Question

Find max force F for which system remains in equilibrium

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F_{max}\leq28\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given:}}} \\ \tt: \implies Coefficient \: of \: friction \: between \: block \: and \: ground( \mu_{1}) = 0.4 \\ \\ \tt: \implies Coefficient \: of \: friction \: between \: block \: and \: block( \mu_{2}) = 0.2 \\ \\ \tt: \implies Mass \: of \: block( m_{1}) = 5 \: kg \\ \\ \tt: \implies Mass \: of \: block( m_{2}) = 2 \: kg \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies Force \: maximum \: so \: that \: the \: system \: will \: be \: in \: equilibrium( F_{max}) =?$

• According to given question :

$\bold{For \: block \: of \: mass \: ( m_{1})} \\ \tt: \implies T = \mu_{1} m_{1}g + \mu_{2} m_{2}g \\ \\ \tt: \implies T = 0.4 \times 5 \times 10 + 0.2 \times 2 \times 10 \\ \\ \tt: \implies T= 20 + 4\\ \\ \tt: \implies T = 24 \: N- - - - - (1) \\ \\ \bold{For \: balanced \: forces \: on\: mass \: ( m_{2})} \\ \tt: \implies F_{max} \leq T + \mu_{2}m_{2}g \\ \\ \tt: \implies F_{max} \leq 24 + 0.2 \times 2 \times 10 \\ \\ \tt: \implies F_{max} \leq 24 + 4 \\ \\ \green{\tt: \implies F_{max} \leq 28 \: N} \\ \\ \green{\tt \therefore Force \: should \: be \: less \: than \: or \: equal \: to \: 28 \: N\:for \:maintaining \: equilibrium}$