How Do Forces Distribute Among Blocks on a Frictionless Surface?

  • Thread starter Thread starter jhson114
  • Start date Start date
  • Tags Tags
    Block
AI Thread Summary
Three blocks of mass 5, 10, and 15 kg are aligned on a frictionless surface, with a 17N force applied to the leftmost block. The force exerted by the middle block on the rightmost block is calculated to be 8.5N, while the force the leftmost block exerts on the middle block is approximately 14.2N. The same force of 8.5N is found when the order of the blocks is reversed, demonstrating the consistency of Newton's Third Law. The total mass of the blocks is 30 kg, leading to a common acceleration that informs the force calculations. Understanding these forces relies on recognizing the equal and opposite reactions as described by Newton's laws.
jhson114
Messages
82
Reaction score
0
Blocks of mass 5,10,15kg are lined up from left to right in that order on a frictionless surface so each is touching the next one. A rightward-pointing force of magnitude 17N is applied to the left-most block. What is the magnitude of the force that the middle block exerts on the rightmost one?
answer: 8.5N

What is the magnitude of the force that the left most block exerts on the middle one?
answer: 14.2N

Suppose now that the left-right order of the blocks is reversed. Now find the magnitude ofthe force that the leftmost block exerts on the middle one?
answer: 8.5N

I already know the answer, but i do not understand the the concept behind solving this. can someone explain to me what kind of equations i need to use and why the question 1 and question 3 have the same answer. THanks :)
 
Physics news on Phys.org
That problem is actually much simpler than it looks. All of the masses must move together so, first, think of the masses as a single 5+10+15 = 30 kg mass. Since F= ma, 17N= 30a so a= 17/30 m/s2.

The right most block has mass 15 kg. Again, F= ma. Since it is moving at a= 17/30,
we must have F= 15(17/30)= 17/2 = 8.5 N. The only way any force is applied to that block is by the middle on pushing it.

The middle block has mass 10 kg. Since it is also moving at a= 17/30, we must have F= 10(17/30)= 17/3 N. But the middle block is push on the right block with for 17/2 N so the right block is pushing on the middle block with force -17/2 N (Newton's third law). Taking F to be the force the left block applies to the middle block, we have total force= F- 17/2= 17/3 so F= 17/2+ 17/3= 51/6+ 34/6= 85/6= 14 1/6 N which is actually slight less than "14.2". You think of this as "the left block pushes the middle block with force 85/6 N. 17/3 of that accelerates it and the other 17/2 is passed on to the right block".

Now the order is switched: the masses of the blocks are 15 kg, 10 kg, 5 kg.
Still, the total mass is 30 kg so the common acceleration of all blocks is 17/30 m/s2. There is 17 N force applied to the right block. Since it has mass 15 kg, it requires only 15(17/30)= 17/2 N to accelerate it. The rest of the 17 N, 17- 17/2= 17/2 N is "passed on" to the next block. The force the left block exerts on the middle block is 17/2= 8.5 N.
 


Sure, I'd be happy to explain the concept behind solving this problem.

First, let's break down the information given in the problem. We have three blocks of mass 5, 10, and 15kg, respectively, arranged from left to right in that order. The blocks are on a frictionless surface, meaning there is no resistance to their motion. A force of 17N is applied to the leftmost block in a rightward direction.

Now, to understand the concept behind solving this problem, we need to remember Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In other words, when one object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.

In this case, the leftmost block is exerting a force of 17N on the middle block. By Newton's Third Law, the middle block will exert an equal and opposite force of 17N back on the leftmost block. This is why the answer to question 1 and question 3 is the same - the force exerted by the middle block on the leftmost block is equal and opposite to the force exerted by the leftmost block on the middle block.

To calculate the magnitude of the force that the middle block exerts on the rightmost one, we can use the same concept. The middle block is being pushed to the right by the leftmost block with a force of 17N. This means that the middle block will also push the rightmost block to the right with an equal and opposite force of 17N.

To summarize, the key concept to understand in solving this problem is Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. By applying this law, we can determine the magnitude of forces exerted by each block on the other. I hope this helps clarify the concept for you.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top