IIT JEE , AIEEE Forum - Physics , Mechanics

m.mangalam

Flexible chain of W hangs between A and B same level

Find: 1)Vector force at each end point

2)Tension at lowest point

sameer_sharma_1989

i think (ii) has the answer

T = W/2sin@

abhishekaich

Assume symmetric reaction forces on both ends. So drawing the FBD of the complete chain. By vertical equilibrium we will have the vertical component of Reaction force at each end = W/2.

Tension (T) acts along the cable hence angle of the cable at the end would be the direction of the net reaction (R) which would be opposite to T. Now vertical component of R is W/2. Hence the horizontal component should be W/2 tan

.

Considering the FBD of half the chain. By horizontal equilibrium we would have T= W/2tan

Also net reaction at the end R would be = W/2 sin

phyana

u have given the angle as theta.

then according to symmetry the force on both a,b will be same. this force is the force of tension produce on the sting due to hanging.
shape of chain is exact "V".
i think small scale integration is needed.
assume length of chain is 2L.
mass per unit length is w/2L
mass of dl= (w/2L)dl

integrate in the limits 0 to L the expression (w/2L)g*sin (theta) *dl.
this is the tension on bothe a,b.
now integrate (2^.5)(w/2L)g*cos(theta).sin(theta).dl in the limits 0 to L

do hese calculations.u will get the answer.

if useful rate me

phyana

u have given the angle as theta.

then according to symmetry the force on both a,b will be same. this force is the force of tension produce on the sting due to hanging.
shape of chain is exact "V".
i think small scale integration is needed.
assume length of chain is 2L.
mass per unit length is w/2L
mass of dl= (w/2L)dl

integrate in the limits 0 to L the expression (w/2L)g*sin (theta) *dl.
this is the tension on bothe a,b.

to get the answer of the second question...........
now integrate (2^.5)(w/2L)g*cos(theta).sin(theta).dl in the limits 0 to L

do hese calculations.u will get the answer.

if useful rate me

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