A block of mass m=2 kg is placed on a plank of mass M=10 kg, which is placed on a smooth horizontal plane. The coefficient of friction between the block and the plank is μ=13. If a horizontal force F is applied on the plank, then the maximum value of F in newton for which the block and the plank move together is (g=10 m/s2)

a=F−fM=F−f10→(2)F−f10=f2F−f=5fF=6ff=F6Not to slidef=fsmaxFmax6=μs2gFmax=13×20×6RFmax=40NNot to topplef×h2=N×h8f=N4=mg4=204=5Fmax 6=5Fmax=30NFmax=30N to move blocks together.

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A block of mass m=2 kg is placed on a plank of mass M=10 kg, which is placed on a smooth horizontal plane. The coefficient of friction between the block and the plank is $μ = \frac{1}{3} .$ If a horizontal force F is applied on the plank, then the maximum value of F in newton for which the block and the plank move together is $(g = 10 m / s^{2})$

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Detailed Solution

$\begin{array}{l} a = \frac{F - f}{M} = \frac{F - f}{10} \to (2) \\ \frac{F - f}{10} = \frac{f}{2} \\ F - f = 5 f \\ F = 6 f \\ f = \frac{F}{6} \end{array}$
Not to slide
$\begin{array}{l} f = f_{s max} \\ \frac{F_{max}}{6} = μ_{s} 2 g \\ F_{max} = \frac{1}{3} \times 20 \times 6 \\ R_{F_{max}} = 40 N \end{array}$
Not to topple
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$\begin{array}{l} f \times \frac{h}{2} = N \times \frac{h}{8} \\ f = \frac{N}{4} = \frac{mg}{4} = \frac{20}{4} = 5 \\ \frac{F_{max}^{}}{6} = 5 \\ F_{max} = 30 N \\ F_{max} = 30 N to move blocks together. \end{array}$

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