Given tension and acceleration determine max weight

AI Thread Summary
The discussion revolves around calculating the maximum weight of a fish that can be pulled upward using a fishing line with a maximum tension of 45 N while accelerating the fish at 1.8 m/s². Participants clarify that the force of gravity acting on the fish must be considered alongside the tension in the line. The correct approach involves setting up the equation mg = -m(1.8 N/kg) + 45 N to solve for the mass of the fish. After manipulating the equation, the maximum mass is determined to be approximately 3.88 kg, leading to a final weight of 38 Newtons. The conversation highlights the importance of understanding the relationship between mass, weight, and acceleration in this context.
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Homework Statement


A fisherman in a boat is using a "10- test" fishing line. This means that the line can exert a force of 45 without breaking (1lb = 4.45N). If he accelerates the fish upward at 1.8m/s/s , what maximum weight fish can he land, in Newtons?


Homework Equations



F=ma

The Attempt at a Solution



Failed Answers : 25, 6

FBD demostrates a line from a fishes mouth. Gravity points down, Force of pull goes up, acceleration is upwards at 1.8m/s/s. I've tried setting the sum of all the forces in the y direction to 45N and then solving for m. That only gave me 25kg, don't know what I was thinking there. Then I tried using F=ma countless times to no avail.
 
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jsewell said:
FBD demostrates a line from a fishes mouth. Gravity points down, Force of pull goes up, acceleration is upwards at 1.8m/s/s.
Good. Express that mathematically.
I've tried setting the sum of all the forces in the y direction to 45N and then solving for m.
45 N is the maximum force of pull (tension in the line), not the net force.
 
Soo...(Force on line - Force of gravity) = mass(1.8m/s/s)

where the we set the Force on the line to 45 Newtons. The heaviest fish you can pull out of the water at that rate will still be acomplished at 45 Newtons of Tension. The question goes on about how many Newtons gravity will be playing at, the weight of the fish. So...

Force of gravity = -[mass(1.8 m/s/s) - force on line]

mg = -ma + 45

mg = -m(1.8m/s/s) + 45

My problem with this is that I have two unknowns (Force of Gravity and Mass) so this equation has me stuck...
 
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jsewell said:
My problem with this is that I have two unknowns (Force of Gravity and Mass) so this equation has me stuck...
They are related. Given a mass, what's the weight? Express everything in terms of mass.
 
well weight is equal to mg. An objects mass in kg and the factor of gravity (9.8 N/kg).

I'll assume my equation was sound and all your asking is to put in terms of mass. This should yeild me...

mg = -m(1.8N/kg) + 45

Idk what you mean friend. I got to thinking and figured that I can determine the objects mass when the line is fully loaded and not accelerating. That figure, 4.6kg would is only useful if the object is stationary or already at a constant velocity. I am lost here
 
jsewell said:
well weight is equal to mg. An objects mass in kg and the factor of gravity (9.8 N/kg).

I'll assume my equation was sound and all your asking is to put in terms of mass. This should yeild me...

mg = -m(1.8N/kg) + 45
Excellent. Now just solve for the mass.

Idk what you mean friend. I got to thinking and figured that I can determine the objects mass when the line is fully loaded and not accelerating. That figure, 4.6kg would is only useful if the object is stationary or already at a constant velocity. I am lost here
Just solve the equation above for the mass. (And then find the weight of the mass, since that's what the question asks for.) That equation will tell you the most mass of fish you can accelerate, since 45 N is the most force you can exert on the line without breaking it.
 
so after some algebraic manipulation I can solve for m

m = 45N
---------
g + 1.8N/kg

setting "g" to 9.8N/kg yeilds

m = 3.88kg

So the weight is 3.88kg x 9.8N/kg

38 Newtons!
 
Answer was correct. Sometimes all you need is a person to tell you your on the right track.

Thanks Doc Al
 
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