A spring, operating within its elastic range, shortens by 60mm when a load of 480N is applied to it. Calculate the total shortening when the load is increased by 120N
- 82mm.
- 45mm.
- 75mm.
Explanation from PART66Online
Within its elastic range a spring obeys Hooke's law, so deflection is proportional to load. The stiffness is 480 N / 60 mm = 8 N/mm. The new total load is 480 + 120 = 600 N, giving a total shortening of 600 / 8 = 75 mm.
R
Richiegeee asked · 29 May 2015
Hi, can someone give me the formula to calculate this?
Community Comments (3)
P
PART66Online
3 Jun 2015
First load = 480N.
Second load = 480N + 120N = 600N
600 / 480 * 60 = 75mm
Second load = 480N + 120N = 600N
600 / 480 * 60 = 75mm
R
Richiegeee
3 Jun 2015
Got it. Thanks
A
aerorodder1
19 Dec 2015
480n/120n=4
60mm/4=15mm.
60mm+15mm=75mm
60mm/4=15mm.
60mm+15mm=75mm
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