Static or Sliding? A Pulley on a Rough Incline
A 2.0 kg block rests on a fixed incline of 30.0° above the horizontal. The coefficients of friction are μs = 0.55 and μk = 0.35. A light string runs parallel to the incline, over a frictionless pulley at the top, and connects to a hanging mass m. Use g = 10.0 m/s² and ignore the pulley's mass and friction. In a first trial the student hangs m = 1.4 kg and observes no motion. She then replaces it with m = 2.0 kg. She claims that the block must move because the hanging weight is larger than the component of the block's weight down the slope. The instructor asks for a force analysis before accepting the claim. In each trial the block starts from rest, the string stays taut, and the incline is long enough for motion. Determine whether motion occurs, the friction force when the system is at rest, and the acceleration and tension when it slides.
1. For m = 1.4 kg, does the system remain at rest? Determine the friction force and its direction.
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Compare the forces along the incline before friction. The hanging weight is (mg = 1.4\times10 = 14\ \text{N}), acting up the incline through the string. The block's weight component is (m_1g\sin30^\circ = 2.0\times10\times0.5 = 10\ \text{N}) down the incline. Without friction the net is 4 N up the incline, so static friction must act down the incline with (f = 4\ \text{N}).
The maximum static friction is [f_{\max}=\mu_s m_1g\cos30^\circ=0.55\times2.0\times10\times0.866=9.53\ \text{N}.] Since (4\ \text{N}<9.53\ \text{N}), friction can hold the system. The block remains at rest; friction is 4 N down the incline; the tension equals the hanging weight, 14 N.
2. For m = 2.0 kg, find the acceleration and the tension.
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Now the hanging weight is 20 N. The frictionless balance would give (20-10=10\ \text{N}) up the incline, which exceeds (f_{\max}=9.53\ \text{N}), so the block slides up the incline. Kinetic friction acts down the incline: [f_k=\mu_k m_1g\cos30^\circ=0.35\times2.0\times10\times0.866=6.06\ \text{N}.] For the two-mass system, [(m_1+m_2)a=20-10-6.06=3.94\ \text{N}.] With total mass 4.0 kg, (a=0.985\ \text{m/s}^2).
For the hanging mass: (m_2g-T=m_2a), so [T=2.0(10-0.985)=18.0\ \text{N}.] Check on the block: (T-10-6.06=2.0\times0.985=1.97\ \text{N}), giving (T=18.0\ \text{N}).
3. What is the largest hanging mass for which the block can remain at rest? Explain the limiting condition.
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At the upper limit the block is about to slide up the incline, so static friction is at its maximum and points down the incline. Force balance along the incline gives [m_{\max}g=m_1g\sin30^\circ+f_{\max}=10+9.53=19.53\ \text{N}.] Thus (m_{\max}=1.95\ \text{kg}). For any larger hanging mass the driving force exceeds the maximum static friction and sliding begins. At (m=2.0\ \text{kg}) the system therefore slides, consistent with the second trial. At (m=1.4\ \text{kg}) it is below the limit and remains at rest.