Compute the reaction in the front wheels

[Joe_1234]

A car and its load weighs 30kN and the center of gravity is 0.6 m from the ground and midway between the front and rear wheels which are 3m apart. The car is brought to rest from a speed of 50 kph in 5 sec by means of the brakes. Compute the reaction in the front wheel.

 

[jvicens]

Based on the information provided, the first step would be to calculate the total mass of the car and its load. Using the formula F=ma, where F is the force, m is the mass, and a is the acceleration, we can rearrange the formula to solve for the mass: m=F/a. In this case, the force (weight) is given as 30kN and the acceleration can be calculated as the change in velocity (50 kph or 13.89 m/s) divided by the time (5 sec). So the mass of the car and its load would be 30kN/13.89 m/s/5 sec = 1.08 metric tons.

Next, we can calculate the center of mass of the car and its load. This is given as 0.6 m from the ground and midway between the front and rear wheels, which are 3m apart. This means that the distance from the front wheel to the center of mass would be 1.5m and the distance from the rear wheel to the center of mass would also be 1.5m.

To determine the reaction in the front wheel, we can use the principle of moments, which states that the sum of the moments (or torques) acting on an object must equal zero for the object to be in equilibrium. In this case, we can calculate the moment of the front wheel by multiplying the reaction force (unknown) by the distance from the front wheel to the center of mass (1.5m). We can also calculate the moment of the rear wheel by multiplying the weight of the car and its load (30kN) by the distance from the rear wheel to the center of mass (1.5m). Setting these two moments equal to each other and solving for the unknown reaction force, we get:

Reaction force in front wheel = (30kN * 1.5m) / 1.5m = 30kN

Therefore, the reaction in the front wheel would be 30kN. This means that the brakes would need to generate a force of 30kN to bring the car to a stop in 5 seconds. It is important to note that this is a simplified calculation and does not take into account other factors such as friction, air resistance, and the distribution of weight within the car. However, it provides a basic understanding of the forces at play in this scenario.