First slide

Mechanical equilibrium force cases

Question

Block B lying on a table weighs w. The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is

Moderate

A

$\frac{w tanθ}{μ}$
B

$μ w tanθ$
C

$μ w \sqrt{1 + \tan^{2} θ}$
D

$μ w sinθ$

Solution

Let weight of A is w'. From the free body diagram, for equilibrium of the system,
$T cosθ = μ N = μ w . . . . . . . . . . (i)$
$T sinθ = w^{'} . . . . . . . . . . (ii)$
where, T = tension in the thread lying between knot and the support.
On dividing Eq. (ii) by Eq. (i), we get
$\frac{T \sin θ}{T \cos θ} = \frac{w^{'}}{μ w} \Rightarrow \tan θ = \frac{w^{'}}{μ w}$
$\Rightarrow w^{'} = μ w \tan θ$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App