Physics Help Forum Problem regarding friction

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 Nov 2nd 2018, 09:48 AM #1 Junior Member   Join Date: Nov 2018 Posts: 3 Problem regarding friction Can someone solve the problem attached to this thread and explain it to me? The coefficient of friction between A and B is 0.1 The coefficient of friction between B and C is 0.2 The coefficient of friction between C and table is 0 Attached Thumbnails   Last edited by Thinkphy6; Nov 2nd 2018 at 10:20 AM.
 Nov 2nd 2018, 10:14 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 963 Is the diagram indicating that the coefficient of friction between C and the table is zero? __________________ ~\o/~
 Nov 2nd 2018, 10:19 AM #3 Junior Member   Join Date: Nov 2018 Posts: 3 Yes, its zero
 Nov 2nd 2018, 12:28 PM #4 Senior Member   Join Date: Aug 2010 Posts: 434 If I am reading this correctly, a 5 N force is applied to A which is sitting on top of B with a coefficient of friction of 0.1 between them. Since A has a mass of 4 kg it weighs 4g N. The friction force between A and B is 0.1(4g)= 0.4g= 0.4(9.8)= 3.92 N. A has a net force on it of 5- 3.92= 1.08 N. It will move with acceleration 1.08/4= 0.27 m/s^2. B is sitting on top of C with a coefficient of friction of 0.2. B has a mass of 6 kg so weighs 6g N. The friction force between B and C is 0.2(6g)= 1.2g= 1.2(9.8)= 11.76 N. That is larger than the 3.92 N A is applying to B will not move relative to C. The 3.92 N force from A is transmitted directly to C so B and C together move with acceleration 3.92/(6+ 8)= 3.92/14= 0.28 m/s^2
 Nov 2nd 2018, 07:28 PM #5 Junior Member   Join Date: Nov 2018 Posts: 3 Thank you for this. But can u explain what really happens to the objects B and C ? I'm trying to understand the process behind all this. Not just the calculations.
 Nov 5th 2018, 06:38 AM #6 Senior Member     Join Date: Jun 2016 Location: England Posts: 963 Friction with B is the downward force from A (its weight) times the coefficient of friction: 4{kg}*0.1*g{m/s/s} = 3.92{N} [g taken to be 9.8{m/s/s}] Friction between B and C is the downward force from B (due to its weight and the weight of A) times the coefficient of friction: (6+4){kg}*0.2*g{m/s/s) = 19.6{N} The force accelerating A (5N) is greater than the force of the friction between A and B (3.92N), so A will accelerate relative to B. The force accelerating B is the friction from A (3.92N), This is less than the force due to the friction between B and C (19.6N) so B will not accelerate relative to C. There is no friction between C and the ground, so B and C will move together as a single unit being accelerated by the 3.92N Force provided by the friction between B and A. benit13 likes this. __________________ ~\o/~

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