Solving a Spring Problem in Engineering Statics

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The discussion revolves around solving a spring problem using Hooke's law, where the spring's lengths and forces are given. The stiffness of the spring is calculated as 944 N/m and 64.7 lb/ft for the stretched state. The confusion arises regarding the term "unstretched length," which is clarified through the application of Hooke's law equations for both tension and compression scenarios. Participants emphasize the importance of distinguishing between tension and compressive forces in understanding the spring's behavior. Ultimately, the problem can be solved by setting up and solving the system of equations derived from the spring's forces and lengths.
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Spring Problem

I need some help or hints on this exercise

When a certain linear spring has a lenth of 180mm, the tension in it is 170N. FOr a length of 160mm, the compressive force in the spring is 130N.

a. What is the stiffness of the spring in SI units? In U.S. Customary units?
b. What is its unstretched length in SI units? In U.S. customary units?

This exercise is from my Engineering statics book in the section talking about units.

a. They are asking for the spring constant k, but for which spring do they want?
I calculated the stiffness of the first spring, 944 N/m and 64.7 lb/ ft just divided the force and the length.

b. the keyword unstretched is confusing me in this part.

Yea the topic is mispelled, couldn't change it
 
Last edited:
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Nope, there is only one spring.
The problem is:
If the spring has the length of 0.18 m, its tension is 170 N.
If the spring has the length of 0.16 m, its compressive force is 130 N.
Let l0 (measured in m) be its normal length (when the spring is unstretched).
Hooke's law: F = -kx. So you have:
\left\{ \begin{array}{l} k(l_0 - 0.16) = 130 \\ k(0.18 - l_0) = 170 \end{array}\right using Hooke's law.
From there can you solve for k and l0?
Viet Dao,
 
Last edited:
VietDao29 said:
Nope, there is only one spring.
The problem is:
If the spring has the length of 0.18 m, its tension is 170 N.
If the spring has the length of 0.16 m, its compressive force is 130 N.
Let l0 (measured in m) be its normal length (when the spring is unstretched).
Hooke's law: F = -kx. So you have:
\left\{ \begin{array}{l} k(l_0 - 0.16) = 130 \\ k(0.18 - l_0) = 170 \end{array}\right using Hooke's law.
From there can you solve for k and l0?
Viet Dao,
Yea, i can solve it now. Just solve the system. I think what was throwing me off was tension and compressive force. I was thinking the spring was being compressed every time, but in fact it is being pulled for for 0.18m and then compressed to 0.16m.

Thanks man
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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