A.2 · IB Physics
Connected masses over a pulley
Change the block mass, the hanging mass and the friction coefficient to see the free-body forces, the acceleration and the string tension, and find when static friction keeps everything at rest.
Moves only if m₂g > μm₁g
a = (m₂g − μm₁g) / (m₁ + m₂)
T = m₂(g − a) = m₁a + μm₁g
g = 9.8 m s⁻²
Model: light inextensible string, frictionless light pulley, one μ used for both static and kinetic friction. Travel stops after 1.00 m.
—
Acceleration a—
Tension T—
Friction f—
Max static friction μm₁g—
Weight of m₂—
Time t0.00 s
Distance s0.00 m
Speed v0.00 m s⁻¹
Try this
- Set m₁ = 2.0 kg and μ = 0.40. Find the smallest m₂ that makes the system move and compare it with μm₁.
- Set μ = 0 and note a and T. Now double both masses: which quantity changes and which stays the same?
- Make m₂ much larger than m₁. What value does a approach, and is T ever as large as m₂g while moving?
Common error: writing T = m₂g for a moving system; apply F = ma to each mass separately, and you will find the tension is always less than the hanging weight when the masses accelerate.