Motion on an incline plane (detailed)

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SUMMARY

The discussion focuses on solving a physics problem involving two masses on an inclined plane connected by a cord over a pulley, factoring in kinetic friction. The first part of the solution involves calculating the acceleration of the system, which was determined to be -3.24 m/s², and the tension in the cord, calculated as 82.1 N. The user expressed difficulty in finding the final velocity after the masses moved 1.00 m, particularly in incorporating the effects of friction into the velocity equation.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of inclined plane dynamics
  • Familiarity with free body diagrams
  • Basic concepts of kinetic friction
NEXT STEPS
  • Learn how to apply the work-energy principle in systems with friction
  • Study the derivation of kinematic equations under non-constant acceleration
  • Explore the effects of friction on motion in inclined planes
  • Review the concept of Free Body Diagrams for complex systems
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Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators seeking to enhance their teaching methods in dynamics.

thomsora
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Hey, folks! I've been working on this one problem for the past little bit. It has been giving me a lot of trouble and I think i just need a pointer or two on how to go about solving it.

http://i.imgur.com/nTGVz.jpg"

Homework Statement



It's not the best image, but it gives the basic idea. I have two masses on an inclined plane connected with a cord over a pulley (kinetic friction included). It wants me to first solve for the acceleration in the system and then find the tension. The second part asks for me to find the final velocity after the masses have moved 1.00m (starting at rest)

mass #1 - 27.2 kg
mass #2 - 12.5 kg
theta - 5.23 deg
kinetic friction - 0.17

Homework Equations



\sumFx = max
\sumFy= may

ax = ayramp (mass #1):
\sumFx = -Ff - T + w1sin(theta) = max
\sumFy= Fn - w1cos(theta) = max

hanging weight (mass #2):
\sum T - w2=max

The Attempt at a Solution



T=m2a+w2
T=12.5a + 122.625

Fn= m1a+ w1
Fn= 27.2a+ 265.721

Sub values in:

-Ff - T + w1sin(theta) = max

Ff = (co-efficient x Fn)

-(27.2a+265.721)0.17 -(12.5a + 122.625) + (27.2 x 9.81x sin of 5.23) = 27.2a

combine values:

44.324a= -143.48

a= -3.24 m/s sq

T= 12.5a + 122.625
T=12.5(-3.24) + 122.625
T= 82.1 N

I solved the first part of this equation by resolving my Fx and Fy into their components and then solving for T and Fn (normal). I then substituted my T and Fn back into my Fx and solved for my acceleration. The answer I got was -3.24 m/s sq.

Now with the second part of this problem i seem to be a little stuck. I am thinking i cannot use the traditional Vfsq= Visq + 2as (correct me if i am wrong). Since there is friction involved (kinetic =0.17) (static = ?) i somehow need to incorporate that as well as my components into my velocity equation but i am unsure on how to do this. If anyone has any tips on how to calculate the velocity it would be much appreciated. I am not looking for *the* answer, i just need some pointers

Thank you!
 
Last edited by a moderator:
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Hi thomsora, welcome to PF.

Your value for the acceleration looks a bit high. Also, I'm not quite following your methodology; it looks to me like some of your force directions are mixed up. For example, on mass one (on the slope), shouldn't the frictional force OPPOSE the downslope force due to gravity of m1 AND the tension applied by the rope? Did you draw a Free Body Diagram for each mass?
 

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