Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

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

0 votes
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),
Substitute (3) into (1) obtains
$F = (M + m) a $
$\;\;\;= (M + m)g \tan\theta$
answered Aug 28, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App