A 1480 kg automobile has a wheel base (the distance between the axles) of 2.9 m. The center of mass of the automobile is on the center line at a point 1.34 m from the front axle.
Find the force exerted by the ground on each back wheel. The acceleration of gravity is 9.8 m/s2.
Answer in units of N
To find the force exerted by the ground on each back wheel, we can start by calculating the weight of the automobile. The weight is equal to the mass of the automobile multiplied by the acceleration due to gravity.
Weight = mass × acceleration due to gravity
Weight = 1480 kg × 9.8 m/s^2
Weight = 14440 N
Now, we need to determine the weight distribution between the front and back wheels. We know that the center of mass of the automobile is at a point 1.34 m from the front axle, which means that the distance between the center of mass and the back axle is 2.9 m - 1.34 m = 1.56 m.
To distribute the weight between the front and back wheels, we can use the concept of torque. Torque is the force applied at a distance from a pivot point. In this case, the pivot point is the front axle and the torque due to the weight of the automobile must be balanced.
The torque due to the weight of the automobile about the front axle can be calculated as follows:
Torque = Weight × Distance
Torque = 14440 N × 1.56 m
Torque = 22550.4 N·m
Since the automobile is in equilibrium, the torque about the front axle must be equal to the torque about the back axle. The torque about the back axle is the force exerted by each back wheel multiplied by the distance between the back axle and the center of mass.
Therefore, we can find the force exerted by each back wheel by rearranging the torque equation:
Force on back wheels = Torque / Distance
Force on back wheels = 22550.4 N·m / 1.56 m
Force on back wheels = 14449.74 N
Hence, the force exerted by the ground on each back wheel is approximately 14449.74 N.