How Do Newton's Laws Apply to a System of Hanging Masses?

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SUMMARY

The discussion focuses on a physics problem involving three masses connected by cords over a frictionless pulley. The masses are defined as mA = 32.0 kg, mB = 40.0 kg, and mC = 18.0 kg. The tension in the cord connecting boxes B and C is calculated to be 187 N using the formula T = ma, where acceleration is derived from the gravitational force acting on the hanging masses. Additionally, the acceleration of box A is determined to be 6.68 m/s², which is essential for calculating its displacement over a time period of 0.250 seconds.

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  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of gravitational force calculations
  • Familiarity with kinematic equations for motion
  • Basic principles of pulleys and tension in cords
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Homework Statement



Three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 32.0 kg, mB = 40.0 kg, mC = 18.0 kg. (box a is on the horizontal suface and b and c hang down.

(a) When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?

(b) How far does box A move in the first 0.250 s (assuming it does not reach the pulley)?[/B]

Homework Equations



F=ma


The Attempt at a Solution


First I found acceleration.
T=Mass of box a * acceleration
F gravity = (mass of b + mass of c) * g
T-F gravity = (Mass of b + mass of c) * -accelleration
acceleration=((mass of b + mass of c) *g)/(mass of a +mass of b + mass of c)
a=(58 * 9.8)/(32 + 40 + 18)=6.68
T=ma
T=32 * 6.68=187 N
and that's as far as I got
 
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A hint for part (b): you know the acceleration of box A as well as the initial velocity and the time period.
 

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