Newton's 3rd Law, pulley, only given friction coefficients

[aatgomez]

Homework Statement

An object consists of 3 connected masses as seen in the figure. None of the surfaces have friction except the surface on m1 which has the coefficients μs=0.4 y μk=0.3.
What is the acceleration of each one of the masses?
It looks like the following image but m1 is the table and m2 is equal to m3 or the block that's hanging from the pulley of the table.

Homework Equations

∑Fx = m1a → τ - fk = m1a
∑Fy = 0 → n- m1g = 0
T= μkm1g+ m1a
∑Fy = m2a → m2g - T =m2a
m2g - (μkm1g+m1a) = m2a
a = m2 - μkm1/m1 + m2

The Attempt at a Solution

I got up to the finally equation but I can't figure out how to solve the equation without mass. I've attached a picture just in case the problem needs some clearing up. Thanks!

 

[Astronuc]

Please separate the work from the relevant equations.

The attached image is a bit confusing since it has two masses, m₂, but one mentions a mass m₃.

The falling mass is 'pulled' by gravity, so there must be 'g' in the final solution. Gravity is 'forcing' the system.

The small mass is pulled by the falling mass through tension in the cord or string.

Opposing the motion of the horizontally moving smaller mass is the friction with the table. If the table is free to slide, then it too will experience acceleration due to the frictional force sliding on top of it.

The accelerations will be related to g by some coefficient which will be a complicated function of the masses.

 

[Doc Al]

Is it safe to say that your first image is not relevant to this problem?

Are there any constraints on the problem? Are things starting from rest? Is the hanging mass falling? Is the system being accelerated?

Please revise your description of the problem to be clear about which surfaces have friction.

 

[Astronuc]

Particularly, one's needs to clarify "None of the surfaces have friction except the surface on m1 which has the coefficients μs=0.4 y μk=0.3." As shown in the attached diagram, m1 has two surfaces - between m1 and the ground, and between m1 and the mass on top. Friction on either of those surfaces, or both, will significantly affect the answer.

 

[HalfLostAndSearching]

I would approach this problem by first applying Newton's 3rd Law, which states that for every action, there is an equal and opposite reaction. In this case, the action is the force of gravity pulling down on m2 and m3, and the reaction is the normal force exerted by the table on m1. This normal force is equal in magnitude to the force of gravity, but in the opposite direction.

Next, I would use the equations for the forces and accelerations in the x and y directions. In the x direction, the only forces acting are the tension in the string and the frictional force between m1 and the table. In the y direction, the forces are the normal force and the weight of m1. Setting up these equations and solving for the acceleration of m1, we get:

∑Fx = m1a → T - μsm1g = m1a
∑Fy = 0 → n - m1g = 0
Solving for n, we get n = m1g. Substituting this into the first equation, we get:

T - μsm1g = m1a
Substituting in the given values for μs and μk, we get:

T - (0.4)(m1g) = m1a
In the y direction, we have:

∑Fy = m2a → m2g - T = m2a
Substituting in the value for T from the first equation, we get:

m2g - (0.4)(m1g) = m2a
Solving for a, we get:

a = m2g - (0.4)(m1g) / (m2 + m1)
This is the acceleration of both m2 and m3, since they are connected by the string and will have the same acceleration.

We can then use this value of acceleration to solve for the tension in the string:

T = μkm1g + m1a
Substituting in the values for μk and a, we get:

T = (0.3)(m1g) + (m1/m2 + m1)(m2g - (0.4)(m1g))
Simplifying, we get:

T = 0.3m1g + 0.6m1g
T = 0