Two Blocks, an Incline and a Static-Friction Check That Cannot Be Skipped
A 4.0 kg block rests on a 30° incline whose surface has coefficient of static friction 0.25 and coefficient of kinetic friction 0.18. It is tied to a 3.0 kg hanging mass by a light inextensible string passing over a frictionless pulley at the top of the incline. Take g = 9.8 m/s² and ignore pulley and string mass. A student is asked whether the system moves and, if so, with what acceleration and string tension. The student writes a = (m₂g − m₁g sin30°)/(m₁ + m₂) = 0.70 m/s² and then 'subtracts friction' to get 0.19 m/s², never checking whether the blocks actually start from rest. The lab instructor marks two things wrong with this: the student assumed sliding before testing the static case, and the answer depends on which friction coefficient applies. The instructor then substitutes a rougher incline (μs = 0.45, μk = 0.35) and separately asks for the minimum hanging mass that would just start the original system moving.
1. For the original surface, decide whether the system moves and find the acceleration and tension if it does.
Read the worked answer
Compare the driving force with the maximum static friction before assuming motion.
Driving force = m₂g − m₁g sin30° = 3(9.8) − 4(9.8)(0.5) = 29.4 − 19.6 = 9.8 N (tending to pull m₁ up the incline). Maximum static friction = μs m₁g cos30° = 0.25 × 4 × 9.8 × 0.866 = 8.49 N.
Since 9.8 N > 8.49 N, the system does start to move, with m₂ descending and m₁ sliding up the incline; kinetic friction now applies: fk = 0.18 × 4 × 9.8 × 0.866 = 6.11 N, directed down the incline.
Net force = 29.4 − 19.6 − 6.11 = 3.69 N; total mass moving = 7.0 kg. a = 3.69/7.0 = 0.53 m/s²
Tension from the hanging mass: m₂g − T = m₂a ⇒ T = 29.4 − 3(0.53) = 27.8 N. Check on block 1: T − m₁g sin30° − fk = 27.8 − 19.6 − 6.11 = 2.1 N = 4 × 0.53 ✓.
The margin over static friction is small (about 1.3 N), which is the physically important point: this system is close to the sticking threshold, so any added load or rougher patch changes the answer qualitatively, not just numerically.
2. The incline is replaced by one with μs = 0.45 and μk = 0.35. What happens, and why is the formula a = (m₂g − m₁g sin30°)/7 invalid here?
Read the worked answer
Driving force is unchanged at 9.8 N. Maximum static friction is now 0.45 × 4 × 9.8 × 0.866 = 15.3 N, which exceeds 9.8 N, so the blocks do not move. The system is in static equilibrium.
Static friction is not equal to μsN; it takes whatever value is needed up to that ceiling. Here it settles at 9.8 N acting down the incline, balancing the net pull. Tension follows from the hanging mass being at rest: T = m₂g = 29.4 N. Check block 1: 29.4 − 19.6 = 9.8 N ≤ 15.3 N ✓.
The formula fails because it presumes sliding and applies kinetic friction. Used here it returns a ≈ (9.8 − 11.9)/7 = −0.30 m/s², a negative acceleration. A negative value with a positive driving force is a signal that the assumed state is wrong, not that the blocks accelerate backwards: friction cannot reverse motion, it can only oppose it, and once static friction can hold the system the correct answer is a = 0 with friction determined by equilibrium, not by μN.
3. Find the minimum hanging mass that would just start the original system (μs = 0.25) moving, and comment on whether the answer is sensitive to the value used for g.
Read the worked answer
At the threshold, the tension just balances the incline-side resistance: m g = m₁g sin30° + μs m₁g cos30°.
m = 4(0.5) + 0.25(4)(0.866) = 2.0 + 0.866 = 2.87 kg (using 4 kg throughout, the g cancels).
So about 2.87 kg, i.e. 0.13 kg less than the 3.0 kg actually hung - which is why the original arrangement sat only 1.3 N above threshold.
Sensitivity to g: g cancels from both sides of the threshold equation, so the predicted mass is independent of the local value of g. What does not cancel is the ratio μs, and there the sensitivity is large: the term μs m₁ cos30° contributes 0.87 kg out of 2.87 kg, so a 10% error in μs moves the threshold by roughly 3%. A student who quotes 2.87 kg to three significant figures from a handbook friction coefficient is over-claiming precision; the honest answer is 'about 2.9 kg, and within 0.1 kg of the mass already on the hook'.