I'm a bit confused about when does Newton's third law come into effect and when it does not. For instance, lets say i was swinging a YoYo in a vertical circular motion, I would obviously have to provide enough centripetal force to cause it to go in a circle in a uniform speed. so let's say we pick a point when the YoYo is all the way at the top and we try to measure the tension on the string. to measure the tension, we use Newton's second law $$\Sigma F = ma => -T-mg = -ma $$ so the tension must equal $ma-mg$. My question is, since the centripetal force is pulling down on the YoYo, why doesn't the YoYo provide a force equal and opposite the direction of the centripetal force that's also pulling up the string? because if it did, i think the formula for tension would be something like $$-T+ma(the opposite force)-mg = -ma$$ can someone please clear this confusion, thanks

  • $\begingroup$ You, who are are holding the string, will be pulled by the YoYo (the Newton third law force) but as your mass is very much greater than that of the YoYo the effect is not that noticeable other than that you can feel the YoYo pulling you. However observer what a hammer thrower has to do when trying to control the force that the hammer is exerting. youtube.com/watch?v=2v0FGyE6zxc $\endgroup$ – Farcher Feb 25 '16 at 7:42
  • $\begingroup$ exactly! so why doesn't that force come into the calculation of the tension. since i'm assuming the yoyo is going to pull me up via the string? $\endgroup$ – GamefanA Feb 25 '16 at 7:44
  • $\begingroup$ You have to distinguish between the forces on the YoYo and the force on the string. When you are describing the motion of the YoYo you are only concerned with the forces on the YoYo. Newton's third law compares two forces each of the forces acting on different objects. in the examples that you deal with the string is assumed to be massless and is only there to transmit a force from you yo the YoYo and vice versa. $\endgroup$ – Farcher Feb 25 '16 at 7:50
  • $\begingroup$ Ok that makes sense. but let's assume that this is a real life experiment done in a vacuum (to avoid air resistance) thus the string is a real object and does have it's own mass. and i wan't to buy a string that can handle enough tension so that when i swing the YoYo it wouldn't break. would I have to consider the opposite force in my calculations for the tension? (if you can provide how would you calculate the tension in this situation i would be grateful) $\endgroup$ – GamefanA Feb 25 '16 at 7:57

When analyzing Newton's 3rd law you need to look at the specific force one body exerts on another.

The rope exerts a force, say with magnitude $T_1$, downwards on the yoyo. Hence the yoyo exerts a force upwards with magnitude $T_1$ on the rope. (reason I'm using T1 here is because I'm going to assume T2 is the tension in the rope where the hand is holding the rope).

Also, when analyzing F=ma on the yoyo, only consider the forces acting ON the yoyo. So the opposite force will not appear in your F=ma equation here, because the opposite force acts on the rope not the yoyo. So this is correct:

$\Sigma F_{onYoyo} = m_{yoyo}a_{yoyo} => -T_1-m_{yoyo}g = m_{yoyo}a_{yoyo}$

(note I'm using the convention that up is positive, down is negative. I leave the acceleration variable with a positive sign assuming by convention it is upwards and positive... it will come out as a negative number and I will then interpret that as a downward acceleration. what you did with signs is fine as long as you are careful.)

When analzying F=ma for the rope, you'd bring in that opposite force:

$\Sigma F_{onRope} = m_{rope}a_{rope} => T_1-T_2-m_{rope}g = m_{rope}a_{rope}$

So you see in the above equation you have a force T1 acting in the opposite direction.

  • 1
    $\begingroup$ thank you sir, you're explanations are awesome. that makes a lot of sense. $\endgroup$ – GamefanA Feb 25 '16 at 8:22

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