Can Forces Acting on an Object Be Zero?

Click For Summary
SUMMARY

The discussion centers on the concept of net forces in Newton's laws of motion, specifically addressing whether forces acting on an object can sum to zero while still being present. Participants clarify that while individual forces, such as gravitational force and normal force, are not zero, their vector sum can be zero, indicating equilibrium. This understanding is crucial for interpreting scenarios where objects remain at rest despite the presence of multiple forces acting upon them.

PREREQUISITES
  • Understanding of Newton's three laws of motion
  • Basic knowledge of forces, including gravitational and normal forces
  • Familiarity with vector addition and equilibrium concepts
  • Ability to analyze free-body diagrams
NEXT STEPS
  • Study Newton's second law in detail, focusing on net force calculations
  • Explore free-body diagram techniques for visualizing forces
  • Learn about static and dynamic equilibrium in physics
  • Investigate real-world applications of forces in equilibrium, such as structures and vehicles
USEFUL FOR

Students of physics, educators teaching Newtonian mechanics, and anyone seeking to deepen their understanding of forces and equilibrium in physical systems.

Nick PG
Messages
5
Reaction score
0

Homework Statement



Just my own personal questions concerning Newtons laws.
For the most part I grasp all three laws to adequately take care of my homework, but I have some conceptual questions that keep bugging me.
when solving diagrams including net forces that are acting on an object, the sum of the forces (say acting in the y direction) are equal to zero. Take for example a block of Mass m at rest on a table, it has a F of gravity and a normal force acting on it, but since the object is at rest it has a force equal to 0.
so, my question is can forces that are acting on an object be equal to zero even though they are acting on the object?

It feels like a dumb question but its been a conceptual question eating at me for a while.
Thank you guys

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
The individual forces that act on the block are not equal to zero. If they were, then they wouldn't be forces. They wouldn't exist. What's important is that they cancel each other out and that the net force is zero. Am I interpreting your question correctly?
 
haha hit the nail on the head there, thank you.
I had myself a little "aha" moment.
Its not that the individual forces (the normal force or force of gravity acting in the y direction) are equal to zero, but the the net result when you add the two is equal to zero.
is that right?
 
and the object is in equilibrium in that direction?
 
That is correct. Remember, Newton's second law isn't that the force on an object is equal to mass times acceleration. It's that the NET force on an object is equal to the mass times the acceleration.

You're probably sitting down right now, so you've got a nonzero gravitational force pulling down on you, and since you're most likely not falling down, there is an equal and opposite normal force on you, and these two forces cancel each other out.

Or if I were to push a block with force F towards the right, and you push it with force F from the left, then the block no longer accelerates, as the forces are canceled out (but there still are forces on it. It's the NET force that we're focused on).
 
haha i feel better now that I understand that, I got so caught up on the homework questions I forgot to look at the sum of the forces as a "net", which could be positive, negative, or zero.
thank you for taking the time to clarify this for me
 
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
61
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
17
Views
2K
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K