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A pulley-and-shaft is connected to a motor which applies torque to spin the pulley as shown below: Now imagine someone jams machine by sticking a broom into one of the holes of the pulley. I would imagine there would be normal stresses at the hole (location of jam) and torsional stress on the...

you mean the "pi" terms? this book suggests using Buckingham pi theorem to find number of pi terms needed.

My book doesn't seem to clearly define "reference dimension", it states "Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T. However, in some instances perhaps only two dimensions, such as L and T, are required, or maybe...

lets say for example, the air drag (ϑ) that wind exerts on a tile is a function of the tile's width (w), height (h), viscosity of air (μ), density of air (ρ) and velocity of the air (V) then ϑ = f(w,h, μ, ρ, V) ϑ : MLT-2 w: L h: L μ : ML-1T-1 ρ: ML-3 V: LT-1 I understand there are 3 basic...

yes, that was my error, I thought I could take the a point as the origin, and just divide the problem into two separate ones, find the acceleration of both relative to the origin and just add up the vectors but that clearly isn't working out. I would need to find the acceleration of B relative...

Homework Statement Homework Equations The Attempt at a Solution First, I find the accelerations of each block, separate the accelerations into Y and X components and then add them as vectors. for B: B goes down the ramp with an acceleration of 3.35 m/s^2 X component : (3.35)(cos 20) =...

since the object is now sliding, kinetic friction is used: Fkinetic = 308.7 N Moment about G : -(308.7N)(0.45m) + (882N)(0.25) - (0.1) T = 0 T = 815.85 N therefore the max weight of C is 83.25 kg (I believe the extra 20 kg difference from my first attempt's calculation is enough to cause a...

Then I believe the acceleration caused by weight C has to be much greater than the deceleration caused by kinetic friction to cause a tipping while sliding?

there has to be a force that causes rotation of the body about an axis, but I'm not sure how to find the magnitude of force that will stop the sliding and cause a tipping.

yes, it will slide because I calculated static friction to be only 352 N.

Homework Statement Homework Equations The Attempt at a Solution When the barrel starts to tip, the normal at point A should be 0 Newtons, and then all weight would be on point B. I'm guessing if the barrel would tip, it wouldn't be sliding across the floor so kinetic friction is not used...

"Forward acting" friction is quite new to me, since usually most problems I've done include friction forces that are opposing the forward movement. But it makes sense, if one was on friction-less ice, one would need friction to move anywhere. Just quite hard to wrap my head around it.

Imagine a car race about to take place. A car is starting at rest and when the timer goes off, the driver steps on the gas. The car starts off so fast, that its front wheels rise upwards so that all the weight of the car is held up by the 2 rear wheels. The car travels this way for a few...

yes, this is much easier than the Yn and Yp stuff. I just need to connect the pieces to the puzzle.

(3,0) + C(1, 1/2)

I understood everything you said but this. this means y' - 4y = 0 therefore y' = 4y?

oh, i understand now, its (3, 0) because 3 - 2(0) = 3 to solve for the last fill in, the best way is to just plug and chug? so it should be C(1, 1/2)

Ok, so it means the difference between the two numbers is 4? so then it should be (3 , -2) + C(1, 1/2) ?

Homework Statement Homework Equations The Attempt at a Solution my answers are in green, and I have no idea how to complete the last part. What is happening? what does C = c + 4 mean?

Homework Statement Homework Equations The Attempt at a Solution So the general solution is the sum of the null solution (Yn) and particular solution (Yp) I believe I just need to write: y = e2t + 5e8t - 5 + C and then find the derivative of both sides y' = 2e2t + 40e8t is this correct?

so for the particular solution, we just set y' = 0, then we have -4y = 2, y = -1/2 and for the null solution, we need to set the right side to 0, then y' = 4y = 4(Ce4t) ?? is it possible to have a constant as the null solution?

Homework Statement Lets say for example, we are given: dy/dx - 4y = 2 or y' - 4y = 2 , y(0) = 4 => M= e^(-4t) e^(-4t) y' - 4e^(-4t)y = 2 e^(-4t) e^(-4t) y = -1/2 [ e^-4t ] + C y = -1/2 + Ce^4t When t = 0, y = 4 4 = -1/2 + C C =...

Oh, i see, using KVL

so that means V1 - V2 = 2volts difference? and since current flows from higher voltage to lower, the current is actually going from V1 to V2?

how does the battery affect the voltage of the two nodes? does it mean the voltage on Node1 is 2 volts higher than node2?

Homework Statement Homework Equations The Attempt at a Solution If I take KCL at node V1 (I'm assuming current traveling from nodeV2 to node V1) Current in = Current out V2 - 2volts / 2 ohms = V1 / 8 ohms so my question is, how does the battery (2 volts in this case) play a role in nodal...

i(t) = 20 +(5-20) e-12.5t the voltage drop across R1 is still (5A)(1ohm) = 5 volts (because current in circuit is still 5A, but will increase to 20A over time) at t=0, but will exponentially increase to 20 volts over time. Since the voltage drop on R1 is 5 volts, and the battery supplies 20...

i thought R2 and the voltage source were shorted. why isn't voltage source shorted? I thought it had to be shorted to find Req

so when t < 0, current through inductor is 5, but change in current is 0, so no voltage drop. when t = 0 , current through inductor increases from 5 to 20, so there will be a voltage drop. when t →∞, current is steady again, therefore voltage drop across inductor is 0 again. when the switch...

After learning a bit about thevenin and norton circuits , this problem has become easier to visual. I understand this now but in this case (after a long period and the capacitor is full charged) , the switch is never opened again after it was closed, so the capacitor will never discharge and...

now solving for UL(t) = L di/dt = L *[ i(∞) - Δi(e^-t/(L/R) = L* [ 20 - 15(e^-t/(L/R))) ] Time constant of when switch is closed is L/R Req is just the 1 ohm resistor 0.08H / 1 ohm = 0.08 sec = L *[ i(∞) - Δi(e^-t/(L/R) = 0.08 [20 - 15(e^-12.5t) ]

I didn't take differential equations so I'm not too good with that.

ok, I think I found it. i(t) = i(∞) - Δi(e^-t/(L/R) = 20 - 15(e^-t/(L/R))

Homework Statement Homework Equations The Attempt at a Solution When the t < 0, the switch is open, the inductor acts like a short therefore the circuit looks like The current is then 20V / 4 ohms = 5 A Since the inductor has been in series with the battery for a long time, it is safe to...

Going through the textbook, I found an example problem is similar to the one in OP. ___________________________________________ So in this problem, once the switch is closed, the voltage source becomes shorted. Then Req is solved. I can see it now, in my problem, once the switch is closed, the...

anywhoo, after reading some problems online, many of them like to have a "chart" or write down what exactly is happening when t→ -∞ , t = 0 , and t →∞ it probably helps organize things a bit so I will try that approach for this problem too, and hopefully refine my understanding a bit more...

oops, made an error IR1(t) = Io ( 1 - e^-t/RC) = ( 7.5x10^-3 A) ( 1 - e^-t/RC) = ( 7.5x10^-3 A) ( 1-e^t/0.005) I got confused, didn't know if Io was the current at t=0 from the current source or the current going through the R1 resistor at t=0. only way the equation makes sense is if Io is...

IR1(t) = [Is - iC1(t)] = [7.5x10^-3 A] - [(0.003A)(e^-t/0.005)] so when t is a high number, all the current flows through R1.

iC1(t) = io (e^-t/RC) = (0.003A)(e^-t/0.005) U1(t) = [Is - iC1(t)] 20k are the answers then. Suppose the question also asks for the IR1(t) and UC1(t) how would the equations look like? The voltage on the capacitor will be exponentially increasing and the current through R1 will be...

very interesting, so whenever there is a current source, and we need to find Req, we need to open the current source. and when we have a voltage source, and we need to find Req, we need to short the volt source. What is the reason for this? I didn't learn thevenin's theorem yet.

ok, having done some reading on source transformation, i think the part that probably confused me a bit more than it should have, was the source was a current source and not a voltage source. So yes, after source transformation, it would be a 150 volt battery in series with 20k and 30k...

when the capacitor is short, the Resistors are in parallel so the Req is 20x30 / 50 = 12k ohms. when the capacitor is fully charged, and is open, only R1 has current flowing through it, so Req is just 20k ohms. I'm still having trouble understanding if the R in RC equal to the Req when t=0 or...

iC1(t) = io (e^-t/RC) = (0.003A)(e^-t/0.003) U1(t) = [Is - iC1(t)] 20k

I have no answer key to the questions, but after asking for clarification, the terminals on the right aren't connected to anything, if that changes anything.

it would be the Req of the two resistors. but in this case, the R in time constant = RC is 30,000 ohms? because it is in series with the capacitor, thus having the same current as the current flowing through the capacitor.

the capacitor, and both resistors would be in series.

I didn't derive this equation, I just thought for a bit and this equation made sense. How would one derive this? I wasn't sure if I had to use the resistors of R2 (the resistor it was in series with, so they have the same current) or Req of the circuit when the capacitor was short. would I be...

Nope. we are low on options haha, sorry.

UR1(t) = [Is - IC1(t)]/ 20kΩ Plugging in IC1(t)] from what I found before: UR1(t) = [Is - io (e-t/RC) ]/ 20kΩ = [Is - (0.003A) (e-t/0.0012) ]/ 20kΩ