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This is second of our many quizzes on Mechanical advantage. Click on this to access more such competitive quizzes on Mechanics.

Mechanical advantage Quiz 2

physics

 In the system shown, friction is absent all over the system.  Force $F$ is sufficient to hold the block of man $6kg$ mechanical advantage of system is<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8cebe4ee-e0a3-4b3a-9a3a-9b4aa1b1ff991006891896620278232.png"/> 

The string used to hang fixed pulleys can be ignored because tension in them is still balanced. The tension is also always equal for the same rope/string.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1be494fc-cc18-45b0-8349-2efaf9a137456222816749064969194.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/24a1dd0d-f24e-4193-bc1d-1388b19dbc9d1137501327544247626.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ee5ca155-e90e-4615-9587-1a11b6775d518749229526980634155.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d6c2a38b-ffe9-417e-acb3-dcce1adf9cdd6642747287485206293.png"/>$ \\ \text{In the given figure for pulley }\left( 1 \right) \\ \text{As the pulley is non moving and whole system is in equilibrium,} \\ F=T \\ \text{For pulley }\left( 2 \right) \\ \text{The same string is wrapped around pulley }\left( 2 \right). \\ \text{ The same string carries equal tension }''T''. \\ \text{As pulley }\left( 3 \right)\text{ is connected by pulley }\left( 2 \right) \\ \text{The force along central rope will be equal i}\text{.e}\text{.}''T'', \\ \text{By Newton’s third law of motion}\text{.} \\ \text{As all are stationary} \\ 3T=6g \\ \text{But }F=T\text{ from pulley}\left( 1 \right) \\ \Rightarrow T=F=2g \\ \Rightarrow\text{Mechanical advantage}=\dfrac{Load}{\text{Applied force}}=\dfrac{6g}{2g} \\ \therefore\text{Mechanical advantage}=3$

 Identify where mechanical advantage is more, if the weight of the load is $100N$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/702c60e7-7afc-405e-b5ea-01ab85c938785782374906801802359.png"/> 

 In both cases mechanical advantage $=\dfrac{\text{Load}}{\text{Applied force}}=\dfrac{100}{50}=2N$. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/46ce618b-bb25-4b29-808f-d19c76f254d23634681009464826170.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fe99ef02-43d0-4fbd-8d54-2e697ca738ad2792933162548952061.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0b5afca4-5efa-4eda-b755-f0fc9f0803438559827158304726657.png"/>$ \\ \text{In case }\left( 1 \right) \\ 2T=100 \\ \Rightarrow T=50N \\ \text{As system is in equilibrium} \\{{F}_{1}}=T=50N \\ \text{In case }\left( 2 \right) \\ N\text{ balances }100\cos 30{}^\circ  \\ \text{While,} \\ T=100\sin 30{}^\circ =100\times \dfrac{1}{2}=50N \\ \text{As the string is not moving and system is in equilibrium,} \\ \therefore F=T=50\,N$

Force experiencing ceiling is more in<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c7b475e9-54b9-4302-8cbc-0c3ecf9bcc55194717592215158618.png"/> 

For a system of pulleys, a same string will have equal tension across it regardless of the pulley. Draw FBD and write force equations keeping in mind that the systems are in equilibrium. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6a7cf18c-23da-4e64-b22c-a1d73ce3f4cc671946767652012630.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/292f58bb-df0d-4af4-95b5-07c3a4fd94a36196756536925164915.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/29dd9b50-04e5-4a66-a55d-7624f55032aa6810086285766046063.png"/>$ \\ $ Drawing separate F.B.D for each pulley in both cases in a horizontal sequence, we get above figures. $ \\ $ The systems are in equilibrium so, $ \\ $ $T=F$ the same string has equal tension throughout it. Thus the longest string will have tension $T$ and $T'$ for each case respectively. $ \\ $ For case $\left( \text{1} \right)$ the force equations from FBD are $ \\ 2T={{T}_{1}} \\ \Rightarrow 3T=500 \\ \Rightarrow T=\dfrac{500}{3} \\ \Rightarrow 2\left( \dfrac{500}{3} \right)={{T}_{1}} \\ \Rightarrow {{T}_{1}}=\dfrac{1000}{3}N \\ $ For case $\left( 2 \right)$ the force equations from FBD are, $ \\ T{{'}_{1}}=2T'+{{T}_{2}} \\ \Rightarrow {{T}_{2}}=2T' \\ \Rightarrow 3T'=500\\ \Rightarrow T'=\dfrac{500}{3}N \\ $ From the first two equations $ \\ \text{T}{{\text{ }\!\!'\!\!\text{ }}_{\text{1}}}=2T'+2T' \\ \text{T}{{\text{ }\!\!'\!\!\text{ }}_{\text{1}}}=4T' \\  \text{T}{{\text{ }\!\!'\!\!\text{ }}_{\text{1}}}=4\left( \dfrac{500}{3} \right) \\ \text{T}{{\text{ }\!\!'\!\!\text{ }}_{\text{1}}}=\dfrac{2000}{3}N \\ $ Thus, the case $\left( 2 \right)$ has ceiling experiencing more force as ${{T}_{1}} > T{{'}_{1}}.$

Is there any possible case in the universe where there is no mechanical advantage? 

For no mechanical advantage, $M.A.=1,$ or ${{F}_{input}}={{F}_{output}}$. 

 Condition of no mechanical advantage is where A=1, or input force is equal to output for the machine. These conditions are possible in the universe.

Two boys use a machine to lift $H$ an object of mass $80\,kg$ to a height of $20\,m$ in $4$ seconds. If the initial velocity of the object was $0\text{ m/s}$, what is the mechanical advantage of the machine used? The final velocity of the object 

Calculate the force applied by the boys by finding out the acceleration of the mass through the second equation of motion, $s=ut+\dfrac{1}{2}a{{t}^{2}}$. Also mechanical advantage $=\dfrac{{{F}_{output}}}{{{F}_{input}}}=\dfrac{\text{Load}}{{{F}_{input}}}$.  

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5044756f-21f4-49e7-93f6-ec4da5df91552453174940653543766.png"/> $ \\ $ As the object gains a height of $y=20m$ in $t=4\sec$ starting from rest $u=0m/s$ $ \\ $ From second equation of motion, $ \\ s=ut+\dfrac{1}{2}a{{t}^{2}} \\ \Rightarrow y=0\times 4+\dfrac{1}{2}\times a\times {{\left( 4 \right)}^{2}} \\ \Rightarrow 20=\dfrac{1}{2}\times a\times 16 \\ \Rightarrow a=2.5m{{s}^{-2}} \\ $ F.B.D of the body $ \\ F-80g=80a \\ \Rightarrow F=80\left( g+a \right) \\ \Rightarrow F=80\left( 10+2.5 \right)=1000N \\ \text{Thus, M}\text{.A}\text{.=}\dfrac{\text{Load}}{{{F}_{input}}}=\dfrac{80g}{1000}=\dfrac{80\times 10}{1000} \\ \therefore A=0.8$