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This is one of our many quizzes on Normal force. Click on this to access more such competitive quizzes on Mechanics

Normal force Quiz 2

physics

From the apparatus shown below in the diagram. What conclusion will you make about the net acceleration of the box $A$ ? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c20fe222-1d0d-4a3c-bb5f-4f49eb9f0559985398372050853841.png"/>

A body will only be at rest if and only if all the forces acting on the body cancel out each other. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/81e07e6e-6daa-4601-8869-4b7967c8459b3655804197791794702.png"/> $ \\ $ The FBD of block $A$ shows that there will be a pseudo force acting on the block. However, the block remains stationary and this is because the forces acting on the block $A$ are balanced. The pseudo force acting backwards is balanced by the normal force acting on the black due to the vertical wall. Since the net force acting on the block $A$ is zero, there is no acceleration of the block. But this is with respect to the trolley. The block is stationary with respect to the trolley but is moving with the same acceleration as the trolley with respect to ground. Therefore, the net acceleration with respect to the trolley is zero but with respect to ground it is $a$.

 If we are continuously pulled downwards by the force of gravity, why does a man standing still on the ground considered at rest? 

 If the net acceleration of a body is zero then that body is said to be at rest. Normal forces act perpendicular to the contact surfaces. 

  The earth constantly pulls everything within its gravitational field. So, if a man is standing still on the ground he is still being pulled downwards by the earth. Even then the net acceleration of the man in the vertical direction is zero because when the man is standing on the surface he is in contact with the earth’s surface. So, there is a vertically upward normal force acting on the man. This normal force is what balances the force of gravity or gravitational force.

 A ball is inside a vertically inverted cone. The largest radius of the cone is $R$ and the mass of the ball is $m$ . If the ball is undergoing a circular motion such that it travels all the way up to the mouth of the cone, what is the height of the cone? Consider the ball’s velocity to be constant and the angle of the cone is $2\theta $. 

 The normal force will act only when the ball is in contact with the inner surface of the cone. Once it reaches the topmost point the ball will leave the cone and there will be no surface in contact so normal will be zero. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/26eb0839-31ce-4407-8bc9-de7dd24143ad1293340734630353488.png"/> $ \\ \text{From the FBD of the ball at the highest point;} \\N=\dfrac{m{{v}^{2}}}{R}\cos \theta +mg\sin \theta  \\\Rightarrow N-mg\sin \theta =\dfrac{m{{v}^{2}}}{R}\cos \theta  \\\Rightarrow R=\dfrac{m{{v}^{2}}\cos \theta }{N-mg\sin \theta } \\\text{From trigonometry height of the cone is given by;} \\h=R\cot \theta \text{ }\to \left( \because \dfrac{R}{h}=\tan \theta  \right) \\\therefore h=\dfrac{m{{v}^{2}}\cos \theta \cot \theta }{N-mg\sin \theta }$

A small mass $m$ starts a circular motion inside a vertical metal circle. The circle is fixed. What is the maximum distance from the starting point that the ball will cover if $N$is the normal force acting on the small mass before losing contact? 

The mass will leave contact when the centripetal force acting radially outwards will cease to zero. At that instance the mass will be under the sole influence of the gravitational force pulling it down. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5db6e53a-9159-45f7-918c-7c5326818d4e7046447732400354952.png"/> $ \\ \text{From the FBD}\text{ of the mass }m\text{ ;} \\mg\cos \theta +N=\dfrac{m{{v}^{2}}}{R} \\\Rightarrow mg\cos \theta =\dfrac{m{{v}^{2}}}{R}-N \\\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{{{v}^{2}}}{Rg}-\dfrac{N}{mg} \right) \\\text{For the point where the mass will lose contact }v=0; \\\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-N}{mg} \right) \\\text{The angle covered by the mass is given by;} \\\alpha =\dfrac{\pi }{2}+\dfrac{\pi }{2}-\theta  \\\Rightarrow \alpha =\pi +{{\cos }^{-1}}\left( \dfrac{-N}{mg} \right) \\\text{The maximum distance covered is the arc suspending angle }\alpha ;\text{So,} \\\therefore \text{arc length }=R\alpha =R\left[ \pi +{{\cos }^{-1}}\left( \dfrac{-N}{mg} \right) \right]$

A rope of mass $m$ is resting on a triangular wedge such that two thirds of the rope is on one side of the wedge and one third of the rope is on the other side of the wedge. The linear mass density of the rope is $\lambda $ . What is the ratio of the normal forces acting on both the parts of the rope? Refer the diagram for more information. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b259e039-0bd1-4ee9-8442-8c3d0e0468d24509868714539275076.png"/> 

The mass will change with length according to the linear mass density. The two parts will have different masses as their lengths are different.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f5538378-f0b1-4ba2-836a-2f9345cef6f96500876552910503953.png"/> $ \\ \text{From the FBD of the rope parts;} \\\text{for }l=\dfrac{2}{3}L\text{ where }L\text{ is the total length of the rope;} \\\Rightarrow {m}'=\dfrac{2\lambda }{3} \\\text{for }l=\dfrac{1}{3}L\text{ where }L\text{ is the total length of the rope;} \\\Rightarrow {m}''=\dfrac{\lambda }{3} \\\text{From the FBD;} \\\Rightarrow {N}'={m}'g\cos \theta =\dfrac{2\lambda }{3}g\cos \theta \text{            }\to \left( 1 \right) \\\Rightarrow {N}''={m}''g\cos 2\theta =\dfrac{\lambda }{3}g\cos 2\theta \text{         }\to \left( 2 \right) \\\text{Divide equation }\left( 1 \right)\text{ and }\left( 2 \right)\text{;} \\\Rightarrow \dfrac{{{N}'}}{{{N}''}}=\dfrac{\dfrac{2\lambda g}{3}\cos \theta }{\dfrac{\lambda g}{3}\cos 2\theta } \\\therefore \dfrac{{{N}'}}{{{N}''}}=\dfrac{2\cos \theta }{\cos 2\theta }=\dfrac{2\cos \theta }{1-2{{\sin }^{2}}\theta }$