(a) Find the tension in the string connecting masses M1 and M2. (b) Find the tension in the string connecting masses M2 and M3.

Question

A hanging mass, M1 = 5 kg, is attached by a light string that runs over a frictionless pulley to the front of a mass M2 = 3.5 kg that is initially at rest on the table. A third mass M3 = 4 kg, which is also initially at rest on the same table, is attached to the back of M2 by a light string. The coefficient of kinetic friction of all contact surfaces is 0.1.

Answer 1

In most text books, they want you to do a force balance on each individual mass, then add the equations together to cancel all the tensions and solve for acceleration. Since the pulley is frictionless and the mass of the rope is negligible, you can treat the entire system of masses as a "caterpillar" with part of its mass hanging off and part of it dragging along: Here is the overall equation for the system:

a = F_Net/m_system = (M₁g - M₂gμ - M₃gμ)/(M₁+M₂+M₃) (This is the a for all the masses)

In order to solve for tensions, you need to apply Newton's 2nd Law for each mass (picking the system that allows the easiest calculation:

M₃ : T_(a) - M₃gμ = M₃a or T(a) = M₃(gμ+a)

You can do the same thing for M₂ or M₁ or (cleverest) (M₂+M₁) in order to solve for T_(b)

M₁+M₂: T(_b) - (M₁+M₂) = (M₁+M₂)a

From M₂, you could use T_(b) - M₂gμ - T_(a) = M₂a which will give same T_(b)

Good luck!