Question Number 22893 by NECx last updated on 23/Oct/17
$${A}\:{wedge}\:{has}\:{two}\:{equally}\:{rough} \\ $$$${faces}\:{each}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the} \\ $$$${horizontal}.{Masses}\:{of}\:\mathrm{5}{kg}\:{and}\:\mathrm{2}{kg} \\ $$$$,{one}\:{on}\:{each}\:{face},{are}\:{connected} \\ $$$${by}\:{a}\:{light}\:{string}\:{passing}\:{over}\:{a} \\ $$$${smooth}\:{pulley}\:{at}\:{the}\:{top}\:{of}\:{the} \\ $$$${wedge}.{The}\:{coefficient}\:{of}\:{friction}\: \\ $$$$\mu,\:{between}\:{each}\:{masses}\:{and}\:{the} \\ $$$${surface}\:{of}\:{the}\:{wedge}\:{is}\:\mathrm{0}.\mathrm{2}.{Find} \\ $$$${the}\:{acceleration}\:{of}\:{the}\:{masses} \\ $$$${when}\:{they}\:{are}\:{released}. \\ $$
Answered by ajfour last updated on 23/Oct/17
$${a}=\frac{\left({M}−{m}\right){g}\mathrm{sin}\:\mathrm{30}°−\mu{g}\mathrm{cos}\:\mathrm{30}°\left({M}+{m}\right)}{{M}+{m}} \\ $$$$=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{7}}\:{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow\:\boldsymbol{{a}}\approx\left(\mathrm{2}.\mathrm{143}−\mathrm{1}.\mathrm{732}\right){m}/{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{4}\:{m}/{s}^{\mathrm{2}} \: \\ $$
Commented by NECx last updated on 23/Oct/17
$${thankz}\:{sir}…..\:{please}\:{can}\:{it}\:{be} \\ $$$${explained}\:{with}\:{a}\:{diagram}.{I}'{ll} \\ $$$${appreciate}\:{that}.\boldsymbol{{T}}{hanks}\:{in}\:{advance}. \\ $$
Commented by NECx last updated on 23/Oct/17
$${thanks}\:{sir} \\ $$
Commented by ajfour last updated on 23/Oct/17
