Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

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

Compound machine Quiz 2

physics

The force $F$ required to move the $2\,kg$ block with an acceleration of $10\,m{{s}^{-2}}$ on the inclined plane is  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fd5a33bb-60a6-4f53-b58f-9f1ef4a21a7c504563279909610922.png"/> 

 For the same string, equal force will be transmitted over the ends of the string across a pulley. Draw the F.B.D and calculate the force.  

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/43572352-ac2b-4ccc-a7c3-7d03ad0f4a4f1132548497149290406.png"/> $ \\ $ The same force $F$ is transmitted over transmitted over the common string $ \\ \Rightarrow F-2g\sin 45{}^\circ =2\left( a \right) \\ \Rightarrow F-2\times 10\times \dfrac{1}{\sqrt{2}}=2\times 10 \\ \Rightarrow F=20\left( \dfrac{1}{\sqrt{2}}+1 \right) \\ \therefore  F=10\sqrt{2}\left( \sqrt{2}+1 \right)\,N$

Calculate the value of force $F$ to produce on upward acceleration of $2\,m/{{s}^{2}}$ in the $3\,kg$ mass. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8d9ba54d-970e-4322-83c9-1332fa9deb733881252732620195377.png"/> 

 Both blocks move with equal acceleration. Draw the F.B.Ds and solve the force equations. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b162c76f-6519-4527-9289-9f7fc416c6846928034535190000027.png"/> $ \\ $ Both the blocks will move with the same acceleration. As string is same $ \\ \Rightarrow {{T}_{1}}=F \\ From\text{ }\left( 1 \right){{T}_{1}}-{{T}_{2}}-2g=2a \\ \therefore F-{{T}_{2}}-20=2\times 2 \\ \Rightarrow F={{T}_{2}}+24..........\left( A \right) \\ $  From $\left( 2 \right), {{T}_{2}}-3g=3a $ $ \\ \Rightarrow {{T}_{2}}-3\left( 10 \right)=3\left( 2 \right) \\ \Rightarrow {{T}_{2}}=30+6 \\ \Rightarrow {{T}_{2}}=36 \\ $  Using this in $\left( \text{A} \right)$ $ \\ \Rightarrow F=36+24 \\ \therefore F=60\,N$

Find force $F$ required to maintain the given system of masses in equilibrium. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c5d7b65d-23ed-4857-a5fb-66e32b3f2f7c7443003201116638166.png"/> 

 A single string always has equal tensions on both of its ends across a pulley. Use this concept and draw the F.B.D.  

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7077c07b-99e8-457f-adf3-afb9560903053138979645450083447.png"/> $ \\ F+T=mg.........\left( 1 \right) \\ \Rightarrow T+T=Mg \\ \Rightarrow 2t=Mg \\ \Rightarrow T=\dfrac{Mg}{2}........\left( 2 \right) \\ \Rightarrow F+\dfrac{Mg}{2}=mg \\ \Rightarrow F=g\left( m-\dfrac{M}{2} \right) \\ \therefore F=g\left( \dfrac{2m-M}{2} \right)$

Given, $O$ is the fulcrum of the lever shown above. If $OA=2\,m$, $OC=2\,m$, $OB=6\,m$, find the minimum mass $M$ that is needed to be placed on point $B$ to stop the $10\,kg$ block from moving down on the inclined plane. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bffcf23b-436c-4753-9b91-9ddc20f005da4090231646783994789.png"/> 

 Calculate the minimum tension in the string which is needed to stop the block from moving down the plane. Then balance all the torque components around the fulcrum to get minimum mass at the point $B$. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1dac6382-87b7-4196-9ddc-20a769c096f22019410341628274145.png"/> $ \\ $  Minimum Tension $T$ needed to stop block and make it stationary $ \\ T=10g\sin 30{}^\circ  \\ \Rightarrow T=10\times 10\times \dfrac{1}{2} \\ \Rightarrow T=50\,N \\ $ Balancing Torque around fulcrum $O$ for the lever, $ \\ {{\tau }_{M}}={{\tau }_{1kg}}+{{\tau }_{T}} \\ \Rightarrow OB \times Mg=OA\times 1+OC\times T \\ \Rightarrow 6\times M\times 10=2\times 1+2\times 50 \\ \Rightarrow M=\dfrac{2+100}{60} \\ \Rightarrow M=\dfrac{102}{60} \\ \therefore M=1.7\,kg$

Find the acceleration of the block suspended in air. $($Take $g=10\,m{{s}^{-2}})$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e5a0ae97-1048-45b4-8338-4a8c5ac2ae6b7260959019397909365.png"/>

 Use constraint relation to find out the different accelerations of the blocks and draw their F.B.D. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b7394fa0-487c-45c1-81af-3aed5f77eb918368086328250096222.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/169978ca-fab3-4e33-a460-83fded0208854999922623875286147.png"/> $ \\ $ Let the length of the string be $\ell$ and $\ell '$. The total string length is constant. By constraint relation $ \\ 2\ell +\ell '=\text{Constant} \\ $  If acceleration of the block on table is $a'$ and acceleration of the block suspended in air is $a$, then $ \\ $ Differentiating twice w.r.t time $ \\ \Rightarrow 2a+a'=0 \\ \Rightarrow a'=2a \\ \text{From }\left( \text{1} \right)\text{ and }\left( \text{2} \right) \\ \Rightarrow T=M\left( 2a \right) \\ \Rightarrow T=2Ma \\ \Rightarrow 2T=Mg \\  \Rightarrow 2\left( Ma \right)=Mg \\ \Rightarrow a=\dfrac{g}{4} \\ \therefore a=2.5\,m{{s}^{-2}}$