Torque and Inertia of objects on Inclined plane

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SUMMARY

The discussion focuses on a physics problem involving two blocks connected by a massless string over a solid disk pulley. The blocks have masses of m1 = 2 kg and m2 = 6 kg, with a pulley mass of M = 10 kg and radius R = 0.25 m. The angle of the inclined plane is 30 degrees, and the coefficient of kinetic friction is 0.36. The user successfully calculated the acceleration of the blocks as 0.31 m/s² but seeks clarification on calculating the tensions in the string, which are given as 7.67 N and 9.22 N in the reference material.

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  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of torque and rotational dynamics (sumTorque = Ialpha)
  • Familiarity with the concepts of friction and inclined planes
  • Basic skills in algebra for solving equations
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  • Study the relationship between torque and angular acceleration in rotational systems
  • Learn how to draw free-body diagrams for complex systems involving pulleys
  • Explore the effects of friction on motion in inclined planes
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to tension, acceleration, and friction in multi-body systems.

cooltee13
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Homework Statement


a block with mass m1= 2 kg, and a block of mass m2 = 6kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = .25m and mass M = 10 kg. These blocks are allowed to move on a fixed wedge of angle pheta = 30 degrees. the coefficient of kinetic friction is .36 for both blocks and pulley. Determine A) the acceleration of the two blocks and B) the tensions in the string on both sides of the pulley.


Homework Equations


well, I already solved part A) I just need help with part B. I used the sumTorque = Ialpha, where alpha = a/R. along with this equation below:


The Attempt at a Solution


well, i wanted to solve for a so i did the following:(note f=coefficient of friction*): -T1Rf+60cos(T2)Rf = 1/2mR^2*(a/R) ====> -T1f +60cos(T2)f = 1/2ma.

M2gf - T1f + 60cosT2f = 1/2Ma
M2gf = M1af + 60cosM2af + 1/2Ma
M2gf = a(M1f + 60cosM2f + 1/2M)
a = M2gf/(M1f + 60cosM2f + 1/2M) = .31 m/s^2

Im not sure how to solve for the Tension though, do i use the same equation??
 
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The answer for B is 7.67 N and 9.22 N.

But I want to know how they got that. Can anybody help please?
 
can anybody help? do I need to explain more?
 
A diagram would be helpful. Are both sides of the wedge at an angle of 30 degrees with the horizontal?

In any case, assuming you've correctly found the acceleration, just apply Newton's 2nd law to each block separately to find the tensions.
 
Im not sure how to put a diagram on this forum, but If i apply Newtons second law, I don't end up with the same Tensions as the book answer.

F = ma
I come up with 6.08 N for T1
 
Show exactly what you did to get the tension. (Until I understand the setup better, I cannot verify your answer for the acceleration.)
 
how do I insert an image so i can show you the picture?
 
One of the icons in the post editing window is for inserting images. Or you can attach a file.
 

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