Newton's Third Law: True or False?

AI Thread Summary
The discussion centers on the interpretation of Newton's Third Law, specifically whether the algebraic sum of forces in a system equals zero. One participant argues that the law implies F1 = -F2, suggesting that the forces are equal in magnitude but opposite in direction, leading to a sum of zero. However, another participant contends that since each force acts on separate bodies, the algebraic sum cannot be determined without clarifying what "the system" refers to. The conversation emphasizes the importance of vectorial addition of forces in an isolated system, ultimately leading to the conclusion that the sum of internal forces must indeed be zero. Clarification on the context of the system is deemed necessary for a definitive answer.
UNknown 2010
Messages
77
Reaction score
0
Hello,
True or False:
- According to Newton third law, the algebraic sum of the forces in the system is equal to zero.

I think the answer is FALSE because, the third law of Newton says that F1 = - F2

and the " - " is for the direction, so that F1 = F2 in the quantity ==> F1 + F2 \neq 0
 
Physics news on Phys.org
Well, from what i understand from Newton's third law is, any force will have a counter force of the same magnitude in the opposite direction. So, if u add both, it would be zero.. however, i am not sure that is what they are asking..
 
The answer is true.
If F1 = - F2, the two forces are equal but one is negative
For example,
1+ -1 = 0
At least that is what I think
 
i think its false ,,,
because each force is acting on separate body,,
so we can not find the the algebraic sum of the forces ,,

 
UNknown 2010 said:
True or False:
- According to Newton third law, the algebraic sum of the forces in the system is equal to zero.
This question is impossible to answer without further information on what "the system" means. We have a template that specifically asks for

Homework Statement

. You should have followed the template, UNknown.

redrose_emf said:
i think its false ,,,
because each force is acting on separate body,,
so we can not find the the algebraic sum of the forces
Sure you can. Just add them vectorially. This is a very important concept. The (vectorial) sum of all internal forces must be zero in an isolated system.
 
D H said:
Sure you can. Just add them vectorially. This is a very important concept. The (vectorial) sum of all internal forces must be zero in an isolated system.

thank you for the correction ,,,
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top