Friction — Question 3

Section 6-1 · Physics pages 776–777

Question 3 Zero acceleration does not identify the friction regime A 2 kg sled is pulled right with 5 N and has zero acceleration. The supporting belt can move independently. Three possible observations are listed below; in every case the ground-frame sled velocity is +2 m/s. There are no other horizontal forces. sled A vs = +2 m/s belt: vb = +2 m/s P = 5 N sled B vs = +2 m/s belt: vb = 0 m/s P = 5 N sled C vs = +2 m/s belt: vb = +4 m/s P = 5 N All velocities are measured relative to the ground. Compare the two contact velocities separately in each case. (a) Find friction in every dynamically possible case. (b) Classify friction for cases A and B. (c) Explain why case C cannot satisfy all the stated observations under ordinary sliding friction.
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Question 3 - Solution Zero acceleration does not identify the friction regime sled A vs = +2 m/s belt: vb = +2 m/s P = 5 N sled B vs = +2 m/s belt: vb = 0 m/s P = 5 N sled C vs = +2 m/s belt: vb = +4 m/s P = 5 N (a) Use the measured acceleration. For any dynamically admissible case, ∑ Fx = P + f = ma = 0 =⇒f = −5 N . The required friction is leftward; the contact regime must also permit that direction. (b) Compare contact velocities. The two admissible relative velocities are vrel,A = 2 −2 = 0, vrel,B = 2 −0 = +2 m/s. A permits static friction fs = −5 N if fs,max ≥5 N. B requires kinetic friction fk = −5 N to balance the pull. (c) Check the direction in C. For C, vrel,C = 2 −4 = −2 m/s =⇒fk > 0. Consequently a = (5 + fk)/2 > 0, contradicting the measured a = 0. Ordinary sliding friction cannot satisfy all of case C’s observations.

Question 3

Zero acceleration does not identify the friction regime

A 2 kg sled is pulled right with 5 N and has zero acceleration. The supporting belt can move independently. Three possible observations are listed below; in every case the ground-frame sled velocity is +2 m/s. There are no other horizontal forces.

Diagram shown in the original SVG worksheet.

  1. Find friction in every dynamically possible case.

  2. Classify friction for cases A and B.

  3. Explain why case C cannot satisfy all the stated observations under ordinary sliding friction.

Question 3 - Solution

Zero acceleration does not identify the friction regime

Diagram shown in the original SVG worksheet.

(a) Use the measured acceleration.

For any dynamically admissible case, ∑Fx=P+f=ma=0⇒f=−5N.\sum F_x=P+f=ma=0\Longrightarrow\boxed{f=-5\ \mathrm N}. The required friction is leftward; the contact regime must also permit that direction.

(b) Compare contact velocities.

The two admissible relative velocities are vrel,A=2−2=0,vrel,B=2−0=+2m/s.v_{{\mathrm{rel}},A}=2-2=0,\qquad v_{{\mathrm{rel}},B}=2-0=+2\ \mathrm{m/s}. A permits static friction fs=−5f_s=-5 N if fs,max≥5f_{s,\max}\geq5 N. B requires kinetic friction fk=−5f_k=-5 N to balance the pull.

(c) Check the direction in C.

For C, vrel,C=2−4=−2m/s⇒fk>0.v_{{\mathrm{rel}},C}=2-4=-2\ \mathrm{m/s}\Longrightarrow f_k>0. Consequently a=(5+fk)/2>0a=(5+f_k)/2>0, contradicting the measured a=0a=0. Ordinary sliding friction cannot satisfy all of case C’s observations.