Question Number 73260 by Lontum Hans last updated on 09/Nov/19
Answered by mr W last updated on 09/Nov/19
$${T}_{{max}} ={m}\left(\frac{{v}^{\mathrm{2}} }{{r}}+{g}\right)=\mathrm{8}\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{10}\right)=\mathrm{224}\:{N} \\ $$$${T}_{{min}} ={m}\left(\frac{{v}^{\mathrm{2}} }{{r}}−{g}\right)=\mathrm{8}\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{10}\right)=\mathrm{64}\:{N} \\ $$
Commented by Lontum Hans last updated on 09/Nov/19
$$\mathrm{nN}{ot}\:{accurate} \\ $$