Question

A block of mass $M$ is at rest on a rough horizontal surface. The coefficient of friction between the block and the surface is$\mu$. A force $F=Mg$ acting at an angle $\theta$ with the vertical side of the block pulls it. In which of the following cases, the block can be pulled along the surface?

• A. $\tan \theta \ge \mu$
• B. $\cot \theta \ge \mu$
• C. $\tan \theta /2\ge \mu$
• D. $\cot \theta /2\ge \mu$

Answer 1

The correct option is D $\cot \theta /2\ge \mu$

As shown in the figure the Normal reaction $N$ on the block and Vertical component of the force $F$ are opposite to weight.

We need only horizontal motion no vertical motion so vertical force should be balanced that is net force in vertical direction should be zero, i.e., $N+FCos\theta -Mg=0$ or $N=Mg-FCos\theta$ , put $F=Mg$ as given in the question. so by putting so we get $N=Mg(1-cos\theta )$
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Now putting another condition to make F able to pull the block horizontally the $Fsin\theta$ should be greater than or eual to $\text{Static friction}$ and we know that maximum value of static friction is $\mu N$
so we have $Fsin\theta \ge \mu N$
putting value of $N$ we get $Sin\theta =\mu (1-cos\theta )$
put $cos\theta =1-2Si{n}^2\frac{\theta }{2}$ and $sin\theta =2sin\frac{\theta }{2}Cos\frac{\theta }{2}$
we will get $Cot\frac{\theta }{2}\ge \mu$ Option D is correct