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Old Nov 2nd 2018, 10:48 AM   #1
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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
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Problem regarding friction-img_20181102_221600.jpg  

Last edited by Thinkphy6; Nov 2nd 2018 at 11:20 AM.
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Old Nov 2nd 2018, 11:14 AM   #2
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Is the diagram indicating that the coefficient of friction between C and the table is zero?
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Old Nov 2nd 2018, 11:19 AM   #3
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Yes, its zero
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Old Nov 2nd 2018, 01:28 PM   #4
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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
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Old Nov 2nd 2018, 08:28 PM   #5
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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.
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Old Nov 5th 2018, 07:38 AM   #6
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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.
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