Question

What is the minimum value of F needed so that block begins to move upward on frictionless incline plane as shown

• A. $Mg\tan \left(\frac{\theta }{2}\right)$
• B. $Mg\cot \left(\frac{\theta }{2}\right)$
• C. $\frac{Mg\sin \theta }{(1+\sin \theta )}$
• D. $Mg\sin \left(\frac{\theta }{2}\right)$

Answer 1

The correct option is A $Mg\tan \left(\frac{\theta }{2}\right)$

Using equilibrium conditions,

$Mg\sin \theta =F+F\cos \theta$

$Mg\sin \theta =F(1+\cos \theta )$

$\Rightarrow F=\frac{Mg\sin \theta }{1+\cos \theta }$

$=\frac{Mg\times 2\sin \frac{\theta }{2}\times \cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$

$=Mg\tan \frac{\theta }{2}$