Question

A block A of mass m which is placed on a rough inclined face of a wedge of same mass being pulled through light string and with force F as shown in Fig. The coefficient of friction between inclined face and block A is $\mu$, while there is no friction between the ground and wedge. If the whole system moves with same acceleration, then find the value of F.

• A. $F<\frac{4mg(\mu cos\theta -sin\theta )}{(2-cos\theta )}$
• B. $F<\frac{2mg(\mu \cos \theta -\sin \theta )}{(2-\cos \theta -\mu \sin \theta )}$
  
• C. $F<\frac{3mg(\mu cos\theta -sin\theta )}{(2-cos\theta )}$
• D. $F>\frac{2mg(\mu cos\theta -sin\theta )}{(2-cos\theta )}$

Answer 1

The correct option is B $F<\frac{2mg(\mu \cos \theta -\sin \theta )}{(2-\cos \theta -\mu \sin \theta )}$

If the whole system moves together, then the acceleration of the system,

$a=\frac{F}{2m}$

$T+\mathrm{mg}\sin \theta =f+\mathrm{macos}\theta$

$\text{But}T=F,\text{therefore,}$

$f=F+\mathrm{mg\; sin}\theta -\mathrm{ma\; cos}\theta$

$\text{for no relative sliding}(f<\mu N)$

$\Rightarrow F+mg\sin \theta -\mathrm{macos}\theta <\mu (\mathrm{mg}\cos \theta +ma\sin \theta )$

$\Rightarrow F+mg\sin \theta -m\left(\frac{F}{2m}\right)cos\theta <\mu (\mathrm{mg}\cos \theta +m(\frac{F}{2m}\left)\sin \theta \right)$

$F<\frac{2mg(\mu \cos \theta -\sin \theta )}{(2-\cos \theta -\mu \sin \theta )}$