Is this from an engineering course or something? I see on the lower left hand corner there in the photo it says "Dynamics 3". Not sure what that means. This looks to me like something out of a machine design class I'm currently taking. There's a whole chapter on balancing with problems that look...
This is fine if you treat Δh as a positive number (this would work with a coordinate system with the downward direction being positive). Perhaps I'm being a little too pedantic about consistency with signs and coordinate systems. The most important thing is that your result for work is positive...
We're dealing with gravity, but the following approach is supposed to work with any force that can be considered conservative (means that work done by that force is path independent). You derive an expression for the work done by that force over some specified path, which begins at point A and...
If you consider the positive direction for the motion to be "up", and the ball is falling "down", then the work could be expressed as -mgΔh as long as we consider the
Δh to be negative due to the choice of orientation for our coordinate system.
However, like what LemmeThink said, you don't...
It's worth pointing out two things about your approach so far:
Do you see that your ΔKE expression implies an initial state of rest (if what you mean by v is vf) ? You did not state that as an assumption, so you might want to make sure that it's an assumption worth making.
Either the PE...
Your results look fine to me. Remember that it doesn't matter if an object starts out with 0 m/s or 1000000 m/s; any given instantaneous acceleration a = (net F)/m will only tell you how the velocity changes with that time at that instant.
No problem. Just one last thing.
Not sure if your edit on your post for the value of W worked. Just to clarify it should be 175 N. Although I will assume that you must have gotten that result if you calculated 17.8 kg as the mass.
Also, as a quick alternative to going through all the math: Examine each FBD carefully. Assuming static equilibrium of the ring at point D, it can be said that F(DE) must be larger in magnitude than F(CD) since only one component of F(DE) balances the force F(CD). A slightly more careful...
Alright, so substitute in for F(AC) in this equation. Use this to write F(BC) in terms of W. Afterwards you can then write F(AC) in terms of W. You will then have expressions for all the forces of tension in the ropes in terms of W. Comparing all of these you should be able to determine which...
I've always wished that I had more time to read up on things that I'm interested in, but I wouldn't go so far as possibly insulting the OP by equating their situation with this "fantasy". Please correct me if I'm wrong, but would you "trade places" with them? You could catch up on all that...
Think of a simple case of centripetal motion at some constant potential energy state (perhaps a "car" going around a circular, horizontal track at constant speed). The centripetal force is responsible for the change in motion, but that doesn't mean the energy changes (constant speed means...
I think you might be a bit hard on yourself. You recognized that you needed help and sought it out from those who are likely to be more knowledgeable. No one got hurt, and no one died as a result of your failure in solving this math problem. In general as well, having the humility to recognize...
Since the surface integral is an integration over a square area in the x-y plane, a single variable integration would also work. I'm not sure how far you've gone in your study of calculus, but if you're new to the subject it's possible you've seen the problem of finding area in the x-y plane via...
Doesn't matter what kind of turning you envision for the pipe. A condition for static equilibrium is the sum of moments about any arbitrary point is zero. If you try to sum all moments of forces with respect to points which are on the line of action of Cy, you will never be able to solve for Cy...
If you push with a force greater than kinetic friction, the object speeds up. If you want to "get it to move at constant speed" no matter if it is a higher or lower speed than you started with, you then must stop the acceleration by going back to a state of equilibrium where again the push/pull...
Both equations (if all the symbols represent the same quantity) imply that V = v*n, is this true?
One equation has a proportionality constant (R) written on a per mol basis, and another on a per unit mass basis, and so they will have different values if both equations are to be consistent with...
This is my first time using ansys Fluent, and I am trying to follow this tutorial for modeling a heat exchanger under various conditions. I have been provided with a "Case 1 Mesh" file of the extension type .cas which is to be used in our simulation involving several or one heating element(s)...
Do you have access to an e-text through your MasteringPhysics? You need to know the equations of 1-D kinematics to help you solve this problem. I highly doubt that many students will get very far through a calculus based physics course without having some kind of standard reference material to...
I wouldn't be able to help you there. I also used a text co-authored by Zill when I took differential ('Differential Equations with Boundary-Value Problems' by Zill and Cullen 7ed.) when I took differential.
I saw a text by Ross recommended in the thread "How to self-study mathematics?" However...
Is this a high school class? There should be a textbook or some sort of reference material you have for the course.
The link you posted previously are all special cases of the formulas I was referring to, specifically arranged and simplified for 'free-fall' constant acceleration. The problem...
You should use the homework template, to begin with.
There are some basic equations that describe objects under constant acceleration. Have you tried applying them to this situation?
That looks like a correct implicit soln; you can also tidy up the RHS of your answer a little by applying some rules of exponents:
\begin{equation} e^{x}e^{-y} = e^{x-y} \end{equation}