First slide

Mechanical equilibrium force cases

Question

A block of mass m is placed on a smooth inclined wedge ABC of inclination $θ$ as shown in the figure. The wedge is given an acceleration a towards the right. The relation between a and $θ$ for the block to remain stationary on the wedge is

Moderate

A

$a = g cosθ$
B

$a = \frac{g}{sinθ}$
C

$a = \frac{g}{cosecθ}$
D

$a = g tanθ$

Solution

According to the question, the free body diagram of the given condition will be
Since, the wedge is accelerating towards right with a, thus a pseudo force acts in the left direction in order to keep the block stationary. As, the system is in equilibrium.
$∴ \sum F_{x} = 0 or \sum F_{y} = 0$
$\Rightarrow R sinθ = m a . . . . . (i)$
$Similarly R \cos θ = m g . . . . . (ii)$
Dividing Eq. (i) by Eq (ii), we get
$\frac{R sinθ}{R cosθ} = \frac{m a}{m g}$
$\Rightarrow tanθ = \frac{a}{g} o r a = g tanθ$
$∴$ The relation between a and 9 for the block to remain stationary on the wedge is $a = g tanθ$ .

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