Question

A car, traveling at , encounters a dip in the road. The radius of curvature at the bottom of the dip is . Each of the car’s four springs has a spring constant of (the spring compresses for every applied to it). Determine the deflection of the springs from their unloaded state when the car is at the bottom of the dip. The weight of the car supported by the springs is . Goal: Find the spring deflection for a car with 4 springs that encounters a dip in the road.

Answer 1

Answer:

spring deflection is  x = (v2 / R + g) m / 4

Explanation:

We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration

 

Let's write the equations on the Y axis of this description

       Fe - W = m $a_{c}$

Where Fe is elastic force, W the weight and $a_{c}$  the centripetal acceleration. The elastic force equation is

       Fe = - k x

     

       4 (k x) - mg = m v² / R

The four is because there are four springs, R is theradio of dip

We can calculate the deflection (x) of the springs

       x = (m v2 / R + mg) / 4

       

       x = (v2 / R + g) m / 4