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Question Number 67655 by ajfour last updated on 29/Aug/19

Commented by ajfour last updated on 29/Aug/19

If the man can provide a maximum  acceleration a to the boxes while  he runs along, find the friction  coefficient of his shoes with  ground. (Assume the ground has  the same friction coefficient  (static or kinetic with the boxes  and the man′s shoes).

$${If}\:{the}\:{man}\:{can}\:{provide}\:{a}\:{maximum} \\ $$$${acceleration}\:\boldsymbol{{a}}\:{to}\:{the}\:{boxes}\:{while} \\ $$$${he}\:{runs}\:{along},\:{find}\:{the}\:{friction} \\ $$$${coefficient}\:{of}\:{his}\:{shoes}\:{with} \\ $$$${ground}.\:\left({Assume}\:{the}\:{ground}\:{has}\right. \\ $$$${the}\:{same}\:{friction}\:{coefficient} \\ $$$$\left({static}\:{or}\:{kinetic}\:{with}\:{the}\:{boxes}\right. \\ $$$$\left.{and}\:{the}\:{man}'{s}\:{shoes}\right). \\ $$

Commented by mr W last updated on 30/Aug/19

F−μ(M+m)g=(M+m)a  μm_0 g−F=m_0 a  ⇒μ=(((M+m+m_0 )a)/((m_0 −M−m)g))=(((m_0 /(M+m))+1)/((m_0 /(M+m))−1))×(a/g)

$${F}−\mu\left({M}+{m}\right){g}=\left({M}+{m}\right){a} \\ $$$$\mu{m}_{\mathrm{0}} {g}−{F}={m}_{\mathrm{0}} {a} \\ $$$$\Rightarrow\mu=\frac{\left({M}+{m}+{m}_{\mathrm{0}} \right){a}}{\left({m}_{\mathrm{0}} −{M}−{m}\right){g}}=\frac{\frac{{m}_{\mathrm{0}} }{{M}+{m}}+\mathrm{1}}{\frac{{m}_{\mathrm{0}} }{{M}+{m}}−\mathrm{1}}×\frac{{a}}{{g}} \\ $$

Commented by ajfour last updated on 29/Aug/19

Very precise, and correct, Sir.  thanks!

$${Very}\:{precise},\:{and}\:{correct},\:{Sir}. \\ $$$${thanks}! \\ $$

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