Tension On Rope, newton's laws

50 kg bucket lifted by rope guaranteed not to break if tension is 500N or less. From rest, it is raised 3m at 3 m/s. Acceleration is constant. Will rope break?

I get all the set up and calculate using Fnet = Ft - Fg where Ft is + upward tension on rope and Fg is - downward tension of bucket. Answer key states that Fnet + Fg = Ft so 75 N + 490 N = 565 N and rope will break

When I calculate using -9.8 m/s2 for "g" I am subtracting 75 and 490! Why? The book has told us previously to use negative 9.8 m/s2 for downward motion.

I'm wondering if you have written the question properly?
What is the acceleration? you say "it is raised 3m at 3 m/s. Acceleration is constant."
So that is a distance at a speed. So there is no expression of acceleration. Except that at one moment it had to instantly go from 0 to 3m/sec.
But ropes stretch through these moments. To much stretching will cause the rope to break. If it starts lifting the acceleration will definitely not have remained constant!

So the height it moves through has nothing to do with the calculation. Are you using the 3 meter height in your calculations? If so what formula did it go into?

Are these the sort of things you are considering?

I used Fnet = Ftension - Fgravity
rearranged formula to be Ft = Fnet + Fg then substituted ma so...

Ft = ma + mg I used the recommended formula for a = v2/2d to sub in for "a"

Ft = 50 kg x (3 m/s squared / 2(3m) for Fnet and getting 75 N

then, 50 kg x -9.8 m/s2 for Fg and getting -490 N

This is from the Glencoe Physics Textbook I am using and it is a sample problem, so they worked it all out as I have done, but I just don't get why they use a positive g when it acts opposite of the upward force on the bucket. They do not use -9.8 m/s2 (it is positive). I am now wondering if the Ftension on rope is gotten by adding the 75 N and taking the -490 N as gravitational force but using the Fnormal as the positive number (opposite force on rope?)

regarding the acceleration, I paraphrased the question. It is, " The bucket started at rest, and after being lifted 3 m it is moving at 3 m/s. Assuming acceleration is constant, is the rope in danger of breaking?"


My understanding of forces on ropes is that if I pull with 30 N to left & an opposite force of 30 N pulls to the right the net force is still 30 N net force right? the net force is not 60N

thanks for your help and Happy New Year

"formula for a = v2/2d to sub in for "a"" Explain?

it is right from the text. it is a derived formula for acceleration. final velocity squared divided by 2 x displacement. i am ok with all the text book stuff. my issue is that the force upward of the bucket being lifted is 75 N and the downward force of gravity is 490 N (and should be negative, right? b/c it acts in the opposite direction of the upward force.

so, do i add the 75 N and the 490 N to get 565N which would break the rope (like you would add vectors that go in opposite directions) OR it is like when you have 30 N pulling left on a rope and 30 N pulling right on a rope being a net force of 30 N like on a spring scale?