Question

Pulling force making an angle $\theta$ to the horizontal is applied on a block of weight $W$ placed on 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. $\frac{W\cos \alpha }{\cos (\theta -\alpha )}$
• C. $\frac{W\sin \alpha }{\cos (\theta -\alpha )}$
• D. $\frac{W\tan \alpha }{\sin (\theta -\alpha )}$

Answer 1

The correct option is C $\frac{W\sin \alpha }{\cos (\theta -\alpha )}$

In vertical direction

$N+Fsin\theta =W$

$N=W-Fsin\theta$

As angle of friction is $\alpha$

then $\mu =tan\alpha$

For motion of block

$Fcos\theta >=f$

or

$Fcos\theta =f$

$Fcos\theta =\mu N$

$Fcos\theta =tan\alpha (W-Fsin\theta )$

$F=\frac{Wtan\alpha }{cos\theta +tan\alpha sin\theta }$

$F=\frac{Fsin\alpha }{cos(\theta -\alpha )}$