(C) The forces acting on the garden roller in the vertical direction are the normal reaction $N$ (upwards),the weight $mg$ (downwards),and the vertica

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(C) The forces acting on the garden roller in the vertical direction are the normal reaction $N$ (upwards),the weight $mg$ (downwards),and the vertical component of the applied force $F \sin 30^{\circ}$ (downwards).Since the roller is in equilibrium in the vertical direction,the net force is zero:$N - mg - F \sin 30^{\circ} = 0$$N = mg + F \sin 30^{\circ}$Given $m = 70\,kg$,$g = 10\,m s^{-2}$,$F = 200\,N$,and $\sin 30^{\circ} = 0.5$:$N = (70 \times 10) + (200 \times \sin 30^{\circ})$$N = 700 + (200 \times 0.5)$$N = 700 + 100$$N = 800\,N$

Class 11 Physics Newton's Laws of Motion and Friction The Common Forces and Equilibrium of Concurrent Forces

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As shown in the figure,a $70\,kg$ garden roller is pushed with a force of $\overrightarrow{F} = 200\,N$ at an angle of $30^{\circ}$ with the horizontal. The normal reaction on the roller is $.......\,N$ (Given $g = 10\,m s^{-2}$)

JEE MAIN 2023 All JEE MAIN

A
$800 \sqrt{2}$
B
$600$
C
$800$
D
$200 \sqrt{3}$

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Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

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The Common Forces and Equilibrium of Concurrent Forces

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As shown in the figure,a $70\,kg$ garden roller is pushed with a force of $\overrightarrow{F} = 200\,N$ at an angle of $30^{\circ}$ with the horizontal. The normal reaction on the roller is $.......\,N$ (Given $g = 10\,m s^{-2}$)

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For the free body diagram shown in the figure,four forces are applied in the $x$ and $y$ directions. What additional force must be applied and at what angle with the positive $x$-axis so that the net acceleration of the body is zero?

Write the condition for the translational equilibrium of a particle.

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$A$ small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed under her pushing force $F$,which is obliquely upward as shown in the figure. Then,

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