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Truck suspensions often have "helper springs" that engage at high loads. One such arrangement is a...

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 5.10 ✕ 105 N/m, the helper spring constant is 3.40 ✕ 105 N/m, and y0 = 0.500 m.

What is the compression of the leaf spring for a load of 4.50 ✕ 105 N?

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

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Answer #1

Before the helper spring engages, the main spring must be compressed 0.5 m.
To do so, you need a force of:
F = (5.1×10^5 N/m) × (0.5 m)
F = 2.55 ×10^5 N

The remaining force that will act on both springs is
F' = (4.50×10^5 N) - (2.55×10^5 N)
F' = 1.95×10^5 N


The compression will be:
1.95×10^5 N = ((5.1×10^5 N/m) + (3.40×10^5 N/m)) × Y
1.95×10^5 N = (8.50×10^5 N/m) × Y
Y = 0.23m

So the total compression is
X = (0.5 m) + (0.23m)
X = 0.73 m < - - - - - - answer (a)

Compressing the main spring by 0.73 m will require
W₁ = (5.1×10^5 N/m) × (0.73m)² / 2
W₁ = 135670.5 J

Compressing the helper spring by 0.056 m will require
W₂ = (3.40×10^5 N/m) × (0.73 m)² / 2
W₂ = 90593J

The total work required is
W = (135670.5 J) + (90593 J) = 226263.5 J.....Answer.

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