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Question

The blocks of mass M and m are arranged as the situation in Fig is shown. The coefficient of friction between two blocks is μ1 and that between the bigger block and the ground is μ2. Find the acceleration of the blocks.
983258_3b17af4f27a94d41a50ae5c73b308fbb.png

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Solution

Along horizontal direction, N1=ma
Along vertical direction, mg(μ1N1+T)=m(2a) (i)
After substitution value of N1 in Eq. (i), we have
Mg(μ1ma+T)=m(2a) (ii)
For the motion of Block M:
Along vertical direction Fv=0,
or N2=T+μ1N1+Mg (iii)
Along horizontal direction,
2T(N1+μ2N2)=Ma (iv)
From equation (i) ans (iii), we get
2T[N1+μ2(T+μ1N1+Mg)]=Ma (v)
or 2T[ma+μ2T+μ1μ2(ma)+μ2Mg]=Ma
Now solving equations (ii) and (iv), we get
a=2mμ(Mm)gM+m[5+2(μ1μ2)]

1029156_983258_ans_90737f9fbe8f47e49125ff6fc28193ad.png

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