Ok, so I guess I am asking what force causes the force of the leg muscle onto the foot and since it is an internal force, what is the reaction force to that force?
Suppose a person is walking on the ground without slipping. For the free body diagram of just the person, only the frictional force is drawn in the horizontal direction. The force exerted by the leg muscles to generate a force against the ground is considered an internal force. What would be the...
I was thinking about doing KVL around the circuit at the right but I noticed when the switch opens, the current through the circuit at the right is not the same throughout
-5 + Ic*2*1-^3 + Ic*10^3 = -Vc
Ic is not the same around the right circuit so I am stuck....
Ok, I wrote my equations wrong initially. But I am wondering what frequency is the impedance an open circuit? The impedance would have to equal infinite?
For the first circuit, Req = ZL + ZC = -j/(w*C) + j*w*L = 0 for short circuit, so w = 0?
For the open circuit case, -j/(w*C) + j*w*L = infinity, so w = infinity?
Is that correct?
Does the work energy theorem: delta W = delta KE apply with external and nonconservative forces as well or does that formula only work if there are only conservative forces and external forces = 0?
Ok, I had a follow up question. How come in the work energy theorem, work = change in kinetic energy; but in the energy conservation law work = change in kinetic energy + change in potential energy?
I was wondering why energy of capacitor does not equal change in kinetic energy PLUS change in potential energy where potential energy is the change in the potential energy of the charges. I believe that should be so because energy conservation = change in kinetic energy plus change in potential...
I am reading a research paper that says weak chemisorption of silver with olefins basically means more effective olefin/paraffin separation. If silver weakly binds to olefins, wouldn't that make a lousy separator?
$$x(y) = b - y\frac{b}{h}$$
$$y(x) = \frac{-h}{b}x +h$$
No they would be different because $$\int_{0}^{h} y*x(y)dy = \int_{0}^{h} y*(b - y\frac{b}{h})dy$$
this does not equal $$\int_{0}^{b} y(x)*y(x)dx = \int_{0}^{b} (\frac{-h}{b}x +h)^2 dx$$
To find the y value of the centroid of a right triangle we do
$$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yxdy}{\int dA}$$
What is wrong with using
$$\int_{0}^{h} ydA = \int_{0}^{b} y*ydx$$ as the numerator value instead especially since ydx and xdy are equal and where h is height of...
Yes, I know that F is the tension but is F the tension of the string in the x direction or is F the tension that is tangent to the string.
My book derives it like this, implying that F is the tension of the string in the x direction
The speed of a wave in simple harmonic motion on a string is $$v= \sqrt{\frac{F}{\mu}}$$ where v= the horizontal velocity of the wave on a string.
Is the F the horizontal force or the resultant force (combination of Fy and Fx)?
Summary: Can these two equations be solved for x like a system of linear inequalities, and how?
##x- 2y \le 54##
##x + y \ge 93##
We start with
##x- 2y \le 54##
##x + y \ge 93##
Multiplying the second equation by 2, we have ##2x + 2y \ge 184##. We cannot seem to cancel the y out with the...
What about "I am wondering how come you are dividing the tensions into 0<ϕ<π/2 and π/2<ϕ<π. I interpreted the tension being 20 N - 60 N at all infinitesimal points around the pulley."
So is ##dl = -R\sin\phi d\phi \vec i + R\cos\phi d\phi \vec j##
or ##-\sin\phi\hat xdx+\cos\phi\hat ydy =...
I am wondering how come you are dividing the tensions into ##0<\phi<\frac{\pi}{2}## and ##
\frac{\pi}{2}<\phi<\pi##. I interpreted the tension being 20 N - 60 N at all infinitesimal points around the pulley.
Also I do not understand why we have to multiply by dl in ##d\vec{T}=20d\vec{l}##...
I just want the translational force of the pulley and am not sure if to calculate the translation force of the pulley I would have to consider all the forces at all the points along the circumference
Below is a pulley with mass and a string around it where the tension of one end of the string is 20 N and the tension of the other end of the string is 60 N. I know there is a net torque due to the differing tensions of the string, so I am wondering what the translational forces are on the...
So you're saying the equation should beJ = ## \int_{t1}^{t2}(-k{x(t)})dt## where k is the spring constant of the ball, and x(t) the ball's compression?
So the force exerted on the ball by the ground equals the force exerted by the ball on the ground, so we were just talking about a sign difference?
So then if the impulse is the force exerted on the ball by the ground, J = ##\int_{t1}^{t2}(k_2{x_2(t)})dt##, where ##k_2## is the spring constant...
According to my class notes, the impulse (change in momentum) for the ball, is just the force the ball exerts on the ground integrated for a certain time duration.
Yes. I am talking about an elastic scenario.
In scenario 2, would the impulse = ##\int_{t1}^{t2}(0.5k{x(t)}^2 + 0.5k_2{x_2(t)}^2)dt## or just J = ##\int_{t1}^{t2}(0.5k{x(t)}^2)dt## where k is the spring constant of the ball, and x(t) is the ball's compression, or should it be ##k_2## the spring...
For conservation of mechanical energy for dropping a ball on the ground, I have 3 scenarios for conservation of energy:
1) For the scenario right before the ball hits the ground ##mgh = 0.5mv^2##
2) For the scenario while the ball hits the ground using the potential spring force of the ball and...
Ok, got it.
My last question was if there were an external force that caused the ball to drop with initial velocity v, would conservation of energy still be applicable since there was an external force? Or would the conservation of energy be applicable only after the external force stopped...
What if 2 balls collide each with mass m, one with initial velocity and one at rest, and both end up with 0 velocity.
The conservation of momentum doesn't make sense like this: mv = 0. How come?
So if there is nothing moving after the collision, momentum is not conserved? Is there an external force (since that is when momentum is not conserved)?
Yes you are understanding me properly. Conservation of energy doesn't work which makes sense, but conservation of momentum does not work either in a case where the ball is dropped with a certain velocity and collides with the floor with 0 velocity, mv = 0... how come?
No, that is not what I mean. In my conservation of energy problem, I am setting the ground with height h = 0; If we have a ball dropped with height h > 0, the initial energy state is mgh. I want the final energy state to be when the ball hits the floor at 0 velocity, so the final energy state is...
Ok, but I would like to understand why it violates conservation of momentum. Is there a net external force? What is the external force? Is it the frictional force? But the frictional force between the 2 masses exists in inelastic collisions as well (which does have momentum conserved)
If we have a ball with mass m dropped from a height h down to the ground, how come we can't set the conservation of energy equation just as the velocity of the ball turns 0.
mgh = 0
If instead the ball were moving with an initial velocity v, would the equation be
##mgh + \frac{1}{2}mv^2 = 0##...