Find acceleration of this system with friction

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SUMMARY

The discussion focuses on calculating the acceleration of a two-body system involving Body A (102 N) and Body B (32 N) with friction coefficients of 0.56 (static) and 0.25 (kinetic) on a 40-degree incline. The key equations utilized include Fnet = MA, Ff = Fnormal(coefficient of friction), and Fnet = T - Ff. The solution requires analyzing three distinct scenarios: Body A at rest, moving up the incline, and moving down the incline, each necessitating separate free body diagrams to accurately determine tension (T) and acceleration.

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  • Understanding of Newton's laws of motion
  • Knowledge of friction coefficients and their application
  • Ability to draw and analyze free body diagrams
  • Familiarity with basic algebra for solving equations
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Homework Statement


Body A weighs 102 N, Body B weighs 32 N. The coefficeints of friction between A and the incline are .56 for static, .25 for kinetic. Angle is 40 degrees. Find the acceleration of the system when block A is at rest... when it is moving up the incline, and when it is moving down the incline.

(Block A is on the ramp attached to a string that goes up to the top of the ramp, down a pulley that attaches block B)


Homework Equations


Fnet=MA
Ff=Fnormal(coefficient of friction)
Fnet=T-Ff
Fnormal=Mgy


The Attempt at a Solution


I have done problems just the same as this very recently, but for some reason, I cannot think of how to get the tension in the string/rope. I don't know why, but it just seems to be escaping me, and after getting T, the rest of the problem is already solved. I have the friction, equal to mgy multiped by the coefficient. So the acceleration would be the T-friction, all divided by the mass of the system.
 
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You have three cases - block A at rest, moving up, and moving down.

Each will have a different free body diagram. You'll need to treat each case separately, rather than trying to lump them together with the same equations.
 


hey heth,
Yeah i just stared at this for some reason and didn't even think about it.
I took a little break and revisited it and got something that looked like this...
(just for A not in motion)

Fnet=mgax+ff-T
Fnet=T-mgb

so mgax+ff-T=T-mgb

solved for T

T=(mgax+mgb+Ff)/2

then take this tension find the unbalanced force, and divide it by the systems mass.
 

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