Static Versus Kinetic Friction on an Incline
Two blocks are connected by a light string over a frictionless pulley. Block 1, mass m1=2.0 kg, rests on a plane inclined at 30° to the horizontal. Block 2, mass m2, hangs vertically. The coefficient of static friction between block 1 and the plane is 0.40; the coefficient of kinetic friction is 0.30. The pulley and string are massless, and the string is parallel to the incline. Take g=9.8 m/s². The blocks are released from rest, and block 1 remains on the incline throughout. A student first sets m2=1.5 kg and claims the system must accelerate because m2g is greater than m1g sin30°. A second student sets m2=2.0 kg and uses the kinetic-friction value without first checking whether the system actually moves. The instructor asks for a force analysis that distinguishes static from kinetic friction and for the tension in each case.
1. For m2=1.5 kg, determine whether the system moves and find the tension and the static friction force.
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Driving difference if block 1 moves up the incline: m2g−m1g sin30°=1.5×9.8−2.0×9.8×0.5=14.7−9.8=4.9 N. Maximum static friction: f_s,max=μ_s N=0.40×(2.0×9.8×cos30°)=0.40×16.97=6.79 N. Because 4.9 N<6.79 N, static friction can balance the driving difference; the system remains at rest, a=0. For block 2 at rest, T=m2g=14.7 N. For block 1, along the incline: T−m1g sin30°−f=0, so f=14.7−9.8=4.9 N directed down the incline; this is within the static limit.
2. For m2=2.0 kg, determine the acceleration and the tension.
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With m2=2.0 kg, driving difference=19.6−9.8=9.8 N>6.79 N, so static friction fails and the system moves: block 2 descends, block 1 moves up the incline. Kinetic friction opposes block 1's motion, down the incline: f_k=μ_k N=0.30×16.97=5.09 N. Net force on the two-block system=19.6−9.8−5.09=4.71 N. Total mass=4.0 kg, so a=4.71/4.0=1.18 m/s². Tension from block 2: m2g−T=m2a, so T=2.0(9.8−1.18)=17.2 N.
3. Explain why using kinetic friction for m2=1.5 kg would give a nonzero acceleration and why that answer is physically invalid.
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Using kinetic friction for m2=1.5 kg gives net=14.7−9.8−5.09=−0.19 N, i.e. a≈−0.054 m/s² if the assumed direction is kept. But a negative net force means the assumed motion direction is wrong; friction would reverse and the actual kinetic-friction regime is not entered. Since the static limit is not exceeded, the correct acceleration is zero. The error is applying f_k=μ_kN before checking whether f_s can hold the system.