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So far i've understood that a single pulley changes the direction of the force, but how does multiple pulleys decrease the amount of force needed?

For example in (a), why is the middle force arrow pointing upwards? shouldn't it point down since the pulley only changes the direction of the force. It makes sense on the right most force arrow since that points down so the left most force arrow points up. Since that left most force arrow points up wouldn't the middle force arrow point down?

Sorry if i'm not clear with the question, I am a bit confused about pulleys even though this is quite basic!

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3 Answers 3

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Tension is an entity that pulls in both directions in a rope. If you were to cut the rope, you would have to manually apply a force to the right on the section to the left in order to keep it in equilibrium; and you would have to manually apply a force to the left on the section to the right in order to keep it in equilibrium. The tension in the rope single handedly accomplishes both these things.

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The change of direction is a fine way to think about a pully, but be aware when a force is pulling on the middle of a string (such as the weight) and not in the end (such as the pull in your examples).

  • A pull in one end causes a tension that follows the string geometry.
  • But a pull on the middle causes a tension in both string halves, pointing opposite. As if each string half was a separate string. When something is hanging on the middle of the string, such as a subpully in your examples, then this is your case.

Think of a non-moving pully with a string over it, which has both ends tied to a hanging block below it. Surely the tensions are upwards in both strings, since they both help in carrying the block's weight.

And here you have the answer to the other question as well: How can the force required become smaller when passing over several pullies? Because the block is carried by both strings in my example, so that the tension in each is half of the weight they carry.

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  • $\begingroup$ Thanks for an answer, but still a little confused. Is the middle force arrow in (a) upwards because of newtons third law (opposite and equal reaction forces)? Since the weight is between the middle force arrow and the left most force arrow, they both point upwards as they are the opposite reaction forces? $\endgroup$
    – Hilkjh
    Jun 24, 2017 at 11:23
  • $\begingroup$ Edit, I think this confusion comes from the difference between a single movable pulley and a single fixed pulley. Why does the single movable pulley have both force arrows up while fixed pulley only has one $\endgroup$
    – Hilkjh
    Jun 24, 2017 at 11:33
  • $\begingroup$ @Hilkjh Not exactly Newton's 3rd law there, but rather Newton's 1st law. The weight must be held up by something. And there are two strings to do that. So they share the load and point upwards both. $\endgroup$
    – Steeven
    Jun 24, 2017 at 14:00
  • $\begingroup$ @Hilkjh In general, do Newton's 1st law on each object and pully, and you can find the force directions in such problems. $\endgroup$
    – Steeven
    Jun 24, 2017 at 14:01
  • $\begingroup$ imgur.com/a/unPMl hi im trying to solve this question but i don't get why the force is half the weight because the pulley system here only has one loop. $\endgroup$
    – Hilkjh
    Jun 25, 2017 at 1:26
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Without the accompanying text the diagrams are misleading. (And diagram (c) is incorrect.)

The tension in a massless string pulls equally in both directions. However, the arrows in this diagram have been drawn to indicate only the forces on the lower block of pulleys. The arrows pulling down on the upper block have been omitted. The diagram illustrates that the lower block is in equilibrium : total upward tension force on this block equals the weight pulling down on it.

The arrows could have been drawn pointing downwards, to illustrate that the upper block is also in equilibrium : total downward tension force on this block equals the upward force provided by the support from which the blocks are hanging.

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  • $\begingroup$ Hi would it be possible if you can help me solve this imgur.com/a/unPMl thanks! $\endgroup$
    – Hilkjh
    Jun 25, 2017 at 8:35

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