Pulley system w/hanging mass, two unknown masses and Mk

[Ummiya]

Homework Statement

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.4 kg mass (m2) that hangs over the side of the shelf 1.4 m above the ground. The system is released from rest at t = 0 and the 2.4 kg mass strikes the ground at t = 0.85 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 2.4 kg mass now strikes the ground 1.3 seconds later.

(a) Determine the mass m1.
(b) Determine the coefficient of kinetic friction between m1 and the shelf.

Homework Equations

Newton's 2nd law: F = ma
Constant acceleration: x(t) = x0 + v0t + 1/2at^2

The Attempt at a Solution

∑Fx (m1) = T - ƒ = m1a
∑Fy (m1) = m1g - N = 0
N = m1g

∑Fx (m1) = T - Mk(m1g) = m1a
T = m1(a + Mkg)
m1 = T / (a + Mkg)

∑Fy (m2) = T - m2g = m2ay
T = m2a + m2g

x(t) = x0 + v0t + 1/2at^2
1.4 = 1/2a(0.85^2)
a = 3.875

T = (2.4)(3.875) + (2.4)(9.8)
T = 32.82 N

m3 = second object placed on m1
∑Fx (m1 + m3) = T - f = (m1 + m3)a
∑Fy (m1 + m3) = (m1 + m3)g - N = 0
N = (m1 + m3)g

∑Fx (m1 + m3) = T - Mk((m1 + m3)g) = (m1 + m3)a

No idea what to do after this. Is my (m1+m3) equation correct? If it helps there's a hint saying:
Draw free-body diagrams of the objects with mass m1, m2, and the m1-additional object system. Apply Newton's second law and the use of a constant-acceleration equation to find a relationship between the coefficient of kinetic friction and m1. The repetition of this procedure with the additional object on top of the object whose mass is m1 will lead to a second equation that, when solved simultaneously with the former equation, results in a quadratic equation in m1. Its solution will allow you to substitute in an expression for μk and determine its value.

 

[BvU]

Well, can you post your diagram ?

 

[BvU]

∑Fx (m1) = T - ƒ = m1a
∑Fy (m1) = m1g - N = 0
N = m1g

∑Fx (m1) = T - Mk(m1g) = m1a
T = m1(a + Mkg)
m1 = T / (a + Mkg)

I think the first two lines are OK. Then I can't follow: you haven't told us what T is, what ƒ is, what k is or what M is.

If they represent what I think, then the third line is wrong (and the dimensions don't match either). So help me out !

Oh, and T > 2.4 g should start some alarm bells ringing !

 

[Ummiya]

Mk is mu. (k for kinetic)
ƒ is the frictional force
T is the tension

None of those are given.

 

[BvU]

Still haven't seen a diagram, or better: two diagrams. One for m₁ and one for m₂. They have two things in common.

 

[Ummiya]

Sorry I didn't see your first reply till now.

 

[haruspex]

∑Fy (m2) = T - m2g = m2ay

Which way is it accelerating?

 

[BvU]

Good. the 'things' in common are T and (not shown in the diagram, but that's understandable) the acceleration (the wire stays at fixed length). Now collect the equations and see if you can come to a set of equations with an equal number of unknowns.