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

Answer: (A)

$F+F\cos \theta =mg\sin \theta$
$F=\frac{mg\sin \theta }{1+\cos \theta }$
$F=\frac{mg2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$
$=mg\tan \frac{\theta }{2}$

$(\because \sin \theta =2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}$ and $1+\cos \theta =2{\cos }^2\frac{\theta }{2})$