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 October 23rd, 2014, 04:38 PM #1 Member   Joined: Oct 2014 From: NJ Posts: 46 Thanks: 2 Why is the value of s = 1/3? a mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 3 inches above the equilibrium position F = ks 24 = k*(1/3) why is it 1/3 inches and not 4 inches?
October 23rd, 2014, 05:03 PM   #2
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Quote:
 Originally Posted by shreddinglicks a mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 3 inches above the equilibrium position F = ks 24 = k*(1/3) why is it 1/3 inches and not 4 inches?
If you are trying to find k then F = -ks (note the - sign!) with s = 4 in is the one to use.

-Dan

October 23rd, 2014, 05:06 PM   #3
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Quote:
 Originally Posted by topsquark If you are trying to find k then F = -ks (note the - sign!) with s = 4 in is the one to use. -Dan
The spring is not getting compressed, it is being pulled so it's a positive value.

I also think that s = 4, but my teachers answer key says s = 1/3. I searched the question on Google and it also says s = 1/3.

 October 23rd, 2014, 05:28 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra The force operates in the opposite direction to the displacement, so it always has a negative sign. The force is also proportional to displacement from the equilibrium position, so I'd expect 3 at the moment. What value are you using for gravity? The force wouldn't normally be 24 if the mass is 24 because F = mg.
October 23rd, 2014, 05:37 PM   #5
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 Originally Posted by v8archie The force operates in the opposite direction to the displacement, so it always has a negative sign. The force is also proportional to displacement from the equilibrium position, so I'd expect 3 at the moment. What value are you using for gravity? The force wouldn't normally be 24 if the mass is 24 because F = mg.
I am using 32. My differential equations teacher likes to use feet per sec instead of meters per sec. The mass is 3/4. But that is not a force.

Do you know where 1/3 comes from? My teachers answer key has 24 = k*(1/3)

Last edited by shreddinglicks; October 23rd, 2014 at 05:41 PM.

 October 23rd, 2014, 06:00 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra I'm a bit confused, because "pounds" is a unit of mass, not of force. So when it says 24 pounds, I assume that is the mass. Also, for which point are you trying to calculate the force? Because the force varies with the displacement $s$.
October 23rd, 2014, 06:08 PM   #7
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Quote:
 Originally Posted by v8archie I'm a bit confused, because "pounds" is a unit of mass, not of force. So when it says 24 pounds, I assume that is the mass. Also, for which point are you trying to calculate the force? Because the force varies with the displacement $s$.
Force is measured in pounds or in the SI system it's Newtons.

Mass is measured in slugs, or in the SI system as the kg.

Pounds is NOT a unit of mass. Pounds is a measurement of weight. By definition weight is the force of gravity on an object. That's why you will weigh less on the moon. Mass is universal. If my scanner wasn't broken I would send you the proof from my engineering and physics textbooks.

According to my physics and engineering books you get the spring constant, "k" by calculating the displacement created by a force on the spring. So the value of, "s" must be 4 inches.

I want the value of, "k" I already know the force is 24 pounds.

My only guess is that the units for K must be, "pounds per feet," instead of, "pounds per inch."

Last edited by shreddinglicks; October 23rd, 2014 at 06:16 PM.

 October 23rd, 2014, 06:17 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Isn't $k$ supposed to be the spring constant? The spring constant is the displacement per unit force on the spring. So that would be $\frac{4}{24}= \frac16$ in/lb. I apologise if I'm wrong about pounds. I've always equated pounds with kilograms (as do most converters).
 October 23rd, 2014, 06:24 PM #9 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey stuff. I am wondering about your system of units. Specifically if you are using feet for s then an extension of 4 inches would be 1/3 feet. -Dan Thanks from shreddinglicks
October 23rd, 2014, 06:25 PM   #10
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Quote:
 Originally Posted by v8archie Isn't $k$ supposed to be the spring constant? The spring constant is the displacement per unit force on the spring. So that would be $\frac{4}{24}= \frac16$ in/lb. I apologise if I'm wrong about pounds. I've always equated pounds with kilograms (as do most converters).
I believe that's upside down 24 = K*4

k = 24/4 = 6

Also, pounds can be converted to kg for the metric system, but that is not the same as mass. You have to convert to mass from weight using f = mg. f in this case being force of gravity.

f = mg is the same as Newton's f = ma. a = 9.8 m/s^2 at sea level.

Last edited by shreddinglicks; October 23rd, 2014 at 06:30 PM.

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