Explanation

A two-particle pulley joins two masses by a light inextensible string over a smooth pulley: one hangs, one sits on an incline. They share one acceleration a and one tension T. Resolve along the string, include friction μN opposite m₁'s motion, then solve fora and T — μ = 0 is the smooth case.

The string is inextensible, so the accelerations have equal magnitude. The pulley is smooth, so T is the same throughout.

The normal reaction on m₁ is N = m₁g cos θ, so the friction force has magnitude F = μN = μ m₁g cos θ. Friction opposes the motion of m₁ along the plane.

• If m₂g > m₁g sin θ, m₂ tends to descend and m₁ goes up the slope, so friction acts down the slope.
• If m₁g sin θ > m₂g, m₁ tends to slide down, so friction acts up the slope.
• If the driving difference is at most F, the system stays at rest and T = m₂g.

Taking downward on m₂ and up the slope on m₁ as positive, and assuming m₂ descends:

• For m₂: m₂g − T = m₂a
• For m₁: T − m₁g sin θ − μ m₁g cos θ = m₁a

Add the equations to eliminate T and solve for a, then substitute back for T. If μ = 0 the friction terms vanish.

Worked example

A particle of mass 2 kg rests on a smooth plane inclined at 30° to the horizontal and is connected by a light inextensible string, passing over a smooth pulley, to a freely hanging particle of mass 3 kg. The system is released from rest. Take g = 9.81 m/s² and find the acceleration and the tension.

1. m₂g = 29.43 N is greater than m₁g sin 30° = 9.81 N, so the 3 kg particle descends.
2. For the hanging particle: 29.43 − T = 3a
3. For the particle on the plane: T − 9.81 = 2a
4. Add: 19.62 = 5a, so a = 3.92 m/s²
5. Substitute: T = 2(3.92) + 9.81 = 17.66 N

The plane is now rough, with coefficient of friction 0.2. The masses and angle are unchanged. Find a and T.

1. N = 2 × 9.81 × cos 30° = 16.99 N, so F = 0.2 × 16.99 = 3.40 N
2. 19.62 N > 3.40 N, so the 3 kg particle still descends.
3. Friction on the 2 kg particle acts down the plane: T − 9.81 − 3.40 = 2a
4. Add to 29.43 − T = 3a: 16.22 = 5a, so a = 3.24 m/s²
5. T = 3(9.81 − 3.24) = 19.70 N

Friction reduces the acceleration and increases the tension. IfF reached the driving difference, the system would remain at rest.