Two blocks with pulley, friction

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SUMMARY

The discussion focuses on calculating the acceleration of two blocks connected by a frictionless pulley system, where the blocks have masses of 8.2 kg and 25 kg, and the pulley has a mass of 5.9 kg with a radius of 0.25 m. The coefficient of friction between the block and the surface is 0.28. The moment of inertia for the pulley is defined as 1/2MR², and the acceleration due to gravity is 9.8 m/s². The user attempts to derive the acceleration using torque and force equations but encounters discrepancies in their calculations.

PREREQUISITES
  • Understanding of Newton's second law (sumF=ma)
  • Knowledge of torque and its relation to rotational motion
  • Familiarity with moment of inertia for a uniform disc (1/2MR²)
  • Basic concepts of friction and its calculation (friction = mew * normal force)
NEXT STEPS
  • Review the derivation of acceleration in pulley systems using free body diagrams
  • Practice problems involving torque and rotational dynamics
  • Explore the effects of varying coefficients of friction on acceleration
  • Learn about the dynamics of multiple connected bodies in motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of pulley systems and friction in action.

nazerofsun
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Homework Statement


Consider the setup shown below. The blocks
have masses 8.2 kg and 25 kg. The pulley
has mass 5.9 kg, and is a uniform disc with
radius 0.25 m. Assume the pulley to be fric-
tionless, but the coefficient of friction between
the block and the surface is 0.28.What is the
acceleration of the blocks? Assume the 25 kg
mass is descending with acceleration a. The
moment of inertia of the disk is 1/2MR2 and the
acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s2.


Homework Equations


Torque= (radius)(force)sin(angle between them)= (inerta)*(alpha)or (inertia)*a/r
sumF=ma
friction= (mew)*(normal force)

The Attempt at a Solution


mo= mass of pulley

I think this problem just has too much algebra =.=
sumF1x= m1gsin(theta)+Friction-T1=m1a
SumF2y=T2-m2g=m2a
sumT= 1/2(mo)ra= rt1-rt2 ( I don't know which one should go first or why)
solving the t and plugging it into torque I receive:

a= g( (1+(mew))m1sin(theta)-m2)
-----------------------------
(1/2mo+m1+m2)

I have a practice problem I can use to verify if i set this problem up but I keep getting like .05 off -.-

Any help would be appreciated!
 
Last edited:
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Show a picture of the set-up and show your work.

ehild
 

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