Spring Definition and 999 Threads

At first I tried using U1 = 1/2kx^2 from -x to zero then U2 from zero to +x but I don't know shall I add U1 and U2 or subtract them?

Hi Guys Please refer to my attached derivation, do you think it is acceptable? There is a torsional spring at the base of the column. in reality the column is going to be a complex structure however the torsional stiffness can be approximated using dummy loads on the structural model.

Included a diagram as well... forgive me... I cannot seem to 'uninvert' the attached picture final velocity is zero if we set the lowest point that the mass reaches as zero, then the final height is zero let H be the 'extra' length that the spring reaches over and above the initial stretch...

A thin uniform rod has mass M=0.510 kg and length L=0.470 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The sequence below shows that the rod is released from an angle θ1=59 degrees, and moves through its horizontal position at (B) and up to (C) where it...

Hi all, I'd like to resurrect an old thread in which the original question is still yet unresolved in my mind. Simply put, in a simple quarter car model, the tyre can never leave the ground as it is modelled as a spring attached to the ground surface. In reality however, subject to input, the...

i have no idea

[Mentors’ note: No template because this post was moved from the technical forums. Everything that the template asks for seems to be present in the body of the post] Suppose there is a spring-mass system arranged as shown in my crude drawing. This occurs on a frictionless surface. The spring...

For P, the force will be 2F and the spring constant will be ##\frac{k}{2}## so the extension will be 4 times, and the energy will be 8E but there are no options showing 8E What is my mistake? Thanks

Assume that we have a block connected to a spring. Also, assume that there is no friction, the spring is massless and ideal. If we were to pull on the block with some force ##\vec{F_{pull}}##, we are going to get the spring force ##\vec{F_{s}}## in the opposite direction. Assume that we are...

I dont need anyone to do the sum directly please. I just need some hints, please dont give me the solution. Im unable to understand how the bob will hit a maximum elongation, there are two forces accelerating the bob: Gravitational force and electrostatic force. There is one force...

I need to determine: 1) The initial acceleration of the disk 2) the speed of the disk when the spring reaches minimum displacement For point one I think I should use the free body diagram and then ##\Sigma F = ma##, I'm taking as positive the right and the upward directions and the counter...

There are some magnet drop experiments in the literature and I want to try an alternative experiment to determine if the magnetic fields affect inertial mass. I am designing a submarine shaped enclosure where I will have either two 2"OD 1/4"ID 1"thick N42 magnets with their opposite poles...

For this problem, For part(a), I am not sure if I am solving it correctly. I define the usual cartesian x-y coordinate system at the base of the wall. This gives ##x = l_0 + q(t) + x_w(t) = l_0 + q(t) + d\sin(\gamma t)## which implies that ##\dot x = \dot q + d \gamma \cos (\gamma t)##...

I know gravity needs to be taken into account and that I can find the stiffness but generally I'm pretty lost

Basically, I thought of a weight as a energy storage. But realised you have to output that energy from the same part that inputs the energy. Now I have done some research and found two ways of storing and discharging mechanical work at the same time. First is the Huygens mechanism(maintaining...

1) I have a spring on the ground with no friction and the spring is not attached from one end. If I apply a force ##F## and the spring is massless, will it stretch? I think that it won't. But if it has a mass ##m##, will it stretch now? Will it be ##x=\frac{F}{k}##? I don't know, but I imagine...

I have tried to answer this using the relevant equations I am provided on my formula sheet, however I get stuck pretty close to the end. I start with 1/2mv^2=1/2kx^2 at the equilibrium position, and kx=mg, x=mg/k. This gets me to v^2=mg^2/k, but I don't know where to go from there. The potential...

Let's say the upper piston goes down by ##y_1## and the lower piston goes down by ##y_2## after the block is suspended \ By volume conservation ##s_1 y_1=s_2 y_2## Let the pressure at the location of the upper piston be ##P_c=\frac{ky_1}{s_1}## Pressure at the lower piston : ##P_a=P_c+\rho...

Hello! I have this problem from an old exam I'm trying to solve. The problem is in Swedish so I've translated it: NOTE that I accidentaly wrote $$C\neq 1$$ in the picture below. The correct problem statement is above. But that part is not what I have problems with. The answer key says "if the...

All types of sprag overrunning clutches have some sort of spring to keep the rollers or sprags in contact with the running surfaces so they can engage and wedge to lock the two surfaces together when turning in the locking direction. But is there any kind of sprag clutch where you can...

I'm leaning towards the same, or maybe increase. I actually have no clue.

At the park

I'm doing a personal experiment where I take a conical spring (that is, a spring with two different diameters on either end), hang it from the ceiling, and measure the period of oscillation for different masses hanging below the spring. I do this for two different orientations of the spring; one...

$$F=kx$$ $$k=\frac F x= \frac {50+50~N} {5+5~ cm}= \frac {100~N} {10~cm}= 10~N/{cm}$$ However, the answer is ##5~N/cm##, because the force on the spring is ##50~N##. I am having trouble understanding why the force isn't ##50~N## + ##50~N##. The diagram looks as though the spring is experiencing...

part d- ii and iii ii) my answer is 300-140/300 *100 ke at y = 300 and spring energy at max compression is 140 iii) e is directly proportional to x^2 so it increases exponentially is my explanation correct?

This was inspired by this:Dropping an extended Slinky -- Why does the bottom of the Slinky not fall?. There is that famous demonstration of dropping a slinky, and the bottom of the slinky does not move until the center of mass reaches the bottom. I was trying to figure out how hard are the...

i do know how to do the working but i dont understand the concept stated above...

A torque meter with a triangular slab extension is inserted into a corresponding triangular slot. The C-shaped arm features a V-shaped dent on which a roller is seated. This roller is held in compression by a spring. The roller's positions are labeled '0' for the initial state and '1' for the...

Is there a typo in this question? Supposing there was no friction, the block would fall until the force of the spring was equal to ##mg = 2 * 9.8 = 19.6##, taking the upward y direction as positive. Since ##F_{spring} = -200y## and ##19.6 = -200(-0.098)##, the block would fall 9.8 cm. It's not...

i'm copying from the book... Hookes Law - F = -kx W = Fdcos∅ since ∅ is 180°, W = -Fd = -Fx W = ∫(-Fxdx) now the book says, from Hookes Law equation "the force magnitude F is kx. Thus, substitution leads to W = ∫(-kxdx)" why are they saying to substitute the magnitude of the force and not the...

Can someone explain that, when using the formula (Fs=1/2 kx^2) why do we use x=0.1m instead of 0.05m? Seems like a simple concept but why isn't it 0.05m (since 0.05m from equilibrium). Thanks.

Here is my attempt at the solution: a) The apparatus may only experience acceleration ##a > g## while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second...

I know that we can answer it using conservation of energy or using N's 2nd law. Using N's 2nd Law: ##F = mv \frac {dv}{dx}## ##Fdx = mvdv## For spring we have : ##F=-kx## ##(mg-kx)dx=mvdv## We'll get same result using above equation. My question: Average spring force from 0 to x is ##-\frac...

Max speed occurs when all energy has been translated from spring into box. E (Potential) = 1/2kx^2 E (Potential) = (1/2)(42 N/m)(0.280 m)^2 = 1.6464 N m Ep = Ek =1/2mv^2 1.6464 N m= 1/2 (1.2 kg) v^2 v = 1.6565 m/s

In my physics lab we determined the spring constant of a steel spring. This turned out to be 20 N/m. However, when I search online, I can't see any uses of springs - I know springs can be used everywhere, but nobody seems to specify their spring constant. Anyone know of any applications?

1) a=0 and v=0 2) 0.7 m/s

I think its critically damped by looking at the graph of the solution.

I approach this by considering the four springs in parallel each with spring constant ##k## as one spring with four times the spring constant ##k' = 4k##. The car is dropped and at the moment its tyres touch the ground I assume that the spring is in its resting position. As the car continues to...

I am completely new to this subject and I am trying to find out how I read data off a displacement vs time graph to find the natural frequencies and mode shapes. Lecturer hasn't provided any materials on graphs, just looking for some help and where to go so I can understand it. Thank you

If I take a spring with clamps and I weight that system accurately. Then I compress the spring and clamp it thus giving it potential energy. If I now weigh the clamped spring I should see an increase in mass because of the added energy. Is this the case and something that could be proved in the...

Hello everyone, I'm a new member, and you might see me around from now on. I'm now on a path to understanding the mathematics behind a complicated mechanical machine. My knowledge is basically what I learned during my school days and also during university courses, and for me, it was mostly...

We put object on weight ang get a mass. What would that mass be if we put a spring between object and weigt, so that the spring woul shrink to half its original size?

Source: A.P. French's Vibrations and Waves I do not recognize the first equation, can someone explain how it came to be? The reasoning behind it. How can force on a body attached to a spring at small displacement x be represented as ? I know recognize F = - kx (restoring force) I realize...

My question is whether I've formed the integral for the work done correctly? It just seems a bit unwieldy to me... If I call the extension of the spring ## x ##, I can see that ## z = \frac l 2 + x ## and ## z^2 = \left( \frac {l} {2} \right)^2 + y^2 ##. Combining them gives: $$ x = \sqrt {y^2...

Thank you guys for taking the time to read this - I'm decently struggling with first year and need some tips on how to properly conceptualize problems and learn what the right approach is on certain problems. Have a wonderful day, again thank you for checking this post out!

There is some discussion currently and I was hoping to get some opinions here. The question is in regard to a Hook's law problem. The text gives the problem as seen below. The text says the answer is 50lb/in. Several people have tried from several different approaches. Factoring the "y" equation...

Determine the amount and type (tensile or compressive) of the spring force so that the resulting force is a vertical force. Also get the resultant force. i find 60N (compressive) and resultant forces is 10800 is that correct?

So here's what I did but it isn't right: W = (Kf + Uf) - (Ki + Ui) (2.6)(9.81)(0.45)(-0.01)=(1/2mvf^2 + 1/2kxf^2) - (1/2mvi^2 + 1/2kxi^2) -0.1 = (1/2(2.6)(vf^2) + 1/2(855)(0.02^2)) - (1/2(855)(0.03^2)) 1.3Vf^2 = 0.114 Vf^2 = 0.09 Vf = 0.3 m/s

The 7/8" Kinetic Recovery Rope like this yankum rope is the most common size used for Jeeps, Broncos, and other SUVs. I apologize for English units but that's how ropes are sold and marketed. I've talked to the biggest rope suppliers and they have no idea how to compute the rope's spring...

It's an explosion problem. When two carts are pulled apart, the bigger one takes longer than the smaller one. So the velocity of the bigger one is small, and the velocity of the smaller one is large, and they are opposite each other. So the momentum before the explosion must be equal to the...