Question Number 24371 by Tinkutara last updated on 16/Nov/17
$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{moving}\:\mathrm{together}\:\mathrm{under} \\ $$$$\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{horizontal} \\ $$$$\mathrm{external}\:\mathrm{force}\:{F}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{block}\:\mathrm{is} \\ $$$$\mathrm{at}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bigger}\:\mathrm{block} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{them},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bigger} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{floor}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 17/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Tinkutara last updated on 16/Nov/17
Commented by ajfour last updated on 16/Nov/17
Commented by ajfour last updated on 16/Nov/17
$${from}\:{balance}\:{of}\:{smaller}\:{block} \\ $$$${along}\:{vertical}: \\ $$$$\:\:{f}={mg} \\ $$$${from}\:{balance}\:{of}\:{bigger}\:{block} \\ $$$${along}\:{vertical}: \\ $$$$\:\:{R}\:=\:{mg}+{f} \\ $$$${Hence}\:\:{R}\:=\:\mathrm{2}{mg}\:. \\ $$
Commented by mrW1 last updated on 17/Nov/17
$${You}\:{are}\:{right}\:{sir}! \\ $$