A chain of mass M and length L is hold vertically on
Smooth horizontal table as shown. If chain is released from
rest then find the normal reaction as a function of
x. x is the distance which the chain
has fallen
& assume no
rebounding of chain after striking the surface
Answers
Answer:
Lets take a small element at a distance x of length dx.
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdp
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=Lmgx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgx
Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgxExplanation:
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