Pulling force making an angle to the horizontal is applied on a b… - Vidyadip

MCQ

Pulling force making an angle $\theta $ to the horizontal is applied on a block of weight $W$ placed on a horizontal table. If the angle of friction is $\alpha $, then the magnitude of force required to move the body is equal to

A
$\frac{{W\sin \alpha }}{{g\tan (\theta - \alpha )}}$
B
$\frac{{W\cos \alpha }}{{\cos (\theta - \alpha )}}$
✓
$\frac{{W\sin \alpha }}{{\cos (\theta - \alpha )}}$
D
$\frac{{W\tan \alpha }}{{\sin (\theta - \alpha )}}$

✓

Answer

Correct option: C.

$\frac{{W\sin \alpha }}{{\cos (\theta - \alpha )}}$

c
(c) $\uparrow: N+F \sin \theta=W \Rightarrow N=W-F \sin \theta$

The block will move if

$F \cos \theta \geq f_{\max }$

$F \cos \theta \geq \mu(W-F \sin \theta)$

$\mu=\tan \alpha=\frac{\sin \alpha}{\cos \alpha}$

$F \cos \theta \geq \frac{\sin \alpha}{\cos \alpha}(W-F \sin \theta)$

$F(\cos \theta \cos \alpha+\sin \theta \sin \alpha) \geq W \sin \alpha$

$F \geq \frac{W \sin \alpha}{\cos (\theta-\alpha)}$

$F_{\min }=\frac{W \sin \alpha}{\cos (\theta-\alpha)}$

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