Atwood Incline Pulley problem with two pulleys

In summary, the conversation revolves around solving a physics problem involving a double pulley system with two weights, m1 and m2. The goal is to find the acceleration, tension, which mass hits the ground first, time needed to hit the ground, and max possible static friction. The solution involves using equations for F=ma and F-tension, with given values for u=.05, m1=15kg, m2=10kg, and g=10m/s^2. The final values for acceleration, time, and velocities are provided, but there is still confusion on how to calculate the max possible friction for the system to still move.
  • #1
Puddles
12
0
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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  • #2
Puddles said:
T = m2*g/2
You're overlooking m2's acceleration. Consider F=ma for m2.
 
  • #3
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
 
  • #4
Yes, those tension equations look right. (Haven't checked the image.)
 
  • #5
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.
 
  • #6
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.
 
  • #7
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
 
  • #8
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.
 

Related to Atwood Incline Pulley problem with two pulleys

1. What is the Atwood Incline Pulley problem with two pulleys?

The Atwood Incline Pulley problem with two pulleys is a classic physics problem that involves a system of two masses connected by a string that is draped over two pulleys. The system is placed on an inclined plane and the goal is to determine the acceleration and tension in the string.

2. How is the Atwood Incline Pulley problem with two pulleys solved?

The Atwood Incline Pulley problem with two pulleys can be solved using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. By setting up force equations for each mass and using the fact that the tension in the string is the same throughout, the acceleration and tension can be determined.

3. What are the assumptions made when solving the Atwood Incline Pulley problem with two pulleys?

The Atwood Incline Pulley problem with two pulleys assumes that the pulleys are massless and frictionless, the string is inextensible, and there is no air resistance. These assumptions allow for a simplified solution to the problem.

4. What is the significance of the Atwood Incline Pulley problem with two pulleys?

The Atwood Incline Pulley problem with two pulleys is a commonly used example in physics to illustrate concepts such as Newton's laws of motion, vector analysis, and the principle of mechanical advantage. It also has practical applications in situations where pulleys are used to lift or move heavy objects.

5. How can the Atwood Incline Pulley problem with two pulleys be modified?

The Atwood Incline Pulley problem with two pulleys can be modified by changing the mass of the objects, the angle of the incline, or the coefficient of friction. It can also be extended to include multiple pulleys and more complex systems. These modifications can provide a deeper understanding of the principles involved in the problem.

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