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Giancoli physics chapter 4 example 4-13 and 4-14

 
 
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Giancoli physics chapter 4 example 4-13 and 4-14
by Jialin Wang - Sunday, 16 January 2022, 8:29 PM
 

I don't understand why the tensions on both sides of a pulley are equal. If a thing has less mass so it has less weight, then it should have less tension to balance it out, and if something has more mass, therefore more weight, shouldn't it have more tension? I don't know why tension on both sides are equal. Why is tension all the same along the same rope?

 
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Re: Giancoli physics chapter 4 example 4-13 and 4-14
by John Lensmire - Thursday, 20 January 2022, 12:21 PM
 

Let's start with a scenario you're not mentioning, but I think will help.

In Case 1 (see the diagram below), we have two weights attached by a rope to the ceiling. Here we are assuming that the rope is attached at the top and nothing can move.


In this case, since the rope is fixed at the top, we can really think of the system as two separate ropes, one on the left side, and one on the right side. Anytime we have ropes (so here we have two ropes) the tensions at both ends of the ropes are always the same. Let's practice here with the forces and Newton's 3rd Law.

  • At point A, gravity is pulling the 10kg block down with a force of $10g$.
  • At point A, the tension of the rope $T_L$ is holding up the block, so $T_L = 10g$ as well.
  • At point C, the tension of the rope is pulling down with a force of $T_L = 10g$ here too.
  • Hence, at point C the left rope is pulling down on the ceiling with a force of $10g$.
A similar thing happens to the right side:

  • At point B, gravity is pulling the 10kg block down with a force of $20g$.
  • At point B, the tension of the rope $T_R$ is holding up the block, so $T_R = 20g$.
  • At point C, the tension of the rope is pulling down with a force $T_R = 20g$.
  • Hence, at point C the right rope is pulling down on the ceiling with a force of $20g$.

Therefore, when the rope is fixed in place on the top, you're correct that we have a larger tension for the heavier rope. Note in total, there is a force of $30g$ pulling downwards on the ceiling.

Now let's look at a second case with a pulley.


It can be a little weird at first, but the magic of the pulley is that it allows us to view the rope as a single rope again. The key difference is now the rope can move freely at point C. This system is NOT STABLE. Here the 20kg block on the right will fall down and lift up the 10kg block. Let $T$ denote the tension in the rope. I'm not going to do all the calculations/forces here, but the key things are as follows:

  • At point A, gravity will pull the block down with a force of $10g$ while the rope will pull upwards with a force of $T$.
  • At point B, gravity will pull the block down with a force of $20g$ while the rope will pull upwards with a force of $T$.
To find the tension $T$, the key idea is that both blocks need to accelerate at the same rate. Without focusing too much on the details here, you will get that $10g < T < 20g$. Hence,

  • At point A, the net force of $T - 10g$ will be UPWARDS, so the 10kg block is pulled up.
  • At point B, the net force of $20g - T$ will be DOWNWARDS, so the 20kg block is pulled down.

Hope this helps!

 
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