetweenfrontwheelsSumoftorques Ffront=4410 N⋅m1.8 mFfront=1.8m4410N⋅m Ffront=2450 NFfront=2450N For each back wheel: Fback=SumoftorquesDistancebetweenbackwheelsFback=DistancebetweenbackwheelsSumoftorques Fback=4410 N⋅m1.8 mFback=1.8m4410N⋅m Fback=2450 NFback=2450N Therefore, the force exerted by the level ground on each front wheel and each back wheel is 2450 N. read less