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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A right triangular wedge of mass M and inclination angle $\theta$ , supporting a small block of mass m on its side, rests on a horizontal frictionless table, as shown in figure. Assuming all surfaces are frictionless, What horizontal force F must be applied to the system to achieve this result?

$\begin{array}{1 1}(M+m)g\cot\theta\\(M+m)g\cos\theta\\(M+m)g\sin\theta\\(M+m)g\tan\theta\end{array} $

Can you answer this question?
 
 

1 Answer

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Answer : $(M+m)g\tan\theta$
In this case M and m are moving with the same acceleration a. Note that a is along x-direction only
For M and m :
$F-N_2\sin \theta=Ma$--------(1)
$mg-N_2\cos\theta=0$$\Rightarrow N_2=\large\frac{mg}{\cos\theta}$------(2)
$N_2\sin \theta=ma$------(3)
From (2) and (3),
$a=g\tan\theta$
Substitute (3) into (1) obtains
$F = (M + m) a $
$\;\;\;= (M + m)g \tan\theta$
answered Aug 28, 2014 by sreemathi.v
 

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