Newton's Law and Frictional Forces

Click For Summary
SUMMARY

This discussion focuses on two physics problems involving Newton's laws and frictional forces. The first problem examines the forces acting on two magnets, identifying the magnetic pull, gravitational forces, and normal forces involved. The second problem analyzes two blocks connected by a massless string over a frictionless pulley, requiring a ranking of the forces acting on each block. Key insights include the importance of recognizing action/reaction pairs as per Newton's third law and understanding the relationship between mass and gravitational force in dynamic systems.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with magnetic forces and fields
  • Knowledge of tension in strings and pulleys
  • Basic concepts of gravitational force and mass
NEXT STEPS
  • Study Newton's third law and its applications in various physical scenarios
  • Explore magnetic force calculations and their implications in physics
  • Learn about Atwood's machine and its relevance to tension and mass
  • Investigate the dynamics of systems involving pulleys and frictionless surfaces
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding forces in dynamic systems.

victorc
Messages
2
Reaction score
0
These are two word problems, so I don't think the template applies to this thread.

1. Two strong magnets are on opposite sides of a tabletop's surface so that the attraction between the magnets is all that is holding up the one underneath. Identify all forces which involve either magnet (either as the agent or the object being acted upon), and all associated action/ reaction pairs completely. Number each pair.

My attempt at this question:
The forces on the magnet above the table include the magnetic pull by the other magnet (pointing down) [as well as the force of gravity (pointing down) and the normal force exerted by the table (upwards)]. The forces on the magnet below the table include the force caused by the magnetic field (upwards) [as well as the force of gravity (downwards)]. The magnetic field created by the two magnets generates a long-range magnetic force on each of the magnets causing the attraction between them.
There are no action/reaction pairs because there is no acceleration involved with either of the objects.

2. Two blocks, A and B, are hanging from opposite ends of a massless string which runs over a massless frictionless pulley. The mass of block A is greater than that of block B. Rank, from largest to smallest, all of the forces acting on either block. (Don't bother to include in this ranking the forces acting on the string.) Put your answer in the usual form: X > Y = Z. Explain why you ranked each force in order.

My attempt at this question:
Forces acting on block B (from largest to smallest):
1. Upwards force caused by tension of the string.
2. Downwards force of gravity (weight).

Forces acting on block A (from largest to smallest):
1. Downwards force of gravity (weight).
2. Upwards force caused by tension of the string.

ps:. I don't understand what the question means with "Put your answer in the usual form: X > Y = Z."

Thanks in advance
 
Physics news on Phys.org
victorc said:
These are two word problems, so I don't think the template applies to this thread.

1. Two strong magnets are on opposite sides of a tabletop's surface so that the attraction between the magnets is all that is holding up the one underneath. Identify all forces which involve either magnet (either as the agent or the object being acted upon), and all associated action/ reaction pairs completely. Number each pair.

My attempt at this question:
The forces on the magnet above the table include the magnetic pull by the other magnet (pointing down) [as well as the force of gravity (pointing down) and the normal force exerted by the table (upwards)].
Very Good.
victorc said:
The forces on the magnet below the table include the force caused by the magnetic field (upwards) [as well as the force of gravity (downwards)].
Good.
victorc said:
The magnetic field created by the two magnets generates a long-range magnetic force on each of the magnets causing the attraction between them.
True, but I'm not sure how this relates to the problem. (Maybe I'm missing something.)
victorc said:
There are no action/reaction pairs because there is no acceleration involved with either of the objects.
Not true. Don't forget Newton's third law (For every force [action] there is an equal and opposite force [reaction]). You've listed three (action) forces in the first part and two in the second.
victorc said:
2. Two blocks, A and B, are hanging from opposite ends of a massless string which runs over a massless frictionless pulley. The mass of block A is greater than that of block B. Rank, from largest to smallest, all of the forces acting on either block. (Don't bother to include in this ranking the forces acting on the string.) Put your answer in the usual form: X > Y = Z. Explain why you ranked each force in order.

My attempt at this question:
Forces acting on block B (from largest to smallest):
1. Upwards force caused by tension of the string.
2. Downwards force of gravity (weight).
True. However, you didn't say why. Hint: m_{A} > m_{B} \wedge a_{A} = a_{B}
victorc said:
Forces acting on block A (from largest to smallest):
1. Downwards force of gravity (weight).
2. Upwards force caused by tension of the string.
True. Some note as before.
victorc said:
ps:. I don't understand what the question means with "Put your answer in the usual form: X > Y = Z."
I think they are asking for something like

F_{A} > F_{B} \wedge \frac{1}{2 T}=\frac{1}{F_{A}} + \frac{1}{F_{B}}

You might want to look at the end of
http://farside.ph.utexas.edu/teaching/301/lectures/node48.html"
where there is a comment about "Atwood's machine".
Also look at
http://www.pha.jhu.edu/~broholm/l8/node3.html"
victorc said:
Thanks in advance
No problem.

73s and clear skies.
 
Last edited by a moderator:
First of all, thank you so much for the reply!

kg4pae said:
Not true. Don't forget Newton's third law (For every force [action] there is an equal and opposite force [reaction]). You've listed three (action) forces in the first part and two in the second.

I identified the magnet above the table as Magnet A, and the one below the table as Magnet B. This is what I added to my response:

"The action/reaction pairs are listed below:
1. Magnetic pull by Magnet A on Magnet B AND magnetic pull by Magnet B on Magnet A.
2/3. Force of gravity on Magnet A/B AND Magnet A/B attracting the Earth.
4. Normal force (normal from table's surface) acting on Magnet A AND gravitational pull on Magnet A." (I'm not sure about this, as action/reaction forces are not on the same body. However, I felt a need to include the normal to Magnet A and gravity was the only force that I reasoned to oppose the normal)

Is there a normal acting on Magnet B?

kg4pae said:
True. However, you didn't say why. Hint: m_{A} > m_{B} \wedge a_{A} = a_{B}

I don't think I understand you hint... perhaps because I don't know what the wedge sign means exactly. Well, this is my revised answer:

"Forces acting on block B (from largest to smallest):
1. Upwards force caused by tension of the string.
2. Downwards force of gravity (weight).
The downwards force of gravity must be smaller than the upwards force of the string because block B moves upwards (it has less mass and F_gravity = mass x a_gravity)

Forces acting on block A (from largest to smallest):
1. Downwards force of gravity (weight).
2. Upwards force caused by tension of the string.
The downwards force of gravity must be greater than the upwards force of the string because block A falls (since it has more mass and F = ma)"

kg4pae said:
I think they are asking for something like

F_{A} > F_{B} \wedge \frac{1}{2 T}=\frac{1}{F_{A}} + \frac{1}{F_{B}}

Oh... I still don't get what the wedge (^) sign means. Is it related to wedge products in exterior algebra?
I first thought the so called "usual form" to be an inequality, X and Y being either gravity or the tension of the string, depending on the block. However, following this reasoning, I don't know what Z would be.

kg4pae said:
You might want to look at the end of
http://farside.ph.utexas.edu/teaching/301/lectures/node48.html"
where there is a comment about "Atwood's machine".
Also look at
http://www.pha.jhu.edu/~broholm/l8/node3.html"

The website was useful to get a broader understanding of the situation, as well as an interesting bit of historical information.

I truly appreciate your time and effort
 
Last edited by a moderator:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
16
Views
3K
Replies
7
Views
1K
Replies
21
Views
890
  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
35
Views
4K