Question

https://upload.cc/i1/2020/03/03/VmZEs6.jpg

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Answer 1

(Only forces involved in the calculations are shown)

Figure out whether the system will Consider each block separately. For simplicity $\beta=40^\circ,\gamma=30^\circ.$

For block A:

$$-T_1+m_Ag\text{ sin}\beta=m_Aa.$$

For block B:

$$T_1-T_2\text{ cos}\gamma-\mu_s(m_Bg+T_2\text{ sin}\gamma)=m_Ba.$$

For block C:

$$T_2\text{ cos}\gamma-\mu_s(m_Cg-T_2\text{ sin}\gamma)=m_Ca.$$

Three unknowns, three equations, solve the system any way you wish. One of the ways is to open the parentheses and add the second equation to the third one, it will give us, along with the first equation:

$$T_1-\mu_sg(m_B+m_C)=a(m_B+m_C),\\ -T_1+m_Ag\text{ sin}\beta=m_Aa.$$

Add them too:

$$g[m_A\text{ sin}\beta-\mu_s(m_B+m_C)]=a(m_A+m_B+m_C),$$

find the acceleration:

$$a=g\frac{m_A\text{ sin}\beta-\mu_s(m_B+m_C)}{m_A+m_B+m_C}=0.62\text{ m/s}^2.$$

See: we solved the problem with static friction, which means that the pulling force created by block A is enough to pull the other two blocks. Therefore, the system will move after release from rest. The actual acceleration, however, can be found if we substitute $\mu_k$for $\mu_s$:

$$a=g\frac{m_A\text{ sin}\beta-\mu_k(m_B+m_C)}{m_A+m_B+m_C}=1.17\text{ m/s}^2.$$

Tension in cable 1:

$$T_1=m_A(g\text{ sin}\beta-a)=61.6\text{ N}.$$

Tension in cord 2:

$$T_2=m_C\frac{a+\mu_kg}{\text{ cos}\gamma+\mu_k\text{ sin}\gamma}=16.9\text{ N}.$$