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A man drags a block of mass 10 kg on a rough horizontal surface. What minimum force (at some angle from horizontal ) he should apply just to move the block ( in N) $(\mu =\frac{1}{\sqrt{3}})$ .
Option: 1
50

Option: 2
75

Option: 3
60

Option: 4
100

Answers (1)

best_answer

Given-

mass of the block, m=10kg

coefficient of friction $\mathrm{ \mu=\frac{1}{\sqrt{3}}}$

Let the applied force be P be applied at an angle α with the horizontal as shown in the figure above

Let the friction force on the block be F.

F.B.D of the block-

For vertical equilibrium,

$\\ \mathrm{R+Psin\alpha=mg}\\ \mathrm{\Rightarrow\therefore R=mg-Psin\alpha}$

and for horizontal motion,

$\mathrm{Pcos\alpha\geq F\Rightarrow Pcos\alpha\geq \mu R}$

Substituting the value of R, we get:-

$\mathrm{Pcos\alpha\geq \mu(mg-Psin\alpha)}$

$\mathrm{\Rightarrow P\geq \frac{\mu mg}{cos\alpha+\mu sin\alpha}}$

For the force P to be minimum $\mathrm{\left ( cos\alpha+\mu sin\alpha \right )}$ must be maximum i.e.,

$\mathrm{\frac{\mathrm{d} }{\mathrm{d} \alpha}[cos\alpha+\mu sin\alpha]=0}$

$\mathrm{\Rightarrow -sin\alpha+\mu cos\alpha=0}$

$\mathrm{\therefore tan\alpha=\mu}$

$\mathrm{\Rightarrow \alpha=tan^{-1}(\mu)}$ = angle of friction.

i.e. For the minimum value of P, its angle from the horizontal should be

equal to the angle of friction.

As $\mathrm{tan\alpha=\mu}$ , so from the figure,

$\mathrm{sin\alpha=\frac{\mu}{\sqrt{1+\mu^{2}}}}$ and $\mathrm{cos\alpha=\frac{1}{\sqrt{1+\mu^{2}}}}$

By substituting these values,

$\mathrm{P\geq \frac{\mu mg}{\frac{1}{\sqrt{1+\mu^{2}}}+\frac{\mu^{2}}{\sqrt{1+\mu^{2}}}}}$

$\mathrm{\Rightarrow P\geq \frac{\mu mg}{\sqrt{1+\mu^{2}}}}$

$\mathrm{\therefore P_{min}=\frac{\mu mg}{\sqrt{1+\mu^{2}}}}$

Substituting the values of $\mathrm{\mu}$ anf m-

$\mathrm{P_{min}=\frac{\frac{1}{\sqrt{3}}\times10\times10}{\sqrt{1+\frac{1}{3}}}}=50 N$

Posted by

Irshad Anwar

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