A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support . Then the tension in the rope at the distance x from the rigid support
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Answered by
162
hi thank you for asking the question
if the chain is held vertically downward two forces will be acting along the surface of chain
one of them is the mass of the chain that is mg acting vertically downward and other is the tension of chain acting vertically upward and canceling the effect of mass pulling it downward
since acceleration is zero so
F(net)=0
so mass of chain is m=(L-x)
T=mg
m=(L-x)/L*m
T=(L-x)/LM*g
hope this answer helps you
if the chain is held vertically downward two forces will be acting along the surface of chain
one of them is the mass of the chain that is mg acting vertically downward and other is the tension of chain acting vertically upward and canceling the effect of mass pulling it downward
since acceleration is zero so
F(net)=0
so mass of chain is m=(L-x)
T=mg
m=(L-x)/L*m
T=(L-x)/LM*g
hope this answer helps you
Answered by
22
Mg + mg (l-x) / A
A is area of cross section of bar
As stress = force / area
Here f = Mg + mg (l-x )
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