Question on pulleys and tension

[demv18]

Homework Statement

Two objects of masses m1 and m2 are connected via a massless string and ideal pulleys. The coefficients of friction between m1 and the horizontal surface are ms and mk
a) what condition needs to be met to set the system in motion?
b)assuming that this condition is (barely) met, describe the motion of the system.
c) what is the tension in the string when the objects are in motion?

Homework Equations

T=ma

The Attempt at a Solution

a) m2 needs to be slightly heavier than m1
b) m2 will be slightly heavier, the tension in the string increases causing m1 to move forward since they're not in equilibrium.
c) ?

This was on a test, but I got it wrong... he wanted actual equations proving everything and i had no idea how to do any of this. I was just really confused. Can anyone explain it to me?

 

[PeterO]

Homework Statement

Two objects of masses m1 and m2 are connected via a massless string and ideal pulleys. The coefficients of friction between m1 and the horizontal surface are ms and mk
a) what condition needs to be met to set the system in motion?
b)assuming that this condition is (barely) met, describe the motion of the system.
c) what is the tension in the string when the objects are in motion?

Homework Equations

T=ma

The Attempt at a Solution

a) m2 needs to be slightly heavier than m1
b) m2 will be slightly heavier, the tension in the string increases causing m1 to move forward since they're not in equilibrium.
c) ?

This was on a test, but I got it wrong... he wanted actual equations proving everything and i had no idea how to do any of this. I was just really confused. Can anyone explain it to me?

m₁ is on the horizontal surface. The Normal Reaction force = m₁g so the maximum possible values of friction will be μ_sm₁g when it is stationary, and μ_km₁g when it is moving [the s and k refer to static and kinetic friction]. Usually [always?] μ_k is less than μ_s.

If the system is stationary, the tension in the string is m₂g since the string is supporting the hanging mass m₂.
Up on the table, the friction force must be balancing that tension, so the tension must be no greater than the maximum possible friction; μ_sm₁g

If the mass m₂ is big enough, the tension will exceed the possible friction, and the system will start to move. ie m₂g > μ_sm₁g

Once the system starts to move, the friction force will drop to μ_km₁g.
This means the masses will accelerate; m₁ across the table and m₂ down.

Net force on the system [in the direction it can move] is m₂g - μ_km₁g
The total mass of the system is m₁ + m₂

Using F = ma we get

a = (m₂g - μ_km₁g) / (m₁ + m₂)

At this point it getting too hard to keep track of all those typing references, so let's pretend a has a value of 2, m₂ = 3 kg and g = 10

Net force on m₂ = 3 * 2 = 6 N
Weight force down = m₂g = 3 * 10 = 30N
SO the tension in the string must be 24N

I will leave it to you to find the expression for tension in the real problem. It is an awful lot easier to write those subscripts with a pen on paper than it is to type them on this computer.

 

[PeterO]

m₁ is on the horizontal surface. The Normal Reaction force = m₁g so the maximum possible values of friction will be μ_sm₁g when it is stationary, and μ_km₁g when it is moving [the s and k refer to static and kinetic friction]. Usually [always?] μ_k is less than μ_s.

If the system is stationary, the tension in the string is m₂g since the string is supporting the hanging mass m₂.
Up on the table, the friction force must be balancing that tension, so the tension must be no greater than the maximum possible friction; μ_sm₁g

If the mass m₂ is big enough, the tension will exceed the possible friction, and the system will start to move. ie m₂g > μ_sm₁g

Once the system starts to move, the friction force will drop to μ_km₁g.
This means the masses will accelerate; m₁ across the table and m₂ down.

Net force on the system [in the direction it can move] is m₂g - μ_km₁g
The total mass of the system is m₁ + m₂

Using F = ma we get

a = (m₂g - μ_km₁g) / (m₁ + m₂)

At this point it getting too hard to keep track of all those typing references, so let's pretend a has a value of 2, m₂ = 3 kg and g = 10

Net force on m₂ = 3 * 2 = 6 N
Weight force down = m₂g = 3 * 10 = 30N
SO the tension in the string must be 24N

I will leave it to you to find the expression for tension in the real problem. It is an awful lot easier to write those subscripts with a pen on paper than it is to type them on this computer.

Since the masses were of a size that motion was only just achieved, m₂g will be equal to the static friction μ_sm₁g

SO the acceleration can be expressed as:

a = (μ_sm₁g - μ_km₁g) / (m₁ + m₂)