Question

A triangular wedge of
mass 'M' is placed on a
smooth horizontal
surface. A block of mass
'm' is placed on the
wedge. The coefficient of
friction between the block
and the wedge is 0.6. The
horizontal acceleration 'a'
of the wedge such that 'm'
moves up the wedge is
​

Answer 1

Given:

A triangular wedge ofmass 'M' is placed on a smooth horizontal surface. A block of mass 'm' is placed on the wedge. The coefficient of friction between the block and the wedge is 0.6.

To find:

Value of horizontal acceleration of the ways such that "m" moves up the wedge.

Calculation:

First of all , look at the free body diagram of mass "m" attached in the photo.

At limiting conditions, (when the block is about to move up), we can say that :

$\therefore \: mg \sin( \theta) + \mu mg \cos( \theta) = ma \cos( \theta)$

$= > \: g \sin( \theta) + \mu g \cos( \theta) = a \cos( \theta)$

$= > \: g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \} = a \cos( \theta)$

$= > \: a = \dfrac{g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}}{ \cos( \theta) }$

Putting values of $\theta$ and $\mu$:

$= > \: a = \dfrac{g \bigg \{ \sin( {60}^{ \circ} ) + 0.6 \cos( {60}^{ \circ} ) \bigg \}}{ \cos( {60}^{ \circ} ) }$

$= > \: a = \dfrac{g \bigg \{ \frac{ \sqrt{3} }{2} + 0.6 ( \frac{1}{2} )\bigg \}}{ \frac{1}{2} }$

$= > \: a = \dfrac{g \bigg \{ 0.86 + 0.3\bigg \}}{ \frac{1}{2} }$

$= > \: a = \dfrac{g \bigg \{1.16\bigg \}}{ \frac{1}{2} }$

$= > \: a = g \times 2.32$

$= > \: a = 2.32 \times 10$

$= > \: a = 23.2 \: m {s}^{ - 2}$

So, the required acceleration is 23.2 m/s².