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 $Mg\tan \left(\frac{\theta }{2}\right)$

From the FBD as shown, we can say that
Along the inclination,
$F+F\cos \theta =Mg\sin \theta$
$F=\frac{Mg\sin \theta }{1+\cos \theta }$
$F=\frac{Mg.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}(\because \sin \theta =2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\text{and}1+\cos \theta =2{\cos }^2\frac{\theta }{2})$
Thus, minimum force required is
$F=Mg\tan \left(\frac{\theta }{2}\right)$