So I have this problem:
and I've drawn free-body-diagrams of pulley b and mass 2
I couldn't help but notice that the tension pulling mass 2, is pulling the pulley B but in opposite direction.
How is this possible?
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1$\begingroup$ Related post by OP: physics.stackexchange.com/q/426027/2451 $\endgroup$– Qmechanic ♦Commented Sep 1, 2018 at 13:57
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$\begingroup$ @Qmechanic why are you wasting your time on newtonian mechanics lol, are theoretical physicists really that free? nvm, you're the moderator, it's your job. :)) $\endgroup$– Daksh M.Commented Sep 1, 2018 at 14:11
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2 Answers
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This is what tension does. Here is a test you can do. Get a friend, a piece of string and a pair of scissors. Have your friend pull the piece of string (between his two hands) so that it is taught. The string is now in what we call tension.
Now, imagine you were to cut the string in the middle, what would happen? We can imaging that the string will fall limply in your friend's hands. Why? Because the left half of the string was being pulled to the right from its rightmost end (to keep it taught), while the right half of the string was being pulled left from its leftmost end. But we ruined it all when we cut the string in half. In other words, the two halves were pulling each other in opposite directions.
Finally, cut the string. Does the experiment match our findings? What would happen if we cut the string at some other location?
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The left hand upward force T2 is the force on block 2 due to the string.
The Newton third law pair to this force is a downward force T2 on the string due to block 2.
The right hand downward force T2 is the force on the pulley due to the string.
The Newton third law pair to this force is an upward force T2 on the string due to the pulley.
The string is assumed to have no mass so the net force on the string due to
- downward force T2 on the string due to block 2
- upward force T2 on the string due to the pulley.
Has to be zero; thus the magnitudes of all forces T2 are the same.
