Finding tension between two fish hung from a fishing line

[Kristin]

1. Jimmy has caught two fish in Yellow Creek. He has tied the line holding the 3.40 kg steelhead trout to the tail of the 2.34 kg carp. To show the fish to a friend, he lifts upward on the carp with a force of 77.4 N. What is the tension of the ropes connecting the steel trout and carp?

I tried calling just the Fg on the Trout as the tension (assuming that the carp is not hanging). Then I tried subtracting the Force of the trout from the force of the pull. Then I tried subtracting the Fg of both fish from the force of the pull, and nothing seems to work. What am I missing?

 

[Astronuc]

Jimmy is applying a force of 77.4 N to the carp, and the weight (2.34 kg * g) of the carp is resisting that force. To the force, Fc, applied to the carp end of the line is what?

The weight of the trout is pulling in the opposite direction of the force applied at the other end of the line, so the tension in the line has to equal the sum of the two forces.

 

[Kristin]

So does that mean that the sum of force of gravity of the two fish, when added to the applied upward force should be the tension in the rope?

 

[Astronuc]

So does that mean that the sum of force of gravity of the two fish, when added to the applied upward force should be the tension in the rope?

Not quite.


You are correct that the force of gravity of the fish are working in the same direction.


Try this ( <---- g = gravity (-z), Up (+z) --->)


(trout) ------------------------------- (carp) - F_u
<-- F_trout - - - - - - - - - - - - - - - - <-- F_carp, F_u --->


Now consider the static situation where F_u balances the weight of the two fish, thus

F_u = F_trout + F_carp = m_T g + m_C g.

The only force on the line though is due to the weight of the trout, and that weight provides the tension, excluding the mass of the line or wire, thus

T = m_T g = F_u - m_C g

OK, now what happens when F_u > m_T g + m_C g?

Then both mass of trout and carp must accelerate!

The tension on the wire is still due to the opposing force of the trout, which is now given by

T = m_T (g + a), where a is the acceleration of the trout and carp.