Will the Block Move? Static Friction on an Incline
A student sets up a 2.0 kg block on a rough plane inclined at 37° to the horizontal. The block is tied to a light, inextensible string that passes over a frictionless pulley at the top of the incline and hangs vertically with a 0.80 kg mass attached. The coefficients of friction between block and plane are μs = 0.45 and μk = 0.12. The student releases the system from rest, predicts that the hanging mass will fall, and measures no motion. A classmate argues the string tension must equal 0.80g regardless of whether the system moves. The student must decide whether the system moves, find the tension and friction, and then test what happens if the incline is waxed so that μs drops to 0.15 while μk remains 0.12. Use g = 9.8 m/s² and sin37° = 0.60, cos37° = 0.80.
1. Determine whether the system remains at rest. Find the friction force and the tension, and evaluate the classmate's claim.
Read the worked answer
Forces along the incline without friction: Block's weight component down the incline = 2.0 × 9.8 × 0.60 = 11.76 N. Hanging weight acting up the incline through the string = 0.80 × 9.8 = 7.84 N. Net driving tendency = 11.76 − 7.84 = 3.92 N down the incline.
Maximum static friction: Normal force N = 2.0 × 9.8 × 0.80 = 15.68 N. f_s,max = 0.45 × 15.68 = 7.056 N.
Since 3.92 N < 7.056 N, static friction can hold the system. The system remains at rest, friction is 3.92 N acting up the incline, and tension is 7.84 N.
The classmate is wrong: tension equals the hanging weight only in equilibrium. If the hanging mass accelerates, the tension is not 0.80g.
2. The incline is waxed so that μs = 0.15 and μk = 0.12. Determine whether motion occurs, and if so find the acceleration and the tension. State the direction of motion.
Read the worked answer
New maximum static friction = 0.15 × 15.68 = 2.352 N. The net driving tendency 3.92 N exceeds this, so the system moves. Because the block's downhill component is larger, the block slides down the incline and the hanging mass rises.
Kinetic friction = 0.12 × 15.68 = 1.8816 N, acting up the incline. Net force on the system = 11.76 − 7.84 − 1.8816 = 2.0384 N. Total mass = 2.0 + 0.80 = 2.80 kg. Acceleration a = 2.0384 / 2.80 ≈ 0.728 m/s².
For the hanging mass moving upward: T − 0.80g = 0.80a, so T = 0.80(9.8 + 0.728) ≈ 8.42 N. Check with the block: 11.76 − T − 1.8816 = 2.0 × 0.728, which gives T ≈ 8.42 N. The student's prediction that the hanging mass would fall is wrong; if motion occurs, it rises.
3. Explain why the tension is not 0.80g when the system accelerates, and describe one measurement that would distinguish static from kinetic friction in this setup.
Read the worked answer
Tension is an internal force that adjusts to the motion. For the hanging mass, Newton's second law gives T − mg = ma when it moves upward, so T = m(g + a), which exceeds mg. If it moved downward, T = m(g − a), which is less than mg. Only when a = 0 does T = mg.
To distinguish static from kinetic friction, slowly add small masses to the hanging side until the block just begins to move; the largest hanging weight that still produces no motion gives the limiting static friction. Then, with the system already moving, measure the acceleration (for example with a ticker-timer or motion sensor) and use the net-force equation to extract the kinetic friction. The kinetic value should be lower than the static maximum.