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Question

(a) Find the acceleration of the blocks shown in Fig.
(b) Find the friction force between all contact surfaces.
983091_18cc9fafdc324ac5a19fad8a1153086b.png

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Solution

Let us first calculate the limitation and kinetic friction between various surfaces. Between 3 kg and ground:
fl1=fk1=μ1(2+3)g=0.1×5g=5N
Between 2 kg and 3 kg:
fl2=fk2=μ2g=0.2×2g=4N
Let us first assume that both blocks move together with common accleration a.
a=5+10fk12+3a=1555=2ms2
Now let us see how much friction force is required between 2kg and 3kg for common acceleration a.
5f2=2a5f2=2×2f2=1N
Since f2<fl2,
1. Both blocks will move together with common accelerationa=2ms2
2. Friction between 2kg and 3 kg =f2 = 1N Friction between 3 kg and ground = fk1= 5N
1027083_983091_ans_ef5102a004284481893680f7c52874b3.png

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