A Pulley, an Incline, and an Acceleration That Is Too Small
A 2.5 kg wooden block rests on a 30-degree incline and is connected by a light string over a small pulley to a 1.5 kg hanging mass. The pulley is a solid aluminum disk of mass 0.20 kg and radius 5.0 cm, mounted on an axle the students assume is frictionless. The incline is the same surface used in an earlier lab, where a coefficient of static friction of about 0.10 was measured. On release, a photogate records the block accelerating up the incline at 0.35 m/s^2. Using g = 9.8 m/s^2, a student computes the frictionless prediction as 0.61 m/s^2, concludes that the pulley bearing must be seizing, and wants to write that in the report. Their partner insists the shortfall is simply kinetic friction on the incline and that the pulley can be ignored because it is small. The report is due tomorrow and no replacement pulley is available.
1. Derive the acceleration for this system including the pulley's rotational inertia, then use the measured 0.35 m/s^2 to extract the kinetic friction force and coefficient. Decide which student is right about the pulley.
Read the worked answer
Write one equation per body, taking the direction of motion as positive:
- Hanging mass: m2 g - T2 = m2 a
- Block along the incline: T1 - m1 g sin(theta) - f = m1 a
- Pulley: (T2 - T1) R = I alpha, with I = (1/2) M R^2 and alpha = a/R, so T2 - T1 = (1/2) M a
Adding the three eliminates both tensions: m2 g - m1 g sin(theta) - f = (m1 + m2 + M/2) a.
The pulley therefore contributes an effective inertial mass of M/2 = 0.10 kg, so the denominator is 4.10 kg instead of 4.00 kg. That is a 2.4% effect: with no friction the prediction would be 2.45/4.10 = 0.598 m/s^2 rather than 0.6125 m/s^2, a shift of only about 0.015 m/s^2.
Now insert the measurement. The driving force minus the gravity component is 14.7 - 12.25 = 2.45 N, and the net force implied by a = 0.35 m/s^2 is 0.35 x 4.10 = 1.44 N. So f = 2.45 - 1.44 = 1.0 N, and mu_k = f / (m1 g cos 30) = 1.0 / 21.2 = 0.048, about 0.05.
The measured shortfall is 0.26 m/s^2, roughly eighteen times the pulley's contribution. The partner is correct: this is kinetic friction on the incline. A bearing torque large enough to matter here would have to be comparable to the tension difference, which a lab pulley on a steel axle does not produce.
2. The block is released from rest. Using mu_s = 0.10, decide whether it should have moved at all, and predict the acceleration if 50 g is added to the hanging mass.
Read the worked answer
Will it start? The driving force is m2 g = 1.5 x 9.8 = 14.7 N. The resistance at rest is the gravity component plus the maximum static friction: 12.25 + 0.10 x 2.5 x 9.8 x cos 30 = 12.25 + 2.12 = 14.37 N. Since 14.7 N exceeds 14.37 N, the block does start, but the margin is only 0.33 N, about 2%.
That margin is worth flagging in the report, because it explains why the result is fragile. Adding just 60 g to the block would raise the resistance to about 14.7 N and the block would sit still; small differences in surface cleanliness, humidity, or how the block was placed would change whether it moved at all. A group that happened to get a slightly rougher patch would report no motion and conclude the apparatus was broken.
Adding 50 g to the hanging mass. m2 becomes 1.55 kg, so the driving force is 15.19 N. With mu_k = 0.048 the kinetic friction is essentially unchanged at about 1.0 N, and the total inertial mass is 4.15 kg. Then a = (15.19 - 12.25 - 1.0)/4.15 = 1.94/4.15 = 0.47 m/s^2.
A 3.3% increase in hanging mass produces a 34% increase in acceleration. That steep sensitivity is not a mistake in the calculation; it is the signature of a system sitting close to its static threshold, where the fixed friction force absorbs most of the driving force and the leftover net force is small.
3. A classmate argues that since mu_k = 0.05 is 'basically zero,' friction can be ignored in the analysis. Evaluate that claim in the context of this experiment.
Read the worked answer
Whether a quantity is negligible depends on what it is being compared with, not on how small it looks on an absolute scale.
Here friction is 1.0 N out of a 2.45 N driving force, so it consumes about 41% of the available force. In terms of the observable, the frictionless prediction is 0.60 m/s^2 and the measured value is 0.35 m/s^2, a 42% reduction. Ignoring friction would put the report roughly 70% high on acceleration, which is far outside any reasonable uncertainty on the photogate timing.
A coefficient of 0.05 is small for a laboratory surface, but the normal force is large because the block sits on an incline and the friction force scales with it: f = mu N = mu m g cos(theta), which is about 21 N of normal force even for a 2.5 kg block. Small coefficient, non-negligible force.
The defensible statement is narrower: friction is large enough to dominate the force balance but small enough that the block still accelerates from rest, which is what makes the measurement possible in a single period.