Question Number 20409 by Tinkutara last updated on 26/Aug/17
![Calculate the force (F) required to cause the block of mass m_1 = 20 kg just to slide under the block of mass m_2 = 10 kg [coefficient of friction μ = 0.25 for all surfaces]](Q20409.png)
$${Calculate}\:{the}\:{force}\:\left({F}\right)\:{required}\:{to} \\ $$$${cause}\:{the}\:{block}\:{of}\:{mass}\:{m}_{\mathrm{1}} \:=\:\mathrm{20}\:{kg} \\ $$$${just}\:{to}\:{slide}\:{under}\:{the}\:{block}\:{of}\:{mass} \\ $$$${m}_{\mathrm{2}} \:=\:\mathrm{10}\:{kg}\:\left[{coefficient}\:{of}\:{friction}\:\mu\right. \\ $$$$\left.=\:\mathrm{0}.\mathrm{25}\:{for}\:{all}\:{surfaces}\right] \\ $$
Commented by Tinkutara last updated on 26/Aug/17
Answered by ajfour last updated on 26/Aug/17
$${T}\mathrm{sin}\:\mathrm{53}°+{N}_{\mathrm{2}} ={m}_{\mathrm{2}} {g} \\ $$$${T}\mathrm{cos}\:\mathrm{53}°=\mu{N}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{m}_{\mathrm{2}} {g}−{N}_{\mathrm{2}} }{\mu{N}_{\mathrm{2}} }\:=\mathrm{tan}\:\mathrm{53}°\:=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\:\frac{\mathrm{100}−{N}_{\mathrm{2}} }{\left({N}_{\mathrm{2}} /\mathrm{4}\right)}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\Rightarrow\:\:\:\mathrm{300}−\mathrm{3}{N}_{\mathrm{2}} ={N}_{\mathrm{2}} \\ $$$${N}_{\mathrm{2}} =\mathrm{75}\:{units}\:\:\Rightarrow\:\:{f}_{\mathrm{2}} =\mu{N}_{\mathrm{2}} =\frac{\mathrm{75}}{\mathrm{4}} \\ $$$${f}_{\mathrm{1}} =\mu{N}_{\mathrm{1}} =\mu\left({m}_{\mathrm{1}} {g}+{N}_{\mathrm{2}} \right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{200}+\mathrm{75}\right)=\frac{\mathrm{275}}{\mathrm{4}} \\ $$$${F}_{{minimum}} ={f}_{\mathrm{1}} +{f}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{275}+\mathrm{75}}{\mathrm{4}}\:=\frac{\mathrm{350}}{\mathrm{4}} \\ $$$$\:\:{F}_{{minimum}} =\:\mathrm{87}.\mathrm{5}\:{N}\: \\ $$
Commented by Tinkutara last updated on 26/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$