Spring Definition and 999 Threads

I am trying to understand the reaction of a steel coil compression spring when pulsed. The spring I am interested in has the following physical characteristics: k (spring constant in pounds per inch) = 2.88 d (wire diameter in inches) = .043 n (number of active coils) = 29 D (mean diameter of...

a) When the system is in motion for the first time, the force causing ##M## to move is contact force with ##m## so: $$\Sigma F=M.a$$ $$N \sin \alpha=M.a$$ $$mg \cos \alpha \sin \alpha =M.a$$ $$a=\frac{mg \cos \alpha \sin \alpha}{M}$$ Is that correct? b) Is acceleration of ##m## the same as...

I completely missed the collisions approach when I first tried to solve this and tried using the work-energy formula. I am wondering if this approach can be made to work? Here is my attempt: So I let the work done on the ball be ##W_b## and work done on spring be ##W_s##. Then $$W_b=\Delta K =...

My attempt at a solution: Is my logic accurate/correct, and is my answer correct? I consider the forces acting to be: Restoring forces in springs parallel, and Force of the current-carrying conductor in the Magnetic field. I imagine a vertical displacement of y upwards ( direction determined...

Hello everybody. I'm trying to figure out how to calculate gas springs for specific doors not with hinge, but lifted up, like on the picture. Do anybody experience with such type of mechanism? Also I'm looking for helpful software, for gas spring calculation (different types). I found screens...

Summary:: Doubt in a spring exercise Text of the exercise "a mass of ##m = 0.4 \ \text{kg} ## is attached to a spring and it oscillates horizontally with period ##T = 1.57 \text{s}##; the amplitude of the oscillation is ##d = 0.4 \text{m}##. Determine the spring constant, the total energy of...

I do not understand how in part a, the units for K can be N/m. If Work is in joules which is kg*m^2/s^2 and we are diving by x^2 which is m^2, then m^2 should cancel out and we should be left with kg/s^2. Kg/s^2 makes more sense because in part b when you find the work done you are multiplying...

I just would like to check if my corrections to each of the wrong options are right. A) & B) elastic potential energy greatest at B, not at A or O, based upon Eq.1 D) Since dU/dt = 0 at O because O is at equilibrium where no change of spring length. E) At A, dU/dt = -kx. At B, dU/dt = -kx = -kx...

Assuming zero spring mass and zero friction, At the greatest value of x, the loss in gravitational potential energy should equal the loss in elastic potential energy. so I did (1/2)kx^2=mgx to isolate x in the formula, x=(2mg)/k then I plugged in my values so: (2*13.6*9.81)/8.8= 30.3218...

Here it writes Wby spring= Uinitial - Ufinal. So is it equal to opposite of the change in potential energy ?

Hello! So here what I did is first calculated the potential energy; $$ E_p = \frac{1}{2} * k * x^2 $$ E_p should be = 0,125 J Now i tried calculating the kinetic energy, I used this formula $$ E_k = \frac{mv^2}{2} $$ to get v I used this formula $$v = x *\sqrt{\frac{k}{m}} $$ v should be =...

Hi all, I'm studying the compression spring design issue that occurred in a machine design application. As illustrated below, spring is bouncing or oscillating after impact to a stopping surface (1 -> 2 -> 3 -> 4) and eventually stop after few bounces. Ideal case for this application is to...

So first I made an equation representing the forces Fnet=kx-12.8v a=1/m(kx-12.8v). Now I am not really sure how to get w from this. I could argue the mass is at its max amplitude when a=0, but that wouldn't help me find w. If I say x(t)=kx-12.8v, then 1/m would be w^2, but this isn't right...

Hi, On a driving force graph ##y = displacement (m)## and ##x = time## where the external force start at t = 0 and the system is in equilibrium at x=0, it's easy to find the driving frequency. $$F = \frac{\omega}{2\pi}, \omega = \frac{2\pi}{T}$$ and we can get ##T## easily with the steady...

The other day when I solved a spring mass damper system in Matlab, I was curious how in olden days would have people solved the equation. We all know the 2nd order differential equation of the system: However if I know the time, damping coefficient, stiffness and mass, will I be able to find...

So first I found what b/2m is and got 0.287129. Then I found what the sqrt part of the equation was and got 1.128713. Then I added them together to find w. Then I divided by 2pi to find frequency and got 0.255, but the answer is 0.180.

Since its critically damped that means k/m=(b/2m)^2, which would mean w=ib/2m. So m=ib/w. My issue now is that I need to find work. I could put w back into x(t) to get Ae^((-b/2m)t+phi). I guess I could make this Acos((-b/2m)t+phi)). But I am kinda lost at this point. Sure, I could find the...

Hi, First of all I hope it doesn't bother if I ask too much question.I found the values of ##u1,u2## for 2 differents posistions ##(r1,r2 )## and I now have to determine the spring constant (k).I'm thinking about using$$ F= -kx $$ with ##F = -\frac{du}{dr}## then $$ U = \int -kr \cdot dr...

i attempted this problem by using conservation of energy, mgh=1/2kx^2 mgD=1/2kD^2 2mg=kD k=2mg/D why is it wrong ? btw the correct answer used mg = kx which is mg/D

I honeslty don't quite know how to start. It seems like the Hooke's coefficent k is independent of the answer to this problem. I would also appreciate any clue of expressing the condition when "balls will collide again". The fact that all balls can keep moving make this rather difficult. It...

I am trying to design a small box with a hinged door/lid that opens with a torsion spring. When closed, the door will latch onto a mechanism where the open button is, and when the button is pressed this mechanism will slide away, removing the latch out of the way so that the door can fly open...

If M is displaced by an amount + x from equilibrium.What happens to the two masses at the point of release for displacements of x and less? Will they remain static because mass m provides whatever it takes to stop mass M from moving till some x where m slips and M oscillates or Will they...

I'm having trouble with this problem, I think I solved it but I don't know if what I did is right... At first when the velocity is 0 and the spring is at its natural length, there's just gravitational potential energy, so $$E_i=mgh$$ And then, when the mass is released and then reaches its...

I attempted using f = 1/(2pi x sqrt l/g) For Earth I found the value of length to be 0.0276m. Then I substituted the value in the equation, putting (1/3)g instead of g, to find the value of f in Mars. My answer is C. I am confused. Please help me.

From Kleppner's Intro to Mechanics (Example 4.7, wording not exact): Two identical blocks a and b each of mass m slide without friction on a straight track. They are attached by a spring with unstretched length l and spring constant k; the mass of the spring is negligible compared to the mass of...

So i know Ek=123.48 from the potential energy that converts into kinetic energy (Ep=1/2kx^2). Now by conservation of momentum, m1v1=m2v2 So m2Sqrt(2Ek/(m1+m2))=m1v1 This is where I am making a mistake I think, but not sure how. Answer is suppose to be 4.44

So first I looked at the forces acting on m1 m1a1=F spring on m2-F spring Then m2 m2a2=F spring-F spring on m1 using 3rd law, m2a2=F spring+F spring on m2 m2a2=F spring+m1a1+F spring m2a2=2F spring+m1a1 Not entirely sure if I've done the above correctly, but I am stuck now because I have...

First I wanted to find the kinetic energy the mass had when it hit the spring (converted from the potential Energy it had) thus Ek=mgh=9.8*2.6*3.5=89.18 Now I know as this Ek changes to 0 the potential energy of the spring as its being compressed will be at its maximum so, Ek=Ep...

First calculated non conservative work from friction using Ff=umg. Non conservative work was -8.82. Initial kinetic energy, 1/2mv^2, was 136.89. Change in potential energy, 1/2k(x)^2, was 8.1216. Ekf-Eki+Change Ep=Work NC Ekf=W NC+Eki-change Ep =-8.82+136.89-8.1216=119.9484 Ekf=1/2mv^2...

ma=-k2l+k1l (ma-k1l)/l=-k (-65.12-35.7)/0.51=-k k=197.69, but answer should be 57.7

Apart from the trivial elements of the motion equation (m z'' = -kz -mg), I am required to find the force produced by the Eddy currents induced by the moving magnet. To do so, I calculated the magnetic flux through the hole plate: For a magnet: Bz=μo m 4π. 2z^2−r^2/(z^2+r^2)^5/2 so Φ = a→ +∞...

1/2 kx^2 = mgh

Hi everyone I've just completed a mechanical engineering degree, and one aspect of the classic 2DoF quarter car model that still bugs me is the representation of the tyre as a linear spring attached to the ground. Does anyone have any experience of modelling the system with the tyre spring...

I have solved this question in Center of mass frame of reference, by using energy conservation. However, I'm not able to write any equation when I'm trying to solve it in Ground frame of reference. Can somebody please let me know how to begin in Ground FoR.

Hello! So my main and first problem about this question is, I do not know what the problem is about. What I mean by that is, in class we talked about pendulums and are given formulas and assignments regarding pendulums. But this problem here does not seem like it has anything to do with...

Summary:: What does I need to consider in order to get the right spring? Hello. I need a compression spring that require 10 lbs of force in order to be compressed 1cm. The springs outer diameter (De) has to be 1cm. The spring will be made out of piano wire. Which values of specification does...

There used to be sold a style bicycle handlebar bags that used what I think is a formed spring steel rod that fit over the handlebars and looped under the handlebar stem that supported a handlebar bag. For whatever reason this style does not appear to be available any more. I think it is a...

First and foremost, I found the max stretch of the spring using the strain energy formula(x=√((2*0.25J)/k)) ). Then, the maximum force exerted(Fmax=k*xmax), in order to find out the seconds needed for the force in [N/s] to reach its maximum value. Now, I got confused about how to find the...

A block of mass 0.2 kg which slides without friction on a θ = 30° incline is connected to the top of the incline by a mass-less spring of relaxed length of 23.75 cm and spring constant 80 N/m as shown in the following figure. (a) How far from the top of the incline does the block stop? (b) If...

the acceleration of the center of mass is ##a_{cm} = F/(M+m)## I considered the forces on the block of mass m(when the system is at maximum extension) I got the equation $$kx - \frac {mF}{(M+m)} = ma_{cm}$$ and from that I got the value of the maximum extension $$x = \frac {2mF} {k(M+m)}$$ which...

a

m1 = top left m2 = bottom left m3 = top right m4 = bottom right My questions: 1. Will all the object (m1, m2, m3,and m4) have same acceleration? 2. Should I assume initial extension of both spring is the same? (only based on the picture) 3. Will the extension of the spring change after the...

Hello! I have some problem getting the correct answer for (b). My FBD: For part (a) my lagrangian is $$L=T-V\iff L=\frac{1}{2}m(b\dot{\theta})^2+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2,\ where\ \boldsymbol{x}=\sqrt{(1.25b-b)^2+(b\sin\theta)^2}-(1.25b-0.25b)$$ Hence my equation of...

Fsp = 90 x 0.12 = 10.8 Ffriction = Magcos(titre) x 0.30 I got the answer 2.09ms2 when the correct answer is 1.11ms2. What am i doing wrong here?

Since there are no external forces, the angular momentum (##L##) and linear momentum (##P##) are conserved. Let's call the left rod ##A## and the right one ##B##. If all the balls were fixed, I'd write ##L_0=L_f## ##L_A+L_B=(I_A+I_B)\omega_f## From this equation I can find the final angular...

Hi I know that the equation of a simple harmonic oscillator is ##\ddot x + \omega^2 x=0##. The thing is that I don't know how to get to that equation in the situations given. In the first situation, I know that ##x) k(x-x_0)=m(\ddot x -x \dot \theta ^2)## ##y) N=m(2 \dot x \dot \theta)## So...

My first attempt was using the period equation of a spring system. I've changed it into k=((2π)^2*m)/T^2, then put Earth's mass into "m" (5.972*10^24), then put the time required for one revolution of Earth around the Sun, 365 days into seconds, 31536000 sec, to "T" So I got (2.371*10^11 kg/s^2)...

Hi, I'm designing a small container with a spring loaded hollow rectangular lid. I want the lid to open when a button is pressed, so I have a torsion spring at the hinge. I want to know if the spring I selected is able to open the lid and also how long it would take to open 90°. Below is a...

First i will use a equation resulted in considering a spring as a continuum limit of massive mass: ω = √(KL/ρ)*kn ρ is the linear density ρ ω = √(KL/ρ)*kn X = A*cos(kn*x)*cos(ω*t) ξ = A (first consequence) X = ξ*cos(kn*x)*cos(ω*t) ∂y/∂x need to be zero in x=L (for strain be zero) kn*L =...