# Search results

1. ### Which location will experience the most stress in this case? (pulley & shaft system)

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...
2. ### Dimensional analysis

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

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...
4. ### Dimensional analysis

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...
5. ### Kinematics of objects on ramp

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...
6. ### Kinematics of objects on ramp

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) =...
7. ### Kinematics of Rigid Bodies

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...
8. ### Kinematics of Rigid Bodies

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?
9. ### Kinematics of Rigid Bodies

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.
10. ### Kinematics of Rigid Bodies

yes, it will slide because I calculated static friction to be only 352 N.
11. ### Kinematics of Rigid Bodies

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...
12. ### Question about friction -- car accelerating so fast its front wheels rise off the pavement

"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.
13. ### Question about friction -- car accelerating so fast its front wheels rise off the pavement

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...
14. ### Null solutions

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

(3,0) + C(1, 1/2)
16. ### What exactly is a null solution and particular solution?

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

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)
18. ### Null solutions

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

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?
20. ### Finding a linear differential equation

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?
21. ### What exactly is a null solution and particular solution?

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?
22. ### What exactly is a null solution and particular 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 =...
23. ### Nodal analysis

Oh, i see, using KVL
24. ### Nodal analysis

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?
25. ### Nodal analysis

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

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...
27. ### Engineering RL circuit problem

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...
28. ### Engineering RL circuit problem

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
29. ### Engineering RL circuit problem

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...
30. ### Engineering RC circuit with current source

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...
31. ### Engineering RL circuit problem

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) ]
32. ### Engineering RL circuit problem

I didn't take differential equations so I'm not too good with that.
33. ### Engineering RL circuit problem

ok, I think I found it. i(t) = i(∞) - Δi(e^-t/(L/R) = 20 - 15(e^-t/(L/R))
34. ### Engineering RL circuit problem

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...
35. ### Engineering RC circuit with current source

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...
36. ### Engineering RC circuit with current source

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...
37. ### Engineering RC circuit with current source

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...
38. ### Engineering RC circuit with current source

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.
39. ### Engineering RC circuit with current source

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...
40. ### Engineering RC circuit with current source

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.
41. ### Engineering RC circuit with current source

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...
42. ### Engineering RC circuit with current source

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...
43. ### Engineering RC circuit with current source

iC1(t) = io (e^-t/RC) = (0.003A)(e^-t/0.003) U1(t) = [Is - iC1(t)] 20k
44. ### Capacitance and Inductance questions

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.
45. ### Engineering RC circuit with current source

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.
46. ### Engineering RC circuit with current source

the capacitor, and both resistors would be in series.

48. ### Engineering RC circuit with current source

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...
49. ### Engineering RC circuit with current source

Nope. we are low on options haha, sorry.
50. ### Engineering RC circuit with current source

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