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A ship of total mass m is anchored in the middle of a river and water is flowing with a constant velocity v. the horizontal component of the force exerted on the ship by the anchor chain is T. If the anchor chain suddenly breaks, determine the time required for the ship to attain a velocity equal to 0.5v ; assuming that the frictional resistance of the water is proportional to the velocity of the ship relative to water. please explain the entire procedure.
Asked by vidyavikram10 | 24 Jul, 2019, 04:52: PM
answered-by-expert Expert Answer
This question was answered earlier (URL 424363).  some explanation is added to make it easily understandable
 
When the ship is anchored in middle of the river, ship is tied using a strong chain to the anchor and the
anchor is fixed tightly in the ground below river water. This makes the ship not to move.
Since river water pushes the ship along the flow direction with force, equal Tension force is developed in the chain.
This tension florce balances the pushing force provided by the water flow of river, hence ship does not move.
 
When the chain breaks, Tension disappears, hence the ship is moved by flowing water
 
If the chain breaks, force F acting on the ship when it is moving with some instantaneous speed w is given by,
 
F = m (dw/dt)  = T - [ k×(w-v) ]  .....................(1)
 
Above eqn.(1) is written using newtons second law of motion, i.e. force = mass×acceleration.
As already explained , pushing force the water flow of river equals the tension force in the chain.
Hence Tension force T appears right hand side. When the ship is moving in water,
it experiences resistive drag force in water. This drag force is given as k times the velocity of the ship
with respect to river water flow i.e., (w-v), where w is speed of ship and v is speed of river water.
 
Hence net force acting on the ship when it is moving is T - [ k×(w-v)]
 
where k is constant,  v is the speed of flowing water in river and  k(w-v) is the resistive force
due to ship movement in flowing water
 
To get speed of ship as a function of time, we need to integrate the differential equation (1)
 
 we have from eqn.(1) ,  begin mathsize 14px style m space cross times fraction numerator d w over denominator T space plus k v space minus space left parenthesis space k cross times w right parenthesis end fraction space equals space d t end style  ..........................(2)
Eqn.(1) is rewritten as eqn.(2), so that RHS of eqn.(1) which is a function of w  and
differential increment of speed dw is on one side so that integration can be performed.

Similarly, differential increment dt is taken on other side to perform integration in time variable.
 
 
begin mathsize 14px style T o space p e r f o r m space i n t e g r a t i o n comma space l e t space u s space m a k e space t h e space f o l l w i n g space s u b s t i t u t i o n comma space space T plus k space v space minus k space w space equals space x space space................. left parenthesis 3 right parenthesis h e n c e space w e space c a n space w r i t e comma space b y space d i f f e r e n t i a t i n g space b o t h space s i d e s space o f space e q n. left parenthesis 3 right parenthesis comma space d x space equals space minus k space d w space space o r space space space space space d w space equals negative space fraction numerator d x over denominator k end fraction space space........... left parenthesis 4 right parenthesis u sin g space t h e space a b o v e space s u b s t i t u t i o n space g i v e n space b y space e q n. left parenthesis 3 right parenthesis space a n d space e q n. left parenthesis 4 right parenthesis comma space space e q n. left parenthesis 2 right parenthesis space i s space r e w r i t t e n space a s space b e l o w minus m over k space cross times fraction numerator d x over denominator x end fraction space equals space d t space space.............. left parenthesis 5 right parenthesis B e f o r e space i n t e g r a t i n g space e q n. left parenthesis 5 right parenthesis comma space w e space n e e d space t o space w o r k o u t space t h e space l i m i t s space o f space i n t e g r a t i o n L e t space tau space b e space t h e space t i m e space t a k e n space b y space t h e space s h i p comma space s t a r t i n g space f r o m space r e s t space t o space a t t a i n space t h e space s p e e d space 0.5 v h e n c e space f r o m space e q n. left parenthesis 3 right parenthesis comma space space w h e n space w space equals space 0 comma space space space x space equals T space plus space k space v a l s o space f r o m space e q n. left parenthesis 3 right parenthesis comma space space w h e n space w space equals space 0.5 v space comma space x space equals space T space plus space 0.5 space k space v N o w space w e space i n t e g r a t e space e q n. left parenthesis 5 right parenthesis space comma space space space open parentheses negative m over k close parentheses integral subscript T plus k v end subscript superscript T plus 0.5 k v end superscript fraction numerator d x over denominator x end fraction space space equals space integral subscript 0 superscript tau d t minus m over k cross times open square brackets ln space x close square brackets subscript T plus k v end subscript superscript T plus 0.5 k v end superscript space equals space tau m over k cross times ln open parentheses fraction numerator T plus k space v over denominator T space plus space 0.5 space k space v end fraction close parentheses space equals space tau space space end style
Answered by Thiyagarajan K | 24 Jul, 2019, 09:32: PM

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