System of two inclines planes and pulleys

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The diagram shows two particles, A of mass 5m and B of mass 3m, connected by a light
inextensible string which passes over two smooth, light, fixed pulleys, Q and R, and under a
smooth pulley P which has mass M and is free to move vertically.
Particles A and B lie on fixed rough planes inclined to the horizontal at angles of arctan(7/24) and
arctan(4/3) respectively. The segments AQ and RB of the string are parallel to their respective
planes, and segments QP and PR are vertical. The coefficient of friction between each particle
and its plane is μ.
1)Given that the system is in equilibrium, with both A and B on the point of moving up
their planes, determine the value of μ and show that M = 6m.
2)In the case when M = 9m, determine the initial accelerations of A, B and P in terms
of g.

Diagram is here: [Broken] http://www.picpaste.com/pics/fizik-3suPlHIQ.1377787280.png [Broken]




Well for starters at 1) if we were to note [itex]\vec{TA}[/itex] and [itex]\vec{TB}[/itex] the vectors for the forces of the strings coming from A and B respectively and considering that the system is in equilibrum, equating the the forces gets us Gp=TA+TB, and considering each inclined plane and the force of friction going downwards (the system is on the point of moving up) we get TA=GA+Ff=mA*g*sin(arctan(7/24))+μ*mA*g*cos(arctan(7/24)). Doing the same thing for B and after the substitution in the first equation I get mP=mA(sin(arctan(7/24))+μ*cos(arctan(7/24)))+mB(sin(arctan(4/3))+μ*cos(arctan(4/3))). Problem is, here I have only one equation and two unknowns: mP and μ. What am I doing wrong? :(
 
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Answers and Replies

  • #2
CAF123
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Well for starters at 1) if we were to note [itex]\vec{TA}[/itex] and [itex]\vec{TB}[/itex] the vectors for the forces of the strings coming from A and B respectively and considering that the system is in equilibrum, equating the the forces gets us Gp=TA+TB, and considering each inclined plane and the force of friction going downwards (the system is on the point of moving up) we get TA=GA+Ff=mA*g*sin(arctan(7/24))+μ*mA*g*cos(arctan(7/24)). Doing the same thing for B and after the substitution in the first equation I get mP=mA(sin(arctan(7/24))+μ*cos(arctan(7/24)))+mB(sin(arctan(4/3))+μ*cos(arctan(4/3))). Problem is, here I have only one equation and two unknowns: mP and μ. What am I doing wrong? :(

Try to subtract the equation you attained for TA from the equation you attained for TB. This will leave the unknown μ.
 
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You mean TA-TB? But what would I equate it to? Is TA-TB=0, in which case, why?
 
  • #4
CAF123
Gold Member
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You mean TA-TB? But what would I equate it to? Is TA-TB=0, in which case, why?

TA=TB since the string is light and inextensible, while the pulleys are smooth. The sole purpose of the pulleys here are to reorient the string.

An extensible string is like a spring and a 'light' string is one we take to be of zero mass. These conditions at the start simplify the analysis.
 
  • #5
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That was what I was missing, thank you!
 

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