How Far Will Springs Compress Under Load?

In summary: They are not asking for x.In summary, the problem involves finding the compression distance of two springs in a parallel system when a load of 2250 N is placed on top. Using Hooke's law, the equivalent spring constant of 4400 N/m is used, and the resulting compression distance is 0.5 m. However, it is unclear if this is the desired x value as the problem does not specify what x represents.
  • #1
Mebmt
12
0
Given: Load of sacks have total weight of 2250 N and each sack weighs 225 N, how far will each of 2 springs in one system be compressed (spring coeff = 2200 N/m) when a sack is put on it?


Energy init +work=final Energy

2250=1/2(4400)x^2
1.02=x^2
x=1.01

Not sure where to go from here.
 
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  • #2
Mebmt said:
Given: Load of sacks have total weight of 2250 N and each sack weighs 225 N, how far will each of 2 springs in one system be compressed (spring coeff = 2200 N/m) when a sack is put on it?

Energy init +work=final Energy
I wouldn't use energy. There is another (even simpler) spring relation that I would use. BTW, I don't know what the problem means by "one system".
 
  • #3
Then using Us=1/2kx^2
I have 1/2(2200)x^2=2250
x=1.43

One system meaning that there are 2 springs working identically to each other.
 
  • #4
Is there a reason why you can't use Hooke's law?



Mebmt said:
One system meaning that there are 2 springs working identically to each other.
I still don't know what that means.
 
  • #5
Using Hooke's law Fs=-kx
2250N=(-2200N/m)x
x=1.02m
1.02m/2 springs =0.5 m
 
  • #6
Mebmt said:
2250N=(-2200N/m)x
x=1.02m
What is the meaning of this x value? Can you interpret it physically?
 
  • #7
The x value is the total amount that the springs would be compressed by a force of 2250 N.
 
  • #8
Mebmt said:
The x value is the total amount that the springs would be compressed by a force of 2250 N.
And is that the condition that the problem is asking for?
 
  • #11
I didn't post the picture of the springs. It shows them as being parallel to each other with a platform on top. Sorry. My bad. I should be using them together. The link helped as well.

Now, using Fspring=-kx
2250=(-)(2200+2200)x
x=0.5 m
 
  • #12
Mebmt said:
Now, using Fspring=-kx
2250=(-)(2200+2200)x
x=0.5 m
I ask again, is that the x that the problem is asking for? Hint: it has the same interpretation as the last time I asked.
 
  • #13
Yes, they are asking for x.
 
  • #14
Mebmt said:
Yes, they are asking for x.
Nope.
 

FAQ: How Far Will Springs Compress Under Load?

What is spring compression distance?

Spring compression distance is the measure of how much a spring is compressed or stretched from its original length when a force is applied to it. It is typically measured in units of length, such as meters or inches.

How is spring compression distance calculated?

Spring compression distance can be calculated using Hooke's law, which states that the distance a spring is compressed or stretched is directly proportional to the force applied to it. The formula is d = F/k, where d is the compression distance, F is the force applied, and k is the spring constant.

What factors affect spring compression distance?

There are several factors that can affect spring compression distance, including the force applied, the spring constant, and the materials and design of the spring. The type of spring, such as a coil spring or leaf spring, can also impact the compression distance.

How does spring compression distance affect a spring's performance?

The compression distance of a spring can affect its performance in various ways. Too much compression distance can cause the spring to lose its elasticity and become permanently deformed. On the other hand, too little compression distance can result in the spring not being able to absorb and release enough energy, leading to a weaker or less efficient performance.

Can spring compression distance be adjusted?

Yes, spring compression distance can be adjusted by changing the force applied to the spring or by altering the spring's design or materials. In some cases, additional components, such as spacers or shims, can also be added to adjust the compression distance of a spring.

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