Two block friction incline problem with rope

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SUMMARY

The discussion centers on a physics problem involving two blocks on an inclined plane, with Block A weighing 60N and Block B weighing 80N, inclined at 30 degrees. The calculated force F required to initiate Block B's movement down the incline is stated as 40.3N, but the user derived a different value of 17.7N through their calculations. The user questions the accuracy of the provided answer and seeks validation of their free body diagrams and equations based on Newton's laws. The consensus among participants suggests that the initial answer may indeed be incorrect.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Ability to create and interpret free body diagrams
  • Knowledge of static and kinetic friction coefficients
  • Familiarity with trigonometric functions in physics applications
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  • Study the application of Newton's 2nd Law in multi-body systems
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Homework Statement


Block A weighs 60N and rests on top of block B which weighs 80N and rests on an incline plane with angle of elevation 30 degrees. Block A is restrained by a mass-less rope as shown. What force F will cause block B to start moving down the plane? Assume all surfaces have a coefficient of friction of 1/3.
The answer is printed as F = 40.3N

http://imagizer.imageshack.us/v2/640x480q90/802/an0g.jpg

According to what's printed, the answer is supposed to be F = 40.3 N
But that's not what I'm getting.

Homework Equations


Free body diagrams for both blocks with y-axis perpendicular to incline plane.
Newton's 2nd Law after free body diagrams to form equations.
Newton's 3rd Law for normal forces and friction forces between blocks.

fs=μNB and fAB=fBA=μNA

Block B
1. μNB+μNA-F-WBcosθ=0 x-direction
2. NB-NA-WBcosθ=0 y-direction

Block A
3. NA-WAcosθ=0 y-direction
4. T-μNA-WAsinθ=0 x-direction

The Attempt at a Solution



I created free body diagrams and set up equations based off Newtons 2nd law. fAB is the force block B is pulling block A with (maximum static friction) and fBA is the maximum force block A is restraining block B with.3. → NA=WAcosθ
2. → NB=(WA+WB)cosθ
1. → F=μ(WA+WB)cosθ+μWAcosθ-WBsinθ
or F= μ(2WA+WB)cosθ-WBsinθ = 17.7N

I solved the systems and finally got F = 17.7N. I'm thinking the answer F = 40.3N is wrong (the professor made a mistake). Or did I make a mistake somewhere? Are my free body diagrams and forces correct? I appreciate any help thanks.P.S. I collect physics problems and try to solve them for fun (like this one). Does anyone know of another good block incline problem?
 
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Your work looks OK to me.
 
I agree, the given answer is wrong. I've not been able to figure out what mistake would lead to that answer. Halving the angle gets close.
If you want a challenging question, try this (warning: I haven't worked it through myself, so it might not do what my intuition says)
Two identical rectangular blocks sit at rest at different heights on an inclined plane. (Make up labels for angles, coefficients of friction, masses, block dimensions...) A third identical block sits squarely on top of the lower block. All surface contacts have the same coefficients of friction, but static exceeds kinetic.
The upper block is nudged down the slope, and kinetic friction is not enough to hold it. Under what circumstances, if any, will the third block end up sitting squarely on the nudged block, at rest relative to it?
 

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