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# M1 question watch

1. why is the tension same on both sides of the string?
2. (Original post by ae86_trueno)
why is the tension same on both sides of the string?
Because it's a smooth bead and so no friction to cause a difference in the tensions either side of the bead. The same goes for smooth pulleys.
3. (Original post by tiny hobbit)
Because it's a smooth bead and so no friction to cause a difference in the tensions either side of the bead. The same goes for smooth pulleys.
I kinda get it but not fully, say if you increase the 8N force to like 20N, wouldn't that increase the T more in right hand section of the string more than the perpendicular one ?
4. Also in equilibrium
5. (Original post by ae86_trueno)
I kinda get it but not fully, say if you increase the 8N force to like 20N, wouldn't that increase the T more in right hand section of the string more than the perpendicular one ?
If the 8N was increased to 20N, the bead Y would move to the left and lots of things would change. The string XY would no longer be vertical and the tension in the string would be different. The tensions on both sides of the smooth bead would still equal each other though.
6. (Original post by ae86_trueno)
I kinda get it but not fully, say if you increase the 8N force to like 20N, wouldn't that increase the T more in right hand section of the string more than the perpendicular one ?
No. Assuming the weight of the bead remains the same, the increase in horizontal force from 8N to 20N will pull the system out of shape such that point Y moves to the left and up a bit (while respecting the fact that the length of the string, XY + YZ, remains unchanged). The system will settle in its new equilibrium position with a new value of tension T in both XY and YZ.

Overlapped with tiny hobbit
7. (Original post by JIMBO789)
Also in equilibrium
The tension on either side of the smooth bead would be equal even if it weren't in equilibrium. Think of the standard pulley questions.

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