Force exerted on block B by Block A when being pushed on a frictionless surface

  • Thread starter Thread starter yashboi123
  • Start date Start date
  • Tags Tags
    Block Force
AI Thread Summary
The force exerted by Block A on Block B is calculated using the formula mB * a, as Block B's mass directly influences the force required to keep it accelerating with Block A. If Block B were absent, Block A would not exert any force on that side, indicating that Block B's inertia is crucial for the interaction. The mass of Block B determines the magnitude of the force due to its inertia. In analyzing such problems, drawing free body diagrams for each block is essential to visualize the forces acting on them. Understanding these principles clarifies the dynamics between the two blocks on a frictionless surface.
yashboi123
Messages
17
Reaction score
0
Homework Statement
Boxes A and B are in contact on a horizontal, frictionless surface (Figure 1). Box A has mass 25.0 kg and box B has mass 7.0 kg . A horizontal force of 100 N is exerted on box A \.
Relevant Equations
F = ma
I was just wondering why you would do
mB * a
to get the force A is exerting on B instead of
mA * a

1695683049741.png
1695683062549.png
 
Physics news on Phys.org
If B were not there, would there be any force on that side of A? If B was super small, would it take much force to keep it accelerating along with A? What if it was very large? So, you see, B’s inertia is the reason there is any force between B and A and B’s mass is what you need to figure out how big the force is.
 
  • Like
Likes PeroK and topsquark
In problems like this always draw the free body diagrams and show the forces on each block separately. Then you can write the equations and they should makes sense.
 
yashboi123 said:
why you would do
mB * a
to get the force A is exerting on B instead of
mA * a
Because A's acceleration is a consequence of the two forces on A, whereas there is only one force on B.
 
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