# Applications of Newton's Laws

1. Nov 9, 2005

### reliquator

I'm on the 3rd unit of my physics unit and we're applying Newton's 3 Laws of motion.

There's this problem:

"A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s^2 using very light fishing line that has a "test" value of 100 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?"

Okay, I thought that using the F = MA formula would work, with 100=4.5 * M, which means that the mass of the fish was greater than 22 and 2/9 KG.

WRONG. The answer guide I'm working from says the mass of the fish > 7.0 kg. Did I use the wrong formula?

2. Nov 9, 2005

### Staff: Mentor

Of course F = ma works. But you've got to consider all the forces acting on the fish, not just the tension in the line. (You forgot gravity.)

3. Nov 9, 2005

### reliquator

Okay, I've got it. 100 = (9.8 + 4.5) * M

M = 6.99 Kg, so the fish has to be heavier than 6.99 Kg. But what I don't get is, why would you add 4.5 (acceleration of the line) with accel due to gravity? Thanks to all in advance.

Edit: Question #2

"A 0.145-kg baseball traveling 30.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove"

Okay, I don't need an answer to this problem but would like to know how to approach it. I'm inferring that you need to find the FORCE applied to the baseball. Its given a mass but no acceleration. Would acceleration be 0, or 9.8?

Edit2: Okay, so I think I figured it out. I have the displacement, vi, vf, and need a. so vf^2 = vi^2 + 2ad, and I get a = -4090.91 m/s^2.

Aight, so I do F= MA, and get force = 593.18 N, am I correct, and if so, should the 593.18 N be negative? Thanks again all.

Last edited: Nov 9, 2005
4. Nov 10, 2005

### Staff: Mentor

To apply Newton's 2nd law, you must use the net force on the fish. The net force is Tension (up) - mg (the weight, down) = T - mg. Set that equal to ma: T - mg = ma.
Looks OK to me. I'm sure the magnitude of the force is all they are looking for. (The sign is arbitrary--it depends on your coordinate system.)