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A block of mass 2kg kept on a rough horizontal force is being pulled by applying a force P. Find minimum force required to just move the block. It is given that the coeficient of the static friction $\mu =\sqrt{3}$ between block and surface and g= 10 m/s².
Option: 1
10.34 N

Option: 2
12.56 N

Option: 3
17.3 N

Option: 4
8.7 N

Answers (1)

best_answer

Given-

mass of the block, m=2kg

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

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

As-

$\\ \mathrm{ (acos\theta+bsin\theta)_{max}=\sqrt{a^2+b^2}}\\ \mathrm{ (cos\alpha+\mu sin\alpha)_{max}=\sqrt{\mu^2+1}}$

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

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

$\\ \mathrm{P_{min}=\frac{\sqrt{3}\times2\times10}{\sqrt{1+3}}}=10\sqrt{3} N\\ \mathrm{P_{min}=17.3N}$

Posted by

Irshad Anwar

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