Question

In the given figure, what maximum force F can be applied on wedge of mass M, so that block of mass m does not move with respect to wedge.

• A.
  $\frac{(M+m)(sin\theta +\mu cos\theta )g}{(cos\theta -\mu sin\theta )}$
• B.
  $\frac{(M+m)(cos\theta +\mu sin\theta )g}{(sin\theta -\mu cos\theta )}$
• C.
  $\frac{(M+m)(sin\theta -\mu cos\theta )g}{(cos\theta +\mu sin\theta )}$
• D.
  $\frac{(M+m)(sin\theta +\mu cos\theta )g}{(sin\theta +\mu cos\theta )}$

Answer 1

The correct option is A

$\frac{(M+m)(sin\theta +\mu cos\theta )g}{(cos\theta -\mu sin\theta )}$
F = (M + m)a
$\Rightarrow$ a = $\frac{F}{M+m}$
Looking m from M $\Rightarrow$



$\Rightarrow$ macos$\theta$ = $f_l$ + mgsin$\theta$
$\Rightarrow$ macos$\theta$ = $\mu$N + mgsin$\theta$
$\Rightarrow$ macos$\theta$ = $\mu$masin$\theta$ + $\mu$mgcos$\theta$ + mgsin$\theta$
$\Rightarrow$m[cos$\theta$ - $\mu$sin$\theta$]a = mg[$\mu$cos$\theta$ + sin$\theta$]
a = $\frac{(\mu cos\theta +sin\theta )g}{(cos\theta -\mu sin\theta )}$
$\Rightarrow$ F = $\frac{(M+m)(sin\theta +\mu cos\theta )g}{(cos\theta -\mu sin\theta )}$