arindameducationusc
  • arindameducationusc
Lets Discuss about Tension... (Tutorial)
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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arindameducationusc
  • arindameducationusc
\(\Huge\color{red}{Tension} \) When a string, thread or a rod is stretched, a force comes into play known as tension. It is \(\Large\color{blue}{pulling} \) in nature.
arindameducationusc
  • arindameducationusc
Consider a block connected with a string such that other end of string is connected with a fixed platform. |dw:1440745098195:dw|
arindameducationusc
  • arindameducationusc
Example=>|dw:1440745678272:dw|

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arindameducationusc
  • arindameducationusc
now, for m1|dw:1440745834828:dw|
arindameducationusc
  • arindameducationusc
now for m2, |dw:1440745871420:dw|
arindameducationusc
  • arindameducationusc
now for pulley, |dw:1440745915192:dw|
arindameducationusc
  • arindameducationusc
now for fixed pulley, |dw:1440745990880:dw|
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{At~wood~machine} \) |dw:1440755955813:dw|
arindameducationusc
  • arindameducationusc
In the figure shown a mass less and an in-extensible chord passes over a mass less and smooth pulley suspended from a fixed platform. The two ends of chord carries two blocks m1 and m2. If we assume m1>m2, then block m1 moves downward and m2 upward with same acceleration a. Consider a portion PQ on the right side of pulley. Let Tp and Tq be the tension ends of portion PQ |dw:1440757592023:dw|
arindameducationusc
  • arindameducationusc
The portion PQ has same acceleration in downward direction. Therefore Tq-Tp=(mass of PQ)*a Tq-Tp=0*a Tp=Tq The tension at every point of chord on the right side of pulley is samee.
arindameducationusc
  • arindameducationusc
now for acceleration, For that let T be the commonn tension at every point of string.
arindameducationusc
  • arindameducationusc
FBD( free body diagram) of m1 |dw:1440758339625:dw|
arindameducationusc
  • arindameducationusc
Equation of motion of m1=> m1g-T=m1a (Eqn1)
arindameducationusc
  • arindameducationusc
FBDof m2|dw:1440758643802:dw| T-m2g=m2a (eqn 2)
arindameducationusc
  • arindameducationusc
Adding equation 1 and 2 (Use mathematics :D) we get a=(m1-m2)g/m1+m2 here m1>m2 //if you are not able to solve let me know, I will solve it.
arindameducationusc
  • arindameducationusc
\[a =\frac{ (m _{1}-m_{2} )g}{ m _{1}+m _{2} }\] here m1>m2 Putting the value of a in equation 1 \[m _{1}g-T=m _{1}\frac{ (m _{1} -m _{2})g}{ m _{1}+m _{2} }\]
arindameducationusc
  • arindameducationusc
we get \[T=\frac{ 2m _{1} m _{2}}{ m _{1} +m _{2}} g\]
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Some~formula~for~certain~cases} \) Well please \(\Large\color{blue}{try~deriving} \) I will just be giving you direct formulas. If you have any problem please contact me, I will definitely help!
arindameducationusc
  • arindameducationusc
Case1) is simple (which I did above) |dw:1440764462862:dw| \[a=\frac{ m _{1} g-m _{2}g}{ m _{1}+m _{2}}\] Always remember \[a=\frac{ Net~pulling~force }{ Total~mass~of~system}\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{Case ~2} \) |dw:1440764771203:dw| \[a=\frac{ m _{1}g-m _{2} g}{ m _{2} }\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{Case~3} \) |dw:1440764887925:dw| \[a=\frac{ m _{2} g-m _{1}g}{ m _{1} +m _{2}}\]