Truck suspensions often have \"helper springs\" that engage at high loads. One s

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 in the figure. 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 5.75 x10^5 N/m, helper spring constant of 3.60 x10^5 N/m, and y0 = 0.500 m. (a) What is the compression of the leaf spring for a load of 4.10 x10^5 N? (answer in m)) (b) How much work is done compressing the springs? (answer in J)

Explanation / Answer

Before the helper spring engages, the main spring must be compressed 0.5 m.
To do so, you need a force of:
F = (5.75×10^5 N/m) × (0.5 m)
F = 2.875×10^5 N

The remaining force that will act on both springs is
F' = (4.10×10^5 N) - (2.875×10^5 N)
F' = 1.225×10^5 N

The compression will be:
1.225×10^5 N = ((5.75×10^5 N/m) + (3.6×10^5 N/m)) × Y
1.225×10^5 N = (9.35×10^5 N/m) × Y
Y = 0.131 m

So the total compression is
X = (0.5 m) + (0.131 m)
X = 0.631 m < - - - - - - answer (a)

(b)

Compressing the main spring by 0.631 m will require
W = (5.75×10^5 N/m) × (0.631 m)² / 2
W = 1.145*105 J

Compressing the helper spring by 0.131 m will require
W = (3.6×10^5 N/m) × (0.131 m)² / 2
W = 308.89 J

The total work required is
W = (1.145*105 J) + (308.89 J) = 1.175*105 J

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