# Forces (Tension)

## Homework Statement

Billy has caught two fish. He has tied the line holding the 4.40kg trout to the tail of the 2.36kg carp. To show the fish to a friend, he lifts upward on the carp with a force of 71.3 N. What is the tension of the rope connecting the trout and carp?

Fy=ma,

## The Attempt at a Solution

m=(mCARP+mTROUT), FA=applied force, FT=force of tension

Carp:
Fy=ma
FA-mg-FT=ma
71.3-(2.36+4.40)(9.8)-FT=(2.36+4.40)a
5.052-FT=6.76a
a=(5.052-FT)/6.76

Trout:
Fy=ma
FT-mg=ma
FT-(2.36+4.40)(9.8)=(2.36+4.40)a
FT-66.248=6.76a
a=(FT-66.248)/6.76

a=a, so:

(5.052-FT)/6.76= (FT-66.248)/6.76
5.052-FT=FT-66.248
5.052+66.248=FT+FT
71.3=2FT
FT=35.65N=35.6N

***My online assignment says that this answer is wrong***

## Answers and Replies

Pengwuino
Gold Member
In the third line, you're looking at the forces acting on the Carp, but the gravitational force acting on the carp isnt the carp + trout mass, it's just the carp. You made the same mistake for the trout as well.

Thanks. That was the right answer, but for the FBD of the carp, why isn't the weight equal to the total weight of both fish? Don't both weights affect the tension of the line?

Pengwuino
Gold Member
The tension from the rope on the carp depends solely on the trout below it. The carp could weigh anything and the tension on something it's pulling will still be dependent solely on the mass of what it's holding. Imagine if instead of the trout hanging from the carp, it hung from the ceiling of a building. Obviously, the tension is simply the mass of the trout multiplied by gravity.

Oh ok. Thanks for explaining that. 