How is the acceleration proportional to the removed force?

Click For Summary

Discussion Overview

The discussion revolves around the relationship between acceleration and force removal in a physics context, particularly focusing on how the acceleration of a mass changes when a force is removed. Participants explore the implications of force direction and vector notation, as well as related examples involving forces exerted by objects in motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant questions the assertion that the acceleration of a mass equals the acceleration of the removed force, suggesting that the acceleration should be the sum of the remaining forces.
  • Another participant proposes that if the initial acceleration is zero and a force F is removed, the resulting acceleration is -F/m, indicating a direction opposite to the removed force.
  • Some participants emphasize the importance of including direction in the acceleration, noting that the correct expression should account for vector notation.
  • There is a suggestion that a previous answer may have contained a typo, but this is contested by another participant who insists that direction vectors are not being considered properly.
  • Further examples are introduced, with participants seeking clarification on how forces are calculated in different scenarios, such as a link being thrown against a wall or resting on a table.
  • Participants discuss the difficulty level of the physics questions in the context of different textbooks, comparing them to the Kleppner and Kolenkow classical mechanics book.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between acceleration and the removed force, with no consensus reached on the correct interpretation. Additionally, there are varying opinions on the difficulty of the physics questions being discussed.

Contextual Notes

Some participants note the potential for confusion due to missing direction vectors in the discussion, and there are references to unresolved questions regarding the calculation of forces in specific examples.

Who May Find This Useful

This discussion may be useful for students studying introductory physics concepts, particularly those grappling with the relationship between force and acceleration, as well as those comparing educational resources in physics.

Mr.Husky
Gold Member
Messages
89
Reaction score
28
16300816139694449597451261552684.jpg

Image above is the question. Below image depicts solution.
16300816976776985517156610342954.jpg

if F1 is removed then the acceleration of that mass must be sum of accelerations of remaining forces. Right??
But answer says that acceleration of that mass is equal to acceleration of F1. I don't understand it. Can someone explain it??
 
Last edited by a moderator:
Physics news on Phys.org
Ganesh Mammu said:
But answer says that acceleration of that mass is equal to acceleration of F1. I don't understand it. Can someone explain it??
If initial acceleration is 0 and you remove a force F, then the acceleration is -F/m (opposite to the removed force).
 
  • Like
Likes   Reactions: Mr.Husky
The sum of the remaining forces must be -F1, otherwise the mass would have been accelerating before F1 was removed.
If you include the direction in the acceleration, it is A=-F/m.
 
  • Like
Likes   Reactions: Mr.Husky
.Scott said:
The sum of the remaining forces must be -F1, otherwise the mass would have been accelerating before F1 was removed.
I don't like their answer. In should be a=-F1/m.
Oh I got it. It is just a typo. Right??
 
A.T. said:
If initial acceleration is 0 and you remove a force F, then the acceleration is -F/m (opposite to the removed force).
Thank you for your reply A.T! Now I understood the situation
 
Ganesh Mammu said:
Oh I got it. It is just a typo. Right??
No. They are not including the direction vector. Note the F1 with and without the direction arrow on top.
 
  • Like
Likes   Reactions: Mr.Husky
.Scott said:
No. They are not including the direction vector. Note the F1 with and without the direction arrow on top.
I forget about the vector notation. Sorry for it.
By the way I also have another example can you explain it??
What I don't understand in the following example is how did they got to know that the total force exerted must be equal to the weight of y part resting on table+ force due to the momentum imparted
16300828031701458655848058732275.jpg
 
By the way the text I am using is prepared by my school and is it worth going through these types of questions (there are about 300 of them per chapter) ??
Can you compare the difficulty of questions of this book to kleppner kolenkow classical mechanics book?
 
Ganesh Mammu said:
What I don't understand in the following example is how did they got to know that the total force exerted must be equal to the weight of y part resting on table+ force due to the momentum imparted
If I throw a link against the wall, it will exert a momentary force on the wall.
If I throw a link onto a table, it will exert a momentary force and then come to rest on the table.
A link resting on the table exerts a downward force on the table.

Regarding the Physics books: It's been decades since I took a Physics course.
 
  • Like
Likes   Reactions: Mr.Husky
  • #10
.Scott said:
If I throw a link against the wall, it will exert a momentary force on the wall.
If I throw a link onto a table, it will exert a momentary force and then come to rest on the table.
A link resting on the table exerts a downward force on the table.

Regarding the Physics books: It's been decades since I took a Physics course.
Now I am slowly understanding the answer.
Can you rate difficulty of the questions generally??
 
  • #11
It would be on par for introductory Physics - either High School or College PH101.
 
  • #12
Can anybody compare the difficulty of the question to that of kleppner and kolenkow. Because I am not sure whether to go on with this book or should I change to kk which explains theory brilliantly.
 
  • #13
.Scott said:
If I throw a link against the wall, it will exert a momentary force on the wall.
If I throw a link onto a table, it will exert a momentary force and then come to rest on the table.
A link resting on the table exerts a downward force on the table.

Regarding the Physics books: It's been decades since I took a Physics course.
Scott can you explain how they got the momentum in the first equation. And also why did they considered momentum rather than downward force??
 
  • #14
Ganesh Mammu said:
Scott can you explain how they got the momentum in the first equation. And also why did they considered momentum rather than downward force??
Thanks Scott for helping me!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 51 ·
2
Replies
51
Views
38K