Answer to Question #104508 in Mechanics | Relativity for havefun7741

Question

[img]https://upload.cc/i1/2020/03/03/VmZEs6.jpg[/img]

Answer 1

(i)

According to the second Newton's low

$-T+m_a\cdot g\cdot \sin40°=m_a\cdot a$

$T-\mu_s\cdot (m_b\cdot g+T_0\cdot \sin30°)-T_0\cdot \cos30°=m_b\cdot a$

$T_0\cdot \cos30°-\mu_s\cdot (m_c\cdot g-T_0\cdot \sin30°)=m_c\cdot a$

Add these equations

$a(m_a+m_b+m_c)=m_a\cdot g\cdot \sin 40°-(m_b+m_c)\cdot g\mu_s$ ;

$m_a\cdot g\cdot \sin 40°-(m_b+m_c)\cdot g\mu_s=$

$=12\cdot 9.8\cdot \sin 40°-(10+5)\cdot 9.8\cdot 0.4=16.8 N$

The block will move.

(ii)

$a=\frac{m_a\cdot g\cdot \sin 40°-(m_b+m_c)\cdot g\mu_k}{m_a+m_b+m_c}=$

$=\frac{12\cdot 9.8\cdot \sin 40°-(10+5)\cdot 9.8\cdot 0.3}{12+10+5}=1.17 m/s^2$

(iii)

$T=m_a\cdot g\cdot \sin40°-m_a\cdot a=12\cdot 9.8\cdot \sin40°-12\cdot 1.17\approx61.6N$

$T_0=\frac{m_ca+\mu_km_cg}{\cos30°+\mu_k\sin30°}=\frac{5\cdot 1.17+0.3\cdot 5\cdot 9.8}{\cos30°+0.3\cdot \sin30°}\approx20.2N$