Newtonian Mechanics: simultaneous equations

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SUMMARY

The discussion focuses on solving simultaneous equations in Newtonian mechanics involving two blocks: a sliding block S with mass M = 3.3 kg and a hanging block H with mass m = 2.1 kg. The equations derived from the system are Tx = ma for block S and T - mg = -ma for block H. The final acceleration of block S and block H is determined to be a = (mg) / (M + m), confirming the correct application of Newton's second law and the relationship between tension and acceleration. The solution process emphasizes the importance of correctly substituting variables and understanding the dynamics of the system.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with simultaneous equations
  • Basic knowledge of tension in a pulley system
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the derivation of Newton's second law in different contexts
  • Learn about tension forces in more complex pulley systems
  • Explore advanced topics in dynamics, such as energy conservation
  • Practice solving simultaneous equations with multiple variables
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of applying Newton's laws in problem-solving scenarios.

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Homework Statement


Figure 5-12 shows a block S (the sliding block) with mass
M 3.3 kg. The block is free to move along a horizontal
frictionless surface and connected, by a cord that wraps over
a frictionless pulley, to a second block H (the hanging
block), with mass m 2.1 kg. The cord and pulley have neg-
ligible masses compared to the blocks (they are “massless”).
The hanging block H falls as the sliding block S accelerates
to the right. Find (a) the acceleration of block S, (b) the ac-
celeration of block H, and (c) the tension in the cord.

Homework Equations


Block S:
Tx=ma (on the X axis)

Block H:

T-mg=-ma (y axis)

The Attempt at a Solution


I understand the theory and question, my problem lies with my math skills. Because of the two unknown variables you can solve them simultaneously, however the sample problem assumes I remember how to do it!

I want to understand how you can mathematically go from the two equations on top to a=(m/M+m)g on the y axis.
Thank you for the help!
 
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To avoid confusion you need to use M and m in your relevant equations.
Also note that what you've called Tx is equal to what you've called T (and of course a is the same also) so you can plug one equation into the other and solve for a.
 
After playing around with the equations I got it. Thanks for the help! I should pay more attention to the variables. :D

I ended up going like:
T-mg=-ma -----> T=-ma+mg
T=Ma -----> (plug in eqquation number 1) -ma+mg=Ma ----->Ma+ma=mg------->a(M+m)=mg----->a=mg/M+m

Would this set up be okay?
 
That looks right. Good work.
 

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