Question

A 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 $\varphi$, the magnitude of the force required to move the body is equal to:

• A. $\frac{Wcos\varphi }{Cos(\theta -\varphi )}$
• B. $\frac{Wsin\varphi }{Cos(\theta -\varphi )}$
• C. $\frac{WTan\varphi }{Sin(\theta -\varphi )}$
• D. $\frac{WSins\varphi }{Tan(\theta -\varphi )}$

Answer 1

The correct option is B $\frac{Wsin\varphi }{Cos(\theta -\varphi )}$

Horizontal component of force is $Fcos\theta$

Vertical component of F is $Fsin\theta$

So, normal reaction $N=W-Fsin\theta$

Friction $f=\mu N=tan\varphi (W-Fsin\theta )$

So, move the body $Fcos\theta >f$

$F=\frac{f}{cos\theta }$

$F=\frac{Wsin\varphi }{cos(\theta -\varphi )}$