Dynamics of a Box attached to a pully

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Homework Help Overview

The problem involves the dynamics of two bodies connected by a pulley system, focusing on the forces acting on a box on an incline. The scenario includes weights, coefficients of friction, and the angle of the incline, with the goal of determining the acceleration of the box under different conditions of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law to analyze the forces on both bodies. There is an attempt to differentiate between mass and weight, and participants raise questions about the calculations leading to discrepancies in the expected results.

Discussion Status

Some participants have provided corrections to the original calculations, and there is an ongoing exploration of the correct application of forces. Multiple interpretations of the problem setup and calculations are being examined, with some guidance offered on distinguishing between mass and weight.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is a focus on ensuring the correct application of physics principles without providing direct solutions.

ciubba
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Homework Statement


Body A in Fig. 6-33 weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are μs 0.56 and μk 0.25. Angle θ is 40. Let the positive direction of an x-axis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down
the incline?

Free body diagram: http://postimg.org/image/fh6livrer/

See attached picture (problem 27)
http://postimg.org/image/5svunp27z/

Homework Equations


Fnet=ma

The Attempt at a Solution



Clearly the answer to part a is 0-- I'm stuck on part b.

For B, F_{Net, y}= -m_B a_y = F_T -m_B g \rightarrow F_T = m_B g - m a_y

For A, F_{Net, x}=ma_x = - \mu _k m_A g cos(40) +F_T - m_A g sin(40)

Plugging in the information,

102 a= -0.25*102*cos(40)+32-32a -102sin(40) \rightarrow a=-.39

However, the book lists the correct answer as -3.9, so I'm off by a power of ten. Where did I go wrong?
 
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Be sure to distinguish between mass and weight.
 
D'oh! It should be

<br /> \frac {102}{9.8} a= -0.25*102*cos(40)+32- \frac {32}{9.8}a -102sin(40) \rightarrow a=-3.82

Thanks!
 
Good.
(I get -3.88 m/s2, but close enough).
 

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