Block on a Rough Incline: Does the System Move at All?
A 4.0 kg wooden block on a rough plane inclined at 25° to the horizontal is joined by a light string, passing over a small pulley at the top of the plane, to a 2.5 kg mass hanging freely. The coefficients of friction between block and plane are μs = 0.45 and μk = 0.35. A student assumes the hanging mass accelerates downwards and writes 2.5g − 4.0g sin25° − μk(4.0g cos25°) = 6.5a, claiming a nonzero sliding acceleration. The demonstrator marks this wrong. When the student adds a further 1.0 kg to the hanger the block does move, and the student is then asked to repeat the calculation and to decide whether the pulley's rotational inertia (I = 1.2 × 10⁻⁴ kg m², radius 2.5 cm) can be neglected. Take g = 9.8 m s⁻² throughout and assume the incline and string are as described.
1. Decide whether the system moves with the original 2.5 kg hanging mass. Show the friction test explicitly and state the tension and the friction force acting on the block.
Read the worked answer
Take 'up the plane' as positive for the block.
- Weight component down the plane = 4.0 × 9.8 × sin 25° = 16.6 N
- Hanging weight = 2.5 × 9.8 = 24.5 N
- Normal reaction R = 4.0 × 9.8 × cos 25° = 35.5 N
- Limiting static friction = 0.45 × 35.5 = 16.0 N
Ignoring friction, the 24.5 N pull wins by 24.5 − 16.6 = 7.9 N, so the block's tendency is to slide up the plane and friction acts down the plane.
Required friction to hold = 7.9 N, which is less than the 16.0 N available. Static friction can supply it, so the system stays at rest and a = 0.
Friction = 7.9 N down the plane; tension = 24.5 N (the hanging mass is in equilibrium). Check on the block: 24.5 − 16.6 − 7.9 = 0 ✓.
The student's error is procedural, not arithmetic: motion was assumed, then the kinetic law was applied. Kinetic friction exists only once sliding has begun; before that the correct statement is f ≤ μsR, and f takes whatever value is needed.
2. Repeat the calculation for a 3.5 kg hanging mass: find the acceleration and the tension, and justify the direction of friction.
Read the worked answer
Hanging weight = 3.5 × 9.8 = 34.3 N. Without friction the net pull up the plane = 34.3 − 16.6 = 17.7 N, which exceeds the 16.0 N static limit, so the system now accelerates with the block moving up the plane.
Friction is now kinetic and acts down the plane: 0.35 × 35.5 = 12.4 N.
- Net force on the system = 34.3 − 16.6 − 12.4 = 5.3 N
- Total mass = 4.0 + 3.5 = 7.5 kg
- a = 5.3/7.5 = 0.71 m s⁻²
Tension from the hanging mass: 34.3 − T = 3.5 × 0.71 = 2.5 N, so T = 31.8 N.
Check on the block: 31.8 − 16.6 − 12.4 = 2.8 N = 4.0 × 0.71 ✓.
Two exam traps are avoided here. First, the tension is not the hanging weight once the system accelerates - quoting 34.3 N is the common error. Second, the friction direction must be re-argued from the tendency of motion, not carried over from the previous part.
3. With the 3.5 kg hanging mass, decide whether the pulley's rotational inertia can be neglected.
Read the worked answer
With kinetic friction and the 3.5 kg hanger, the massless-pulley result is approximately 0.707 m/s squared. The given inertia contributes an effective mass I/r squared = 0.192 kg, so the acceleration becomes approximately 0.690 m/s squared, about 2.5 percent lower. The tension difference is I a/r squared, approximately 0.132 N in the inertial-pulley model. These values quantify the approximation, but no acceptable error tolerance or uncertainty in the friction coefficient has been supplied. Therefore it is not justified to declare the pulley negligible unconditionally. Neglect it only if an explicitly chosen accuracy requirement accepts the approximately 2.5 percent acceleration change, and state that modeling assumption. Comparing I/r squared with total moving mass provides the relevant scale check.