Wedge & Pulley — Two Connected Blocks
Two blocks on inclined faces share a string over a pulley — who slides which way?
About this simulation
Two blocks rest on the two faces of a wedge, joined by a string over a pulley at the apex. The system accelerates toward whichever side wins the tug of gravity-component versus friction. Students set each side's incline angle, mass and friction coefficient and watch the acceleration, string tension and direction resolve — a compact lab for Newton's second law on connected bodies.
How to use it
- Set the Left angle, Left mass and Left friction for block m₁, then the Right angle, Right mass and Right friction for block m₂.
- Compare the 'pull' and 'grip' bars drawn inside the wedge: the blocks move only when pull is bigger than grip.
- Watch the arrow on the descending block and read the lines at the top: the acceleration and which block descends, the rope tension T, or 'a = 0 — friction holds' when nothing moves.
- Read the acceleration a at the top right, then try the 'Frictionless' preset to see how much faster the system moves.
- Use 'Balanced sides' (equal masses and angles) and 'Right side wins' to test which side controls the motion.
What to notice
- Each block is pulled down its face by m g sin(angle), so the net pull is the difference of the two components, not the difference of the masses.
- Friction on both faces works against the motion, so the total grip μ₁m₁g cos α + μ₂m₂g cos β must be beaten before anything moves.
- Both blocks share one string, so they have the same magnitude of acceleration and the whole system can be treated as one body of mass m₁ + m₂.
- The rope tension lies between the two weight components: it slows the descending block and drags the rising one up.
The formulas
- a = g(m₁ sin α − m₂ sin β − μ₁m₁ cos α − μ₂m₂ cos β) / (m₁ + m₂)
- Acceleration when block 1 slides down; α and β are the incline angles, μ₁ and μ₂ the friction coefficients, g the acceleration due to gravity.
- T = m₂(a + g sin β + μ₂ g cos β)
- Rope tension found from block 2 as it is hauled up its face with acceleration a.
Questions
How do you find the acceleration of two blocks on a double inclined plane connected by a string?
Take the direction in which the heavier pull acts as positive. The driving force is m₁g sin α − m₂g sin β, and friction on both faces opposes it with μ₁m₁g cos α + μ₂m₂g cos β. Divide the net force by the total mass m₁ + m₂. If the result would be negative, friction is strong enough and the system stays at rest. The acceleration a shown at the top right of the simulation follows exactly this rule.
Why can two blocks stay at rest on a wedge even when one is heavier?
Motion depends on the components of weight along each face, m g sin(angle), not on the masses alone. A heavy block on a gentle face can pull no harder than a light block on a steep one. Even when the pulls differ, static friction on both faces can supply up to μN each to balance the difference. In the simulation, the 'Friction just holds' preset shows grip just bigger than pull.
How is the tension in the string calculated for a wedge-pulley system?
Find the acceleration of the system first, then write Newton's second law for one block on its own. For the block moving up its face, the tension must overcome its weight component and friction and also give it acceleration: T − m₂g sin β − μ₂m₂g cos β = m₂a. Using the other block gives the same tension, which is a useful check. The simulation does not print T, so work it out from the a it shows and check it with both blocks.
Does a frictionless pulley change the tension on either side of the string?
No. For a light string over a frictionless, massless pulley, the tension is the same all along the string. The pulley only changes the direction in which the tension acts, so it pulls each block up its own face. That is why one value of T appears in the equations of both blocks.
