Properties of Friction — Question 3

Section 6-2 · Physics pages 796–797

Question 3 Onset and steady sliding measure different coefficients Take g = 10 m/s2. Use ideal dry friction, |fs| ≤µsN and |fk| = µkN. A block on an adjustable incline begins to slide when the angle is increased just beyond 30◦. During a separate steady downward sliding test, it moves at constant speed at angle 20◦. Assume the same coefficients in both tests. block adjustable slope Schematic slope; its numerical angle is specified separately for each trial. (a) Infer µs and µk. (b) If the already-sliding block is on a 25◦incline, find its acceleration. (c) Could a block initially at rest remain at rest at 25◦? Explain the history dependence.
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Question 3 - Solution Onset and steady sliding measure different coefficients block adjustable slope (a) Use the correct equality for each test. At impending slip and at steady sliding, respectively, mg sin 30◦= µsmg cos 30◦=⇒µs = tan 30◦= .57735 , mg sin 20◦= µkmg cos 20◦=⇒µk = tan 20◦= .36397 . (b) Use kinetic friction for the moving block. For a block already moving downhill, use kinetic friction: a = g(sin 25◦−µk cos 25◦). a = 10[sin 25◦−(tan 20◦) cos 25◦] = .9275 m/s2 downhill . (c) Test the static inequality independently. A resting block instead requires mg sin 25◦≤µsmg cos 25◦⇐⇒tan 25◦= .4663 < .57735. Rest is feasible. At the same angle, an already sliding block accelerates because the kinetic coefficient is smaller; the contact regime depends on the motion history.

Question 3

Onset and steady sliding measure different coefficients

Take g=10g=10 m/s2^2. Use ideal dry friction, |fs|≤μsN|f_s|\le\mu_sN and |fk|=μkN|f_k|=\mu_kN. A block on an adjustable incline begins to slide when the angle is increased just beyond 30∘30^\circ. During a separate steady downward sliding test, it moves at constant speed at angle 20∘20^\circ. Assume the same coefficients in both tests.

Diagram shown in the original SVG worksheet.

  1. Infer μs\mu_s and μk\mu_k.

  2. If the already-sliding block is on a 25∘25^\circ incline, find its acceleration.

  3. Could a block initially at rest remain at rest at 25∘25^\circ? Explain the history dependence.

Question 3 - Solution

Onset and steady sliding measure different coefficients

Diagram shown in the original SVG worksheet.

(a) Use the correct equality for each test.

At impending slip and at steady sliding, respectively, mgsin⁡30∘=μsmgcos⁡30∘⇒μs=tan⁡30∘=.57735,mg\sin30^\circ=\mu_smg\cos30^\circ\Longrightarrow\boxed{\mu_s=\tan30^\circ=.57735}, mgsin⁡20∘=μkmgcos⁡20∘⇒μk=tan⁡20∘=.36397.mg\sin20^\circ=\mu_kmg\cos20^\circ\Longrightarrow\boxed{\mu_k=\tan20^\circ=.36397}.

(b) Use kinetic friction for the moving block.

For a block already moving downhill, use kinetic friction: a=g(sin⁡25∘−μkcos⁡25∘).a=g(\sin25^\circ-\mu_k\cos25^\circ). a=10[sin⁡25∘−(tan⁡20∘)cos⁡25∘]=.9275m/s2 downhill.\boxed{a=10[\sin25^\circ-(\tan20^\circ)\cos25^\circ]=.9275\ \mathrm{m/s^2}\text{ downhill}}.

(c) Test the static inequality independently.

A resting block instead requires mgsin⁡25∘≤μsmgcos⁡25∘⇔tan⁡25∘=.4663<.57735.mg\sin25^\circ\leq\mu_smg\cos25^\circ \Longleftrightarrow\tan25^\circ=.4663<.57735. Rest is feasible. At the same angle, an already sliding block accelerates because the kinetic coefficient is smaller; the contact regime depends on the motion history.