For 2) wouldn't there also be an equal and opposite force on block B by block A?

AI Thread Summary
The discussion centers on the forces acting on three blocks on a frictionless table when pushed by a hand. The force applied by the hand on block A must account for the total mass of all blocks, as they accelerate together, leading to a calculated force of 54.6 N. For block B, the net force is determined by the forces exerted on it by blocks A and C, which must be considered separately. The confusion arises from miscalculating the forces acting on block B, emphasizing the need to apply Newton's second law correctly. Understanding the interaction between the blocks is crucial for solving the problem accurately.
Gooner5
Messages
2
Reaction score
0
Member advised to use the formatting template for questions posted to the homework forums
Three blocks, each of mass 13 kg are on a frictionless table. A hand pushes on the left most box (A) such that the three boxes accelerate in the positive horizontal direction as shown at a rate of a = 1.4 m/s2.

1)What is the magnitude of the force on block A from the hand?

2)What is the net horizontal force on block B?


For number 1 I was able to get the answer by adding up all the blocks and multiplying that by the acceleration but I am confused by why this is right. Before I did this I thought that the answer would be the mass of A * 1.4. Why does the force by the hand on block A include the other blocks as well? Is it because the whole system is accelerating and for that to happen the force has to take into account the other blocks as well?

For number 2 I found the force done on the system to be (13+13+13)*1.4 = 54.6N. I then found that the force that object C does on block B and the force that object A does on block B would 13*1.4= 18.2N. I then added the force done on the system by the hand with block A and subtracted that by the force by C on B to get 54.6, but this is not the right answer. What should I be looking for in this problem?

Thanks
 
Physics news on Phys.org
For 1, the hand doesn't care whether it's three separate blocks that happen to be adjacent, or three blocks glued together, or a single block mass 39kg. Alternatively, consider the force pair (action and reaction) at each interface. Each block accelerates according to the net force on it.

For 2), all block B "knows" is that it experiences two forces, one from block A and one from block C, of opposite sign. The net force on B is simply the sum of those two. Alternatively, consider that for each block separately ΣF=Fnet=ma, where the sum is over the forces acting directly on the block. You tried using the first part of that, ΣF=Fnet, to find Fnet (but summed the wrong forces); the alternative is to use the second part, Fnet=ma.
 
Last edited:
For 2) wouldn't there also be an equal and opposite force on block B by block C?
 
Gooner5 said:
For 2) wouldn't there also be an equal and opposite force on block B by block C?

Do you care about that?

When using Newtons law F=ma what does F stand for? Hint... the answer is NOT just "Force".
 
Gooner5 said:
For 2) wouldn't there also be an equal and opposite force on block B by block C?
I made a typo in my post. Now corrected.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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