In the diagram shown, the normal reaction force between 2 ~kg and 1...

In the diagram shown, the normal reaction force between

2 \mathrm{~kg}

and

1 \mathrm{~kg}

is (Given

g=10 \mathrm{~ms}^{-2}

) (Consider the surface, to be smooth): In the diagram shown, the normal reaction force between 2 ~kg and 1 ~kg is (Given g=10 ~ms^ -2 ) (Consider the surface,

Options

A $25 \mathrm{~N}$
B $39 \mathrm{~N}$
C $6 \mathrm{~N}$
D $10 \mathrm{~N}$

Correct answer

A. $25 \mathrm{~N}$

Step-by-step solution

Total force along inclined plane,

\begin{aligned} f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\ f & =-60+6 g \sin 30^{\circ}+18 \\ f & =-60+6 \times 10 \times \frac{1}{2}+18 \\ f & =12 \mathrm{ N} \text{ (downwards) } \\ \text{acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{ m} / \mathrm{s}^2 \end{aligned}

As per question, the normal force between

2 \mathrm{ kg} and

\mathrm{ kg}

is,

\begin{aligned} \mathrm{N}-18-10 \sin 30^{\circ} & =m a \\ \mathrm{ N}-18-10 \times \frac{1}{2} & =1 \times 2 \\ \mathrm{ N} & =2+23 \\ \mathrm{ N} & =25 \mathrm{ N} \end{aligned}

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