If a chain of 30 identical short springs linked end-to-end has a stiffness of 420 N/m, what is the stiffness of one short spring?
Okay, I don't understand how you would get N/m if they are linked together. Can you explain how you would get the answer? Please.
Certainly! To solve this problem, we can use the concept of series springs to find the stiffness of one short spring.
In a series arrangement, the total stiffness is inversely proportional to the sum of the reciprocals of the individual stiffness values. So, if we represent the stiffness of one short spring as k, the total stiffness of the chain of 30 springs linked together would be:
1/total stiffness = 1/k + 1/k + 1/k + ... + 1/k (30 times)
Since all the springs are identical, we can simplify the above expression:
1/total stiffness = 30/k
Now, we are told that the total stiffness of the chain of 30 springs is 420 N/m. Therefore, we can substitute this value into the equation:
1/420 N/m = 30/k
To solve for k, we can rearrange the equation:
k = 30/(1/420 N/m)
= 30 * 420 N/m
= 12600 N/m
So, the stiffness of one short spring is 12600 N/m.
To summarize, we found the stiffness of one short spring by using the concept of series springs, where the total stiffness is the sum of the reciprocals of the individual stiffness values. We then rearranged the equation to solve for the stiffness of one short spring and obtained 12600 N/m as the final answer.