How Do You Solve a Physics Problem Involving Multiple Blocks and Tensions?

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SUMMARY

The discussion focuses on solving a physics problem involving three blocks (A, B, and C) connected by ropes, with specific weights and friction coefficients. The blocks weigh 25.0 N each, and the coefficient of kinetic friction is 0.35. The problem requires drawing free body diagrams, calculating the tension in the ropes, and determining the weight of block C. The user attempts to apply Newton's Second Law but encounters difficulties in finding the acceleration of block C.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Ability to draw free body diagrams
  • Knowledge of friction coefficients and their application
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of free body diagrams in physics
  • Study the application of Newton's Second Law in multi-body systems
  • Learn about the effects of friction on motion
  • Practice solving problems involving tension in ropes and pulleys
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of multi-body problems involving tension and friction.

raychelle93
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Homework Statement


"Blocks A, B, and C are placed as in the figure and connected by ropes of neligible mass. Both A and B weigh 25.0 N each, and the coefficient of kinetic friction between each block and the surface is 0.35. Block C descends with constant velocity. a) Draw separate free body diagrams shwoing the forces acting on A and B. b)Find the tension in the rope connecting blocks A and B. c) What is the weight of block C? d) If the rope connecting A and B were cut, what would be the acceleration of C?"



Homework Equations





The Attempt at a Solution


I've gotten as far as trying to find "A" :b)
Let ' T1 ' & ' T2 ' be the tensions in the rope connecting blocks AB and BC respct.

Apply ; Newton's 2nd Law
Writing the equation of motion for mB
T1 - mBg Sin 36.9 0 - mk mBg Cos 36.9 0 - T2 = mB a -----------(i)
Writing the equation of motion for mA

T2 - mk mA g = mA a ----------------------------(ii)
Writing the equation of motion for mC

mC g - T1 = mC a ----------------------------(ii)

Add equ(i); equ (ii) & equ(iii) ; we have
to calculate the value of a & put it in equation (i) & (ii) respt, to get the value of T1 & T 2
>>Can't find a though :( Help! :cry:
 
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Hi raychelle93, welcome to PF.
Your second equation for mass A is wrong. Check it.
 

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