How Does Newton's Third Law Apply in a Frictionless Two-Block System?

AI Thread Summary
In a frictionless two-block system, when a force of 1.2 N is applied, both blocks accelerate together at 0.4 m/s² to the right. The horizontal force exerted by the larger block on the smaller block is also 0.4 N, directed to the right, illustrating Newton's Third Law, which states that forces are equal in magnitude and opposite in direction. The confusion arises from misinterpreting the direction of forces; while the smaller block pushes down on the larger block, the larger block pushes back with an equal force to the right. The coefficients of friction are not relevant in this scenario since the blocks do not slide relative to each other. Understanding these principles clarifies the application of Newton's laws in this system.
ScullyX51
Messages
35
Reaction score
0

Homework Statement


Two blocks are accelerated across a horizontal frictionless surface as shown. The coefficient of static friction between the two blocks is 0.7, and the coefficient of kinetic friction is 0.5, use M= 1.0 kg. When F=1.2 N, frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. In this case, what is the acceleratiob of the two block system? (The picture is of mass 2m on the bottom with mass M on top of it, and F pointing to the right)
2) What is the horizontal component of the force the large block exerts on the small block
3) Suppose that F is increased. What is the maximum acceleration that mass 2M can have, without mass M slipping off?

Homework Equations



F=ma

The Attempt at a Solution


I got the first portion of the problem right. I just solved for f=ma, and plugged in the f=1.2. I figured that the coefficients of friction given were not relevant here because the system is accelerating. For an answer I got a=.4 m/s2 to the right.
2) For the second part of the question: the horizontal component the large block exerts on the small block, I put .4 to the left. I thought that according to Newton's third law the force is equal in magnitude and opposite in direction, but it is saying this answer is wrong. It says the correct answer is .4 to the right. Why is it the same?
 
Physics news on Phys.org
I don't understand how you are applying Newton's 3rd law in this case.

Realize that both blocks have the same acceleration, which equals 0.4 m/s^2 to the right. To find the force on the top block (which is exerted by the bottom block), apply Newton's 2nd law.
 
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