Atwood Incline Pulley problem with two pulleys

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Homework Help Overview

The problem involves an Atwood incline pulley system with two pulleys, where weight m1 is on an incline and weight m2 is suspended. The original poster seeks to determine acceleration, tension, the order in which the masses hit the ground, time to impact, and maximum static friction, based on a scenario with specific mass values and angles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply principles of mechanics to analyze the forces acting on both masses, including tension and friction. They express uncertainty about how to handle the double pulley system and the calculation of maximum static friction.
  • Some participants question the original poster's assumptions regarding the acceleration of m2 and suggest considering the forces acting on it more carefully.
  • Others provide feedback on the tension equations derived by the original poster, indicating they appear correct but do not verify the accompanying image.
  • There is a discussion about the impact of the coefficient of friction on the system's acceleration and the implications for the motion of the masses.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and equations. While some guidance has been offered regarding the tension equations, there remains uncertainty about the maximum static friction and its effect on the system's behavior. The original poster has acknowledged confusion and is seeking further clarification.

Contextual Notes

The original poster initially omitted specific values for the coefficient of friction and mass, which were later provided. The discussion reflects a need to reconcile these values with the calculations presented.

Puddles
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Homework Statement

Given that a weight m1 is attached to a string going over a perfect pulley attached to another pulley suspending weight m2, and that weight m1 is on an incline with an angle of 37 degrees, find acceleration, tension, which mass hits the ground first, time needed to hit the ground, and max possible static friction. Note that m2 < m1. m1 is 2m from the bottom of the incline and m2 is 1m off the ground. Note that my teacher didn't actually write out a problem for us, so I'm trying to write this out as coherently as possible based on the drawing of a system we were given. So if something about this doesn't make sense please tell me.

Okay, so for a simple Atwood incline problem I kinda know what to do, but what about with double pulleys? It shouldn't be that confusing because I just have to use 2T for the weight accelerating upwards (m2) since it's attached to a pulley and not just the string attached to m1, right? I'll attach an image to see if that clears up what I mean. I've put down my work for most of the problem and I'm wondering if someone could help me check it or tell me how I can confirm I've done a problem like this right and self-check it. Also I completely don't know how to find the max possible static friction for a system to still work, can anyone explain this to me?

rh7h9j.jpg


The attempt at a solution

T = m2*g/2
T = m2*g/2 = m1*g*sin 37°+m1*a
g cross
m2 = 1.2m1+2m1a
a1 = (m2-1.2m1)/2m1
2ℓ = 4.00 = a*t^2
t = √4/a = √8m1/(m2-1.2m1)
Vf1 = a1*t
a2 = 2h/t^2 =2(m2-1.2m1)/8m1 = (m2-1.2m1)/4m1 = a1/2

Where m1 = 10 kg and m2 = 15 kg,

a1 = (m2-1.2m1)/2m1 = (15-12)/20 = 3/20 = 0.15 m/sec^2
t = √8m1/(m2-1.2m1) = √8*10/(15-12) = 5.164 sec
Vf1 = a1*t = 0.15*5.164 = 0.775 m/sec
a2 = a1/2 = 0.15/2 = 0.075 m/sec^2
Vf2 = a2*t = Vf1/2 = 0.387 m/sec
 
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Puddles said:
T = m2*g/2
You're overlooking m2's acceleration. Consider F=ma for m2.
 
haruspex said:
You're overlooking m2's acceleration. Consider F=ma for m2.

Yup, I did. I went back and re-resolved F-tension for both parts.

For F-tens. (m2) I have (m2a+m2g)/2

For F-tens. (m1) I have (m1g sin(37))-(u m1g cos(37))-(m1a)

Does this look better? I think I've got it now but I'm just checking.

2a4wj92.jpg
 
Yes, those tension equations look right. (Haven't checked the image.)
 
haruspex said:
Yes, those tension equations look right. (Haven't checked the image.)

Okay, I'm a little stuck. I got an acceleration a1 of 2.72, thus a2 is 1.36, t then equals 1.213, v1f = 3.3 m/s and v2f = 1.65 m/s. m1 will hit the ground first. However, I still don't understand how to calculate max possible friction for the system to still move, I just looked online for help but I'm not really getting it.
 
Puddles said:
Okay, I'm a little stuck. I got an acceleration a1 of 2.72, thus a2 is 1.36, t then equals 1.213, v1f = 3.3 m/s and v2f = 1.65 m/s. m1 will hit the ground first. However, I still don't understand how to calculate max possible friction for the system to still move, I just looked online for help but I'm not really getting it.
I don't understand how you got a value for the acceleration without substituting a value for u. I don't see any value given. At the max value of u (or arbitrarily close to it), the acceleration becomes arbitrarily close to 0 and the time goes off to infinity.
 
Oh, wow. I forgot to put up the givens for my problem. I was waiting to substitute them in, sorry. u=.05, m1=15kg,m2=10kg, g=10m/s^2
 
Puddles said:
Oh, wow. I forgot to put up the givens for my problem. I was waiting to substitute them in, sorry. u=.05, m1=15kg,m2=10kg, g=10m/s^2
The acceleration you quote in post #5 seems too much. Please post your working.
 

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