A rope of linear mass density 4kg/m is pulled with a force of 20N . Find tension at the mid point of rope ?
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Answer:If the mass of the rope is M
Then, the acceleration is
a=\dfrac{F}{M}a=
M
F
Now, consider a portion of mass m of the rope from is left end to x distance from the right end
If the linear mass density of the rope is assumed constant
Then,
\lambda =\dfrac{M}{L}=\dfrac{m}{L-x} λ=
L
M
=
L−x
m
m=M\left( 1-\dfrac{x}{L} \right) m=M(1−
L
x
)
Now, considering this portion of rope
T=ma T=ma
T=M\left( 1-\dfrac{x}{L} \right)\dfrac{F}{M} T=M(1−
L
x
)
M
F
T=F\left( 1-\dfrac{x}{L} \right) T=F(1−
L
x
)
Hence, the tension is F\left( 1-\dfrac{x}{L} \right)F(1−
L
x
)
Explanation:
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