# Rope tension question

1. Oct 24, 2007

### Grapz

1. The problem statement, all variables and given/known data
Jimmy has caught two fish in Yellow Creek. He has tied the line holding the 3.0 kg steelhead trout to the tail of the 1.5 kg carp. To show the fish to a friend, he lifts upward on the carp with a force of 60 N. What is the force of the trout on the carp?

2. Relevant equations
F_a on b = - F_b on a
Fnet = ma

3. The attempt at a solution

So i draw the free body diagrams. I draw one for the carp, and i have The applied force of 60 N upwards, and force of gravity and F_carp on trout both are downwards.

I do Fnet = 60 / 4.5 = a = 13.333

then i do Fnet for the trout = 60 - mg - F_carp on trout = 3(13.333)

I get that F_carp on trout is = -9.39

BUT i test my answer out by drawing the free body diagram for the carp. Which should have two forces on it, the force of the trout on the carp which is equal but opposite of F carp on trout, points upwards, and gravity pointing downards.

So i do Fnet carp = F_trout on carp - mg = 1.5 ( 13.333)

If i sub in the numbers, 9.39 - (1.5)(9.8), i don't end up with (1.5)(13.333)

Can someone tell me what id id wrong, thanks

EDIT: I think i know whats wrong, my fbd is wrong

Last edited: Oct 24, 2007
2. Oct 24, 2007

well, F= (m1 + m2)(g + a) , where F is the force with which the fisherman lifts up on the system and m1 & m2 are the masses of the carp and trout. Thus the acceleration of the whole system is given by a = F/(m1+m2) - g = 60/4.5 - 9.8 = 3.5333333 m/s^2.
The tension in the rope is then the force required for the trout to achieve this acceleration. Using F = ma, we get T = 3 x 3.533333 = 10.6 N , which is equal to the force of the carp on the trout. The force of the trout on the carp is just the opposite of this, -10.6 N.

i had trouble locating an exact error in your formulation, but i hope this helps.

3. Oct 24, 2007

### Grapz

Why is the acceleration of the whole system a = f / (m1 + m2) - g

i don't get the - g part :/

4. Oct 24, 2007

This is just a rearrangement of the formula F = m(g + a) , which is just a standard equation for the force required to accelerate a mass in a direction opposite gravity. In this case m is the combined mass of the two fish.
Step by step : F = m(g+a) ==> F/m = g+a ==> a = F/m - g

5. Oct 24, 2007

### Grapz

I see, can someone show me the free body diagram of the two fish?

6. Oct 24, 2007

i can draw up some quick schematics in paint if you like, but there should be an upward arrow representing the force with which the fisherman pulls upward, a downward arrow representing the weight of the two fish, and two equal and opposite arrows representing the forces the two fish exert on each other. The first arrow represents a force of 60 N; the second, 44.1 N (= weight of the fish); third and fourth = +/- 10.6 N.

7. Oct 24, 2007

### Grapz

ahhhh

thank you i get it noww

8. Oct 25, 2007

### Grapz

I have another question :x. The acceleration of the whole system is 3.5333, that i agree. The calculation for T = 3x 3.5333 is the part i am not too sure about. Indeed the net force is 10.6 N. But i do not think tension is the only force acting on the trout. There has to be gravity acting downwards. So then it must be T - mg = 3x3.53333. So T = 3x.3.533333 + 3x9.8 = 40

so then tension is 40

9. Oct 25, 2007

### Staff: Mentor

You are correct.