Truck suspensions often have "helper springs" that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as in Figure P7.17. The helper spring engages when the main leaf spring is compressed by distance y0, and then helps to support any additional load. Consider a leaf spring constant of 6.20 105 N/m, helper spring constant of 3.60 105 N/m, and y0 = 0.500 m.
(a) What is the compression of the leaf spring for a load of 3.65 105 N?
m
(b) How much work is done compressing the springs?
To solve this problem, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is compressed or stretched.
(a) To find the compression of the leaf spring for a load of 3.65 * 10^5 N, we can use the equation:
F = k * y
Where F is the force applied, k is the spring constant, and y is the compression of the spring.
Given:
Leaf spring constant (k1) = 6.20 * 10^5 N/m
Helper spring constant (k2) = 3.60 * 10^5 N/m
Compression of the main leaf spring (y0) = 0.500 m
Load (F) = 3.65 * 10^5 N
Since the load exceeds the compression value of the main leaf spring, it will compress and engage the helper spring.
First, we need to find the compression of the main leaf spring beyond y0, which can be calculated as:
Delta_y = F / k1
Delta_y = 3.65 * 10^5 N / 6.20 * 10^5 N/m
Delta_y ≈ 0.5882 m
Next, we need to find the compression of the helper spring, which can be calculated as:
y2 = Delta_y - y0
y2 = 0.5882 m - 0.500 m
y2 = 0.0882 m
Therefore, the compression of the leaf spring for a load of 3.65 * 10^5 N is approximately 0.0882 m.
(b) To calculate the work done compressing the springs, we can use the equation:
Work = (1/2) * k * y^2
However, since we have two springs in series (main leaf spring and helper spring), we need to calculate the work done on each separately and add them up.
For the main leaf spring:
Work1 = (1/2) * k1 * y1^2
For the helper spring:
Work2 = (1/2) * k2 * y2^2
where y1 is the compression of the main leaf spring (y1 = y0) and y2 is the compression of the helper spring.
Using the given values and the equations above, we can compute the work done.
Work1 = (1/2) * 6.20 * 10^5 N/m * (0.500 m)^2
Work1 ≈ 0.77 * 10^5 N
Work2 = (1/2) * 3.60 * 10^5 N/m * (0.0882 m)^2
Work2 ≈ 1.46 * 10^3 N
Total work done = Work1 + Work2
Total work done ≈ 0.77 * 10^5 N + 1.46 * 10^3 N
Total work done ≈ 0.78 * 10^5 N
Therefore, the total work done compressing the springs is approximately 0.78 * 10^5 N.