Answer to Question #104508 in Mechanics | Relativity for havefun7741

Question #104508
[img]https://upload.cc/i1/2020/03/03/VmZEs6.jpg[/img]
1
Expert's answer
2020-03-05T09:37:02-0500


(i)


According to the second Newton's low


"-T+m_a\\cdot g\\cdot \\sin40\u00b0=m_a\\cdot a"


"T-\\mu_s\\cdot (m_b\\cdot g+T_0\\cdot \\sin30\u00b0)-T_0\\cdot \\cos30\u00b0=m_b\\cdot a"


"T_0\\cdot \\cos30\u00b0-\\mu_s\\cdot (m_c\\cdot g-T_0\\cdot \\sin30\u00b0)=m_c\\cdot a"



Add these equations


"a(m_a+m_b+m_c)=m_a\\cdot g\\cdot \\sin 40\u00b0-(m_b+m_c)\\cdot g\\mu_s" ;


"m_a\\cdot g\\cdot \\sin 40\u00b0-(m_b+m_c)\\cdot g\\mu_s="


"=12\\cdot 9.8\\cdot \\sin 40\u00b0-(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\u00b0-(m_b+m_c)\\cdot g\\mu_k}{m_a+m_b+m_c}="


"=\\frac{12\\cdot 9.8\\cdot \\sin 40\u00b0-(10+5)\\cdot 9.8\\cdot 0.3}{12+10+5}=1.17 m\/s^2"


(iii)


"T=m_a\\cdot g\\cdot \\sin40\u00b0-m_a\\cdot a=12\\cdot 9.8\\cdot \\sin40\u00b0-12\\cdot 1.17\\approx61.6N"


"T_0=\\frac{m_ca+\\mu_km_cg}{\\cos30\u00b0+\\mu_k\\sin30\u00b0}=\\frac{5\\cdot 1.17+0.3\\cdot 5\\cdot 9.8}{\\cos30\u00b0+0.3\\cdot \\sin30\u00b0}\\approx20.2N"








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