Calculating Acceleration and Tensions in a Blocks and Pulley System

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SUMMARY

The discussion focuses on calculating the acceleration and tensions in a blocks and pulley system involving a block of mass M = 4.4 kg on a ledge and a hanging block of mass m = 2.1 kg. The coefficient of kinetic friction between the ledge and the block is 0.25, and the pulley has a radius of 8.4 cm and mass of 0.62 kg. Key equations include Newton's second law and torque equations, specifically T - Mg(coefficient of kinetic friction) = Ma and mg - T = ma. The moment of inertia for the pulley is calculated as 0.00656 kg·m², which is essential for determining the system's angular acceleration and tensions.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with torque and moment of inertia concepts
  • Knowledge of friction coefficients and their application in mechanics
  • Basic principles of pulley systems and angular motion
NEXT STEPS
  • Calculate the linear acceleration of the blocks using a = g((m-M)/(M+m))
  • Explore the relationship between linear acceleration and angular acceleration in pulley systems
  • Study the effects of varying the coefficient of kinetic friction on system dynamics
  • Investigate the role of different pulley shapes and masses on tension calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone involved in solving problems related to pulleys and friction in dynamic systems.

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Homework Statement



In the system shown below, there is a block of mass M = 4.4 kg resting on a horizontal ledge. The coefficient of kinetic friction between the ledge and the block is 0.25. The block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging block of mass m = 2.1 kg. The pulley is a uniform disk of radius 8.4 cm and mass 0.62 kg. Find the acceleration of each block and the tensions in the segments of string between each block and the pulley.

Homework Equations



T - Mg(coefficient of kinetic friction) = Ma
mg - T = ma
a = g((m-M)/(M+m))

torque = inertia * alpha
alpha = linear acceleration / radius

The Attempt at a Solution



inertia = 0.5 * 0.62 * 0.084^2 + 0.62 * 0.084^2 = 0.00656

For the acceleration, I think the main equation I'll need to use is torque = inertia * (linear acceleration / radius), but I'm not sure how to find torque.
 
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Assume the string to have two tensions. One for the region between the pulley and the hanging block. One for the region between the pulley and the resting block.
 
So would that be:

T - Mg(coefficient of kinetic friction) = mg - T
 
There is only one T term there. And did you take into account the moment of inertia of the pulley?

Assume tension T_{1} for the region of the string between the hanging block and the pulley. Assume tension T_{2} for the region of the string between the resting block and the pulley.

Use Newton's 2nd law for the two blocks, and an equation connecting angular acceleration, torque and moment of inertia for the pulley.
 
the FBD will be like:

attachment.php?attachmentid=35911&stc=1&d=1306330573.jpg


note that tensions are different
and also linear acceleration of blocks will be (angular acc of pulley)*(radius of pulley)Now wrote eqn's
 

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