Homework Help: Pushing Block Question

1. Oct 22, 2004

jhson114

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?

What is the magnitude of the force that the left most block exerts on the middle one?

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?

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 :)

2. Oct 22, 2004

HallsofIvy

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.