Can this info get a spring constant and a drop height

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SUMMARY

The discussion focuses on calculating the spring constant and drop height for a 450 lb cart with a 16-inch travel distance. Using Hooke's Law (F=k*x), the spring constant (k) can be determined based on the displacement (x) caused by an additional weight of 125 lbs, which results in a 3-inch sink. The relationship between the spring's configuration and the cart's mechanics is crucial for accurate calculations, emphasizing the need for detailed information regarding the spring's attachment and movement.

PREREQUISITES
  • Understanding of Hooke's Law (F=k*x)
  • Basic knowledge of mechanical systems and forces
  • Familiarity with spring mechanics and configurations
  • Ability to interpret displacement and force relationships
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  • Research how to calculate spring constants using Hooke's Law
  • Explore the mechanics of coil springs in cart systems
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Mechanical engineers, physics students, and anyone involved in designing or analyzing spring systems in vehicles or carts.

griokin
if an 450lb cart has 16 inches of travel and is fully extended under its own weight, it sinks 3 inches with me, 125 lbs addition, can this info get a spring constant and a drop height where the springs would bottom out with me in it? if any other info is needed, assume the best case dealy-bob and put it in with an answer.
 
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What do you mean by "16 inches of travel". I don't have a clear picture of the situation you're describing.

Hooke's law is F=k*x, where x is the displacement, k is the spring constant, and F is the force.
 
the 16 inches of travel could mean several things. from the information given thus far, i am going to assume the cart can drop 16 inches before it bottoms out. now we must look at how the sping is connected. if the spring is a coil inbetween the weel and cart, a direct relationship can be used here. but if the spring is connected to an A-arm, we can say the relationship of the angle [the] that the arm is allowed to move to the arm is 2Rarm (sin[the] /2)= 16 inches or [the] /2=arcsin (8 inches / Rarm ) ; and now we must consider how and where the spring is attatched to find out exactly how much force is required. So to answer the question, you did not give enough information to calculate it.
 
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