A couple of question regarding tension and Newton's laws

AI Thread Summary
The discussion revolves around two scenarios involving tension in strings and Newton's laws. In the first scenario, the tension varies based on the speed of pulling the strings, with a quick pull causing the lower string to tear and a slow pull leading to the upper string tearing, explained through impulse and momentum principles. The second scenario involves calculating the tension in an upper string when a 1 kg mass supports a 2 kg mass in an elevator, with participants debating the correct approach to find the tension, which is suggested to be 14.7 N. The complexity arises from the elevator's acceleration, affecting the tension calculations. Understanding the differences in tension forces during different pulling speeds and the implications of acceleration in an elevator is crucial for solving these problems.
guardians
Messages
7
Reaction score
0
1. A led cylinder hangs on a string. By another string, another cylinder is hung on the previous cylinder. If you pull fast enough, the lower string tears, if you pull slowly, the upper one tears. How does one explain this?

I tried to explain this with the help of Newton's second law, but I seem to be having some problems with the tension forces. Could someone offer an explanation?

2. On an elevator ceiling hangs an object with mass m1=1kg. On that object, another one with mass m2=2kg is hung. What is the tension force in the upper string, if the tension force in the string between the two objects is 9,8 N?

Again, the tension forces seem to be giving me some trouble. I think that the tension in the string between the two objects should be 2*g, which it isn't in the problem. Could someone help?
 
Physics news on Phys.org
guardians said:
I tried to explain this with the help of Newton's second law, but I seem to be having some problems with the tension forces.

Hi guardians! :smile:

You know you need to show us more work than that.
2. On an elevator ceiling hangs an object with mass m1=1kg. On that object, another one with mass m2=2kg is hung. What is the tension force in the upper string, if the tension force in the string between the two objects is 9,8 N?

It's an elevator … maybe it's accelerating! :wink:
 
No problem :)

1. So, if the same force is applied, but with different time of application - I'm starting to go into impulse, and change of momentum. If I assume that the impulse is the only thing that is influenced - I don't get how that would produce the desired effect... But since this question is in the lesson about Newton's second law, and Newton's second law is basically the same as the impulse-momentum law.

2. Yes, since it is an elevator, I tried to assume that it is moving, but I got a result that's different from the one suggested in my book (14,7 N). What I did is basically equate F=m2g downwards and F(elastic)=9,8 upwards to m*a, where a is the acceleration of the lift, and thereby the whole ball system. Than I use that acceleration to get the upper tension, which again is a couple of easy calculations. If someone would be willing to check that the result is indeed 14,7 N, I can write out the whole solution (in case I am doing something wrong inbetween).
 
guardians said:
2. Yes, since it is an elevator, I tried to assume that it is moving, but I got a result that's different from the one suggested in my book (14,7 N). What I did is basically equate F=m2g downwards and F(elastic)=9,8 upwards to m*a, where a is the acceleration of the lift, and thereby the whole ball system. Than I use that acceleration to get the upper tension, which again is a couple of easy calculations. If someone would be willing to check that the result is indeed 14,7 N, I can write out the whole solution (in case I am doing something wrong inbetween).

You don't need to calculate a … just leave it as a.

Then Ttop/Tbottom = … ? :smile:
 
guardians said:
1. So, if the same force is applied, but with different time of application - I'm starting to go into impulse, and change of momentum. If I assume that the impulse is the only thing that is influenced - I don't get how that would produce the desired effect... But since this question is in the lesson about Newton's second law, and Newton's second law is basically the same as the impulse-momentum law.
This 'upper string-lower string' problem 1 gets tricky, and since I'd have trouble explaining it, here's a couple of hints to get you moving in the right direction. When the string is pulled abruptly, Newton 2 (or, as you correctly note, the impulse-momentum law) applies, as long as the strings are extensible. But when the string is pulled slowly, Newton 1 in effect, because the movement is at constant speed, or at any rate, with minimal acceleration. Which string experiences the greater tension force for this equilibrium condition?

Note that for the 'quick pull' part, if the strings were inextensible (rigid), you wouldn't get any acceleration since there would be no movement, and you're back to Newton 1 for the rigid string case, and the upper string would always break first regardless of whether the pull was slow or fast. That's why you need to consider the deformation of the strings to get movement and acceleration in the 'quick pull' case under the impulsive force.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top