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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 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 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/~  Tags friction, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post jasonlewiz Kinematics and Dynamics 3 Jan 17th 2010 10:24 AM JoanF Kinematics and Dynamics 8 Dec 29th 2009 12:01 PM chrissn13 Kinematics and Dynamics 1 Feb 12th 2009 12:56 AM Morgan82 Kinematics and Dynamics 2 Jan 8th 2009 02:28 AM Morgan82 Kinematics and Dynamics 1 Nov 2nd 2008 03:05 PM