A light block of mass m and a heavy block of mass M are attached to the ends of a rope. A student holds the heavier block and lets the lighter block hang below

Question

A light block of mass m and a heavy block of mass M are attached to the ends of a rope. A student holds the heavier block and lets the lighter block hang below it. Then she lets go. Air resistance can be neglected.

1. What is the tension in the rope while the blocks are falling, before either hits the ground?


2. Would your answer be different if she had been holding the lighter block initially?

Answer 1

Hi Essie!

 

One great aspect of Newton's 2nd law, among many, is that we get to define the "system" that we are working with.  Newton's 2nd law tells us that the net external force on the system equals the system's mass times its acceleration.  But we can make the system whatever we want.  Only forces outside our system affect acceleration (this fact can be explained via Newton's third law, but we need not do that at the moment).

 

For this case, let's start by wrapping our "system bubble" around both masses.  Then, any tension in the rope between them would be internal, inside the system, and we need not include it in our Newton's 2nd law expression.  The only force external to our system that we care about in that case is gravity.  Thus, Newton's 2nd law becomes:

 

F_net = F_g = m_sysa  (thinking of downward as positive in this case)

 

where m = mass of system, which in this case is (M+m)

 

We have an expression for the force of gravity on an object, mass * g.  In this case, then, F_g on our system equals (M+m)g.  So we can put this information into Newton's 2nd law to get:

 

(M+m)g = (M+m)a

 

Thus, the blocks must be accelerating downward at a = g, as you would expect for free fall.

 

Now, let's change our system to be just, say, the lighter mass.  We are allowed to change our system, even in the middle of a problem, however we see fit.  In the situation described, the lighter mass is below.  So, as it falls, gravity acts on it downward, and any tension would be upward.  So, we would have (with down positive again):

 

F_net = F_g - T = ma

 

where m is the mass of the lighter block.  We already know a = g, and F_g = mg, so this becomes:

 

mg - T = mg

 

The only way this is true is if T = 0.  So there is no tension in the string.

 

*****************************************************************************************

 

What this means is that both objects are in free fall under gravity, and thus accelerate at the same rate.  There cannot be an additional force of tension in the rope, because it would pull down on one block and up on the other, giving the two blocks different accelerations, which cannot happen in free fall.

 

Note that nothing about this argument changes if the order of the blocks is reversed.

 

This may momentarily seem counter-intuitive, compared to the results for two similarly connected blocks pulled horizontally (over a frictionless surface) by some force.  In that case, both blocks also would accelerate at the same rate, but there WOULD be tension in the string (or whatever) between them.

 

The difference is that, in the horizontal case, the force pulling the blocks only directly acts on the first one.  The only way the second block also accelerates is if the force is transferred to the second block through the string between them (the tension).  The second block would not accelerate if the first block were pulled and there were no string between the blocks.  

 

However, gravity works differently.  Gravity reaches out to both blocks separately, and accelerates them regardless of how they are connected by a string.  Therefore, there is nothing for the string to do (and in fact, nothing it can do), and both blocks would fall under gravity whether connected by a string or not.

 

I hope this helps some!  If you have any questions at all, just let me know.