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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]

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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