Question

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 shown in the figure below. When the main leaf spring is compressed by distance y₀, the helper spring engages and then helps to support any additional load. Suppose the leaf spring constant is 5.05 10⁵ N/m, the helper spring constant is 3.60 10⁵ N/m, and y₀ = 0.500 m.

(a) What is the compression of the leaf spring for a load of 5.90 10⁵ N?
m

(b) How much work is done in compressing the springs?
J

Answer 1

Part A.

Using Force balance in vertical direction:

F_net = Fs_leaf + Fs_helper - W = 0

Fs_leaf + Fs_helper = W

Now spring force is given by: Fs = k*x

k = spring constant

x = compression in spring

W = Weight of load = 5.90*10^5 N

Fs_leaf = k1*x1

k1 = spring constant of leaf spring = 5.05*10^5 N/m

x1 = compression in leaf spring = ?

Fs_helper = k2*x2

k2 = spring constant of helper spring = 3.60*10^5 N/m

since helper spring helps to support any additional load, So compression in helper spring will be:

x2 = compression in helper spring = (x1 - y0)

So,

k1*x1 + k2*(x1 - y0) = W

x1 = (W + k2*y0)/(k1 + k2)

x1 = (5.90*10^5 + 3.60*10^5*0.500)/(3.60*10^5 + 5.05*10^5)

x1 = 0.890 m = compression in leaf spring

Part B.

Work-done in compressing the springs will be:

W = work-done by leaf spring + work-done by helper spring

W = (1/2)*k1*x1^2 + (1/2)*k2*x2^2

W = (1/2)*k1*x1^2 + (1/2)*k2*(x1 - y0)^2

Using known values:

W = (1/2)*5.05*10^5*0.890^2 + (1/2)*3.60*10^5*(0.890 - 0.500)^2

W = 227383.25 J

W = 2.27*10^5 J

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