Respuesta :
The stiffness of one short spring is 13500 N/m
Solution:
We are given a chain with number of spring N = 30 and linked end to end ( in series) and stiffness of this chain is 450 N/m
We have to find the stiffness of one short spring
The springs are identical, which means they have same stiffness
The stiffness of one spring in series is given as:
[tex]k_i = N \times k_s[/tex]
Where,
N is the number of springs
[tex]k_i[/tex] is the stiffness of one spring
[tex]k_s[/tex] is the stiffness of this chain
Substituting,
[tex]N = 30\\\\K_s = 450[/tex]
Therefore,
[tex]k_i = 30 \times 450\\\\k_i = 13500[/tex]
Thus the stiffness of one short spring is 13500 N/m
If a chain of 30 identical short springs linked end-to-end has a stiffness of 450 N/m , The stiffness of one short spring is 13500 N/m
Given : Stiffness = 450 N/m,
To find : the stiffness of one short spring
According to the given question,
- A chain with number of spring N = 30
- Chain is linked end to end
- Stiffness of this chain is 450 N/m
We knows that,
The springs are identical, by which means they have the same stiffness
Hence, The stiffness of one spring will be given as:
[tex]\rm k_i=N \times k_s[/tex]
Where,
- N = number of springs
- [tex]\rm k_i[/tex] = Stiffness of one spring.
- [tex]\rm k_s[/tex] = Stiffness of given chain.
On substituting the values in the formula we will get,
N = 30
[tex]\rm k_s[/tex] = 450
Then,
[tex]\rm k_i =30 \times450\\\\k_i = 13500[/tex]
Therefore, The stiffness of one short spring is 13500 N/m
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