SPRING is a freeware GIS and remote sensing image processing system with an object-oriented data model which provides for the integration of raster and vector data representations in a single environment. It has Windows and Linux versions and provides a comprehensive set of functions, including tools for Satellite Image Processing, Digital Terrain Modeling, Spatial Analysis, Geostatistics, Spatial Statistics, Spatial Databases and Map Management.
SPRING is a product of Brazilian National Institute for Space Research (INPE), who is developing SPRING since 1992, and has required over 200 man-years of development and
includes extensive documentation, tutorials and examples. More than 70,000 users from 60 countries have downloaded the software, as of January 2007.
tried writing the x position as
x = Acos(wt) (ignoring the phase)
so that d2x / dt2 = -w2x
Substituting that into the individual motion equations would get the required result for the individual masses, but I am not sure how to combine the equations to get the reduced mass
m=6.00Kg
K=145N/m
a=1.80m/s
Fp=Fs
ma=-kx
(6)(1.8)=-(145)x
x=-0.0745m
I’m just wondering why I see some people make Fs=kx instead of -kx? isn’t the force of a spring a vector?
Summary: I am confused in when to use 1/2 e squared k
we studied in class that normally, the extension of the spring is directly proportional to the tension applied
and we did this question :
what is the work that has to be done for a spring to have an extension of length e
as the area of the...
I don't understand the question. If it want the mass to stop and reverse its direction, then does that not means ##vo## can be anything? (obviously not 0 since it will make the system not moving at all).
Trying to derive the length of a spring hanging under its own weight. I was trying to approach it like a series of small springs free length ##l## connected in series, in hope to use a limit as ##l \to 0 ## to get the final result, but either I'm bungling it, or it just doesn't work.
I'm...
yooo.
Some help on the following problem would be much appreciated.
I don't get how to solve the two equations I obtained for the COIs A and phi.
calculated: ##\omega == 7rad/s## and ##\gamma = 0.396s^-1##
for part C
we have two initial conditions:
at t = 0 > ##0 = Acos(\phi)##
at t = 1s >...
What I've done so far is find the spring force through
##F_s = -kx##
##F_s = -111*16.7##
## = -1853.7N##
My conclusion was that since this is the spring force, the tension force must be just the negative of that so ##1853.7N## because the net force has to balance out, but I am horribly...
I don't understand what I have done wrong in part (c) I have the initial velocity for the second part of the motion and have the final velocity zero and then the net work done is W_mg + W_Fs and the actual answer for x is 2.37m
Could I get some help/tips please, thanks in advance.
Here is my...
(a) By setting up a coordinate system with the x-axis pointing to the right and the y-axis pointing downward we have ##\begin{cases}-kx_{eq}+T_1+F_{s}=0\\ -RF_{s}+rT_1=0\\ r_p (T_2-T_1)=0\\ -T_2+mg=0\end{cases}\Rightarrow x_{eq}=\frac{mg}{k}\left(1+\frac{r}{R}\right)## which coincides with the...
Given the pendulum setup below:
Details:
##m## is the mass of the bob
## r ## is the instantaneous length of the spring
## \theta ## is the angle the bob makes w.r.t vertical
## I ## is the bobs mass moment of inertia about pivot
## l_o## is the free length of the spring.
##k## is the spring...
A longer pendulum swings slower. So changing the length l of the pendulum changes the period T, which affects the timekeeping accuracy. But the problem is talking about the body on the spring, not the string. So the second formula cannot be applied here directly and I don't know how to progress...
How can I find the maximum bending moment and maximum deflection for a spring?
It would be very helpful if you could explain the specific procedure and formula in an easy-to-understand manner.
that's all, thank you very much.
I'm reading an article about the order-chaos-order sequence of a spring pendulum [Ref 1], as I'm reading it I'm trying to reproduce the graphs and results through Mathematica.
However, I am new to this software.
I will list below some of the most important equations mentioned in the paper.
"In...
i would like to get some help and to understand why my answer is incorrect , here is how i did the first and second part.
about the first part i did it right but i don't understand what I am doing wrong in the second part i tried -k and i get -23.997 and i also try +k and i get 23.997 but they...
Summary: How to express ωₙ in terms of only mass (m) and stiffness (k)? I tried doing it with F=kx but it is out of my ability to simplify it to only m and k.
Here is my approach:
I was doing the exercise as follows:
I am not sure if you agree with me, but i disagree with the solution given.
I was expecting that the kinect energy of the mass ##m## (##T_2##) should be $$T_2 = \frac{m((\dot q+lcos(\theta)\dot \theta)^2 + (lsin(\theta) \dot \theta)^2)}{2}$$
I could be...
In question 1, the spring constant from the two formulas was not the same. When we used the first formula, we got that the spring constant was 7.83 N / m. The second formula we got that the spring constant was 8,03 N / m.
In questions 2 and 3 I do not know and am unsure about how to answer...
My wife just lifted a stack of dinner plates in a cupboard and found about a thousand carpenter ants in a 3" ball. The plates were on a dishtowel that served as a cupboard liner and the ants were under the dishtowel.
There was zero detritus, zero rot, zero poop, zero sawdust, no eggs, no...
Part A) So from a force diagram we can see that the only two forces acting in our system are the spring force(positive y axis) and the weight of the rocket(negative y axis), which means the spring force is equal and opposite to the weight force.
The weight is simple enough ##12* 9.8=117.6N##...
Hi everyone,
Any alternatives you know for metal spring? The idea is to find a replacement for a metal spring that has the same efficiency in bouncing an object that's hung by the spring. Any input of a kind would be valuable.
TIA
Ei = 1/2 K (x)^ 2
K = .0152N/m
x = .0375 m
Ei = 1.06x10^-5
Ef= 1/2mv2 + mgh
m = .164kg, v is unknown, h is .0375sin(8.3)=.00541, Ef set equal to Ei
1.06x10^-5=1/2(.164kg)(v^2)+ (.164kg)(9.8)(.00541)
v = .3254m/s
I have gotten this answer multiple times but it is not correct. I am going...
I'm building a power hammer for a buddy's forge and am working through the design phase. We are going to use a 1200lb leaf spring to actuate the upper anvil. Pretty much got the mechanical design down, but having some issues with the selection of motor and control.
The leaf spring we selected...
A backpack is attached to a spring scale which is attached to the ceiling of an elevator. The elevator is moving downwards with an acceleration of 3.8 m/s^2. The scale reads 60 N (Fscale). What is the mass of the backpack?
The solution to this problem says that Fscale - w (weight) = Fnet.
This...
EDIT for clarity: I solved the question, just asking for if the explanations make sense and if the mechanical energy is considered to be conserved before and after the collision due to reasons listed below the photo.
I hope this image is readable (grr, scanner is janky).
I'm guessing the...
1) By the Work-Energy Theorem, ##W=K_f-K_i=\frac{1}{2}I_{0}\omega^2=\frac{L^2}{2I_0}.##
2) By assuming that the initial length of the spring is ##0##, calling its final length ##S## and ##T## the tension in the rope connecting the pulley and mass ##m_p## I have: ##\begin{cases}(kS-T)r=0\\ m_p...
I'm surprised that this question only occurred to me recently. If a have an electrically charged mass attached to a spring and set it oscillating, the resulting production of electromagnetic waves must cause a kind of "friction", a force resisting the motion of the charged mass, so that its...
This is the problem I am working on at the moment. The question states that the bar is at rest in this state. At a 60 degree angle to the horizontal and supported by the vertical spring at B.
Small oscillations are introduced and I am required to find the equation of motion and the natural...
I have used the work energy theorem like all source have shown me an have arrived at the right answer
where work one by all the forces is the change in kinetic energy
-1/2kx^2 - umgcosΘx +mgsinΘx = 0 is the equation
which becomes
-1/2kx -umgcosΘ+ mgsinΘ = 0
where k= spring constant
u=...
The hypothesis is that the force of a real spring can be described as $$F = -kx + \alpha x^2$$ with x being the spring deformation and k its constant. The \alpha x^2 would be the force lost by the spring as x becomes too big. To test that, a system was build with block of mass m suspended by a...
In the initial position the spring is previously compressed, then the block adds a force, and the spring is again deformed. I think the energy balance is incorrect; the potential energy of the spring is repeated.
Hi,
does anyone know how to calculate the current spring constant of a variable pitch spring when under compression. Since some of its coils get inactive when compressed the stiffness is increasing and consequently “k” is changing as well, is there an equation I can use to calculate the new...
I've tried establishing a systems of equations with -(1/2)kx^2+mpgx+(1/2)mpvp^2=0, but this leads to an annoying quadratic that when solved does not give one of the listed answers. My thought was that the spring's potential energy is going to be equal to the gravitational potential energy and...
I just need someone to check my calculations. There are no specific numbers, I just need a general equation. Look at my attached solution. I think I did everything right but the worrying detail is that gravitational energy is not included and I am not sure whether it should be.
Thanks in advance.
I averaged the masses and times (which i took the time given and divided by 10 because in the problem it says you measure the time it takes to complete 10 oscillations) then plugged them directly into the T=(2(pi)((m/k)^1/2) and got the wrong answer. This is really confusing me because I don't...
Here is the image the answer is say 100 but why. Why isn't it 0. Is it because of the pulley but even with the pulleys the tensions in both ropes are 100 N.
Hi,
I'm trying to calculate the deflection of a leaf spring. I've found several formulas but the one I'm currently using looks like this: $$f=k \frac{F L^{3}}{6EI}$$ where: ##k## - coefficient depending on the way leaves change their shape in the width direction (in my case it's ##1## because...
Question:
A string spring is connected between two bodys with a rope above them.
M1 = 25Kg
M2 = 50KG
Distance between them is 100m.
I answered a bit and got to the point where the distance between the two masses are 110m ( the mass below got 10m lower and is on balance, I mean, acceleration = 0...
the mass will drop 1 m before it comes in contact with the spring. I’m stuck afterward. Please help.
The total energy of 1 m is mgh= 1kgx9.8m/ss x 1m = 9.8J
9.8J = .5 x 20N/m x x^2 ---> x = .99 m
the spring is compressed by .99 m ?
This is a spring problem
From this, it says I need to answer in terms of kinematic friction which to me doesn't make much sense. I also looked at similar questions online to the "in terms of" problems and they don't use all four variables in their derived equation. Do I not need to use all...