The weight will be in balance when:
{Tsinα=FTcosα=mg\begin {cases} T \sin \alpha = F \\ T \cos \alpha = mg \end{cases}{Tsinα=FTcosα=mg Then T=mgcosα=20Ncos(30∘)T = \frac{mg}{cos \alpha} = \frac{20N}{\cos (30^\circ)}T=cosαmg=cos(30∘)20N , so
T = 23.09N
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments