A mass hung from two attached springs

  • Thread starter Thread starter shawli
  • Start date Start date
  • Tags Tags
    Mass Springs
AI Thread Summary
The discussion focuses on calculating the total extension distance of two springs supporting a mass. The initial approach incorrectly treated the springs as a single unit, leading to an incorrect formula for extension. The correct method involves treating each spring separately due to their different spring constants, resulting in the formula x = mg(1/k1 + 1/k2). Clarification was provided that the forces on each spring are not the same, which is crucial for accurate calculations. Understanding the individual behavior of each spring is essential for solving the problem correctly.
shawli
Messages
77
Reaction score
0

Homework Statement



A light spring with constant k1 is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant k2. An object of mass m is hung at rest from the lower end of the second spring.

a) Find the total extension distance of the pair of the springs.

Homework Equations



F= ma
Fs = kx

The Attempt at a Solution



I treated the two springs as a single device and made the following equation for the forces acting on the mass:

Fnet = 0
Fs1 + Fs2 - mg = 0
k1x + k2x = mg
(k1 + k2)x = mg
x = mg/(k1 + k2)

But this answer is incorrect. (Correct answer: x = mg(1/k1 + 1/k2).

Would someone be able to clarify what is happening in this system? I seem to be missing something...
 
Physics news on Phys.org
Seems like you need to treat each spring as a separate system. They have different spring constants so will put out different forces.

k1x = mg gives x1=mg/k1 similarly x2 = mg/k2
add x1 and x2 gives you total distance the spring is stretched.

Just my $0.02, this is my first year taking calculus based physics as well.
 
Hm I see. I was wrong to assume the forces on each spring would be the same...

Thank you! And I hope your physics course(s) are going well :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top