Question #104790
[img]https://upload.cc/i1/2020/03/06/CsXpzH.jpg[/img]


where d8 is 4
1
Expert's answer
2020-03-11T14:18:58-0400


(ii) Consider block A only and write Newton's second law for it:


Ox:T cos30+μkNA=0,Oy:T sin30+NAmAg=0.Ox:-T\text{ cos}30^\circ+\mu_kN_{A}=0,\\ Oy: T\text{ sin}30^\circ+N_{A}-m_Ag=0.

Express the normal force from the second equation:


NA=mAgT sin30,T cos30+μkmAgμkT sin30=0, T=μkmAgcos30+μk sin30=14.5 N.N_{A}=m_Ag-T\text{ sin}30^\circ,\\ -T\text{ cos}30^\circ+\mu_km_Ag-\mu_kT\text{ sin}30^\circ=0,\\ \space\\ T=\frac{\mu_km_Ag}{\text{cos}30^\circ+\mu_k\text{ sin}30^\circ}=14.5\text{ N}.


Consider only block B and find its acceleration from Newton's second law:


Oy:NBmBgNA=0,Ox:mBa=μkNBμkNA+N. a=Nμk(NB+NA)mB= =Nμk[mBg+2(mAgT sin30)]mB=0.36 m/s2.Oy: N_B-m_Bg-N_A=0,\\ Ox: m_Ba=-\mu_kN_B-\mu_kN_A+N.\\ \space\\ a=\frac{N-\mu_k(N_B+N_A)}{m_B}=\\ \space\\ =\frac{N-\mu_k[m_Bg+2(m_Ag-T\text{ sin}30^\circ)]}{m_B}=0.36\text{ m/s}^2.

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