https://youtu.be/jGLjpODKXew
PHYS 1101: Lecture Thirteen, Part Six
From that example, let me point out and pull out a very important point that you can extend to many problems that involve ropes, rods, or wires. If the rope is reasonably lightweight, the mass is smaller compared to other masses in the problem, say the objects that this rope is connecting, then you can think of that rope as simply extending the contact between these two connected objects.
In essence then, the tension in the rope is the magnitude of this force pair, the Newton’s Third Law pair, between the objects. Let me show you the force breakdown of this problem and Car 3 and Car 2, but let’s do that where we explicitly also include the forces on this rope or the coupling bar.
Remember, for Car 3 and Car 2, Car 3, yeah, and Car 2, I have the same vertical forces. Then I know, horizontally, that I had tension 3 to the right. But now, before I just immediately draw the same tension on Car 2, let’s be more careful. Technically, at the edge of Car 3, I’m in contact with a coupling bar. I’m not directly Car 2. So, T3 comes from this object.
Well, T3 is the force of the coupling bar on Car 3. So, the Third Law pair is the force of Car 3 on the coupling bar. It’s at this end of the coupling bar that technically is the match, the Third Law pair, to this force.
Okay. If this car is accelerating to the right, so is this coupling bar. If it’s really lightweight, though, the force to the right, yeah, technically, it has to be a little bit bigger than the force to the left because I do need a net force. But, remember that the acceleration is fnet divided by the mass. If I have something extremely lightweight, it takes a very small net force to give me that acceleration.
So, if this is lightweight, the rope, the coupling bar, the force to the right is very close to only slightly bigger than the force to the left. To the precision of three digits in solving the problem, probably, it’s reasonable that the value at the right end of this coupling bar is the same as the value on the left end. So, I’ve got T3 at both ends.
This being just slightly bigger than this, I’m approximating them that they are the same, because it’s so light, is consistent with still having an acceleration of this object to the right as I do the Car 3 and Car 2.
Now that I’ve bootstrapped my way to this end of the rope and convinced myself its T3, now this force is the action/reaction pair to the force on Car 2 directly. I know these are an action/reaction pair and have to be exactly equal magnitude, opposite direction. Then, so for Car 2, we know we have a larger tension, T2, to the right.
So, notice the justification in my statement that’s in the big red box above. The statement is, “If this rope is lightweight, I can consider that this coupling bar simply extends the connection between Car 3 and Car 2. I can think of the direct, as though these are directly in contact with each other. The coupling bar just links that together. So, the direct contact of T3 at this car is the action/reaction pair, effectively, to the direct T3 contact force at that end of Car 2.
The next important point to emphasize is that pulleys will be involved often in this problem. A pulley simply serves to change the direction of the force caused by a rope. If you think about it, the tension in a rope is a measure of how snappy the line feels to you, if you will. If you were to go in to pluck the rope here, it would snap back the same amount of tension if you plucked it over here. The pulley has simply redirected that force.
Take a scenario like this, where let’s say that the set is, this big block s, is very heavy compared to the man. So, you can picture what will happen, that the block is going to accelerate down, pulling the man up or the boy up. He’s going to accelerate up.
Let’s do a quick force diagram on each of these two objects. I’ve already argued or convinced you that the tension that the boy experiences pulling him up has to be the action/reaction pair, effectively, to this set. This rope simply extends what I can think of as the direct contact or interaction between these two objects. So, if this force is pulling up, if I go to the other end, it’s the same tension in the line. But again, a force always pulls up. It’s going to pull up on this block s.
That’s the same scenario as the force on the coupling bar or the rope itself, as being in the opposite directions and therefore, on the objects connected to them being in opposite directions. It’s the same, except the direction is changed only by the pulley.
So, let’s draw it out. For the boy pulling himself up or hanging from this rope, I have the tension up, and I have his weight, the force due to gravity, down. That’s the same tension everywhere in the line. So, the length of this force has to be the same length that’s pulling up on this block. Then, I have to have the force due to gravity down on the block.
Okay. Did I draw these in a way that’s consistent with Newton’s Third Law? Yes. The interaction between the objects has been extended, so to speak, by this rope. But, these forces are equal and opposite. They end up looking in the same direction, only because the pulley has changed or rotated the direction. If you were to straighten it out, the tensions do point opposite directions.
Then, what’s the acceleration? I know the acceleration for the block on the right is down. I argued it has more mass. It’s heavier. The acceleration for the boy on the right is going to be up. I have more force up than down on the boy. On the block, I have to have more force down than I have up.
Quiz question 14 has you putting some of these ideas together. You have the scenario shown on the left, two unequal masses, capital M being larger than the smaller mass m, and they’re connected by a light rope that goes over this lightweight pulley.
When the system is released, your intuition, I’m sure, tells you or confirms that the big mass is going to move down, and the light mass is going to move up. Which of the force diagram choices on the right, A, B, C, or D, matches the magnitudes and the direction of forces that you expect must be the case for the forces for each of these two masses?