Question #104508

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

Expert's answer

(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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