System of two inclines planes and pulleys

In summary, the diagram shows two particles, A and B, connected by a light inextensible string and pulleys Q, R, and P. The particles are on fixed rough planes, inclined at angles of arctan(7/24) and arctan(4/3), with the segments of the string parallel to their respective planes. The coefficient of friction between each particle and its plane is μ. The system is in equilibrium, with both particles on the point of moving up their planes. By equating the forces on the particles and considering the inclined planes and friction, it is determined that μ can be found by subtracting the equations for TA and TB. In the case when M = 9m, the initial
  • #1
deea11235
3
0
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: http://www.picpaste.com/pics/fizik-3suPlHIQ.1377787280.png

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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  • #2
deea11235 said:
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.
 
  • #3
You mean TA-TB? But what would I equate it to? Is TA-TB=0, in which case, why?
 
  • #4
deea11235 said:
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
That was what I was missing, thank you!
 

1. How does a system of two incline planes and pulleys work?

A system of two incline planes and pulleys works by using the principle of mechanical advantage to reduce the amount of force needed to lift an object. The incline planes allow the weight of the object to be distributed over a larger distance, while the pulleys reduce the amount of force needed by redirecting the force in different directions.

2. What is the purpose of using a system of two incline planes and pulleys?

The purpose of using a system of two incline planes and pulleys is to make it easier to lift heavy objects. By using this system, the amount of force needed to lift an object is reduced, making it more manageable for humans to lift and move heavy loads.

3. How does the mechanical advantage of a system of two incline planes and pulleys affect the amount of force needed?

The mechanical advantage of a system of two incline planes and pulleys refers to the ratio of the output force (the force needed to lift the object) to the input force (the force applied to the system). The mechanical advantage of this system can be increased by adding more pulleys, which reduces the amount of force needed to lift the object.

4. What are some real-life applications of a system of two incline planes and pulleys?

A system of two incline planes and pulleys has many practical applications, such as in cranes, elevators, and construction equipment. It is also commonly used in simple machines, such as a block and tackle, to lift heavy objects.

5. What factors can affect the efficiency of a system of two incline planes and pulleys?

The efficiency of a system of two incline planes and pulleys can be affected by factors such as friction, the angle of the incline planes, and the weight of the object being lifted. Friction can reduce the efficiency of the system, while a steeper incline or a lighter object can increase the efficiency.

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