The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Pulleys+Tension

If I have a pulley with a 1kg weight attached to one end via a string and I pull at an angle of 30° below the x-axis so that the weight is stationary, why is the tension in the string the same for both sides?

surely if i hold the string at different angles there'll be slight differences in forces needed to keep the system in equilibrium?
Not sure how to explain it well really but I'll try my best.

The string will accelerate with the mass. On the other side no matter the angle it will accelerate by the same amount whatever its direction is. f=ma so since the acceleration is the same then so is the tension force along it on the other side. If I held the string at 0 30 60 and 90 degrees to the other side it will still accelerate by the same amount in the direction that its in so you need equal force in that direction.
Reply 2
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
Not sure how to explain it well really but I'll try my best.

The string will accelerate with the mass. On the other side no matter the angle it will accelerate by the same amount whatever its direction is. f=ma so since the acceleration is the same then so is the tension force along it on the other side. If I held the string at 0 30 60 and 90 degrees to the other side it will still accelerate by the same amount in the direction that its in so you need equal force in that direction.


but the mass on both sides isn't the same....
Original post by thefatone
but the mass on both sides isn't the same....


Yeah there's only one mass on one of the sides. That causes the string to accelerate. It accelerates by the same amount on both sides along itself. So the hand on the left side (if mass is on the right) stops it accelerating on the left side which also causes the right side to stop accelerating. So the tension along the string on the opposite side due to holding it should be equal to the force caused by the mass which creates the tension in the string.
Reply 4
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
Yeah there's only one mass on one of the sides. That causes the string to accelerate. It accelerates by the same amount on both sides along itself. So the hand on the left side (if mass is on the right) stops it accelerating on the left side which also causes the right side to stop accelerating. So the tension along the string on the opposite side due to holding it should be equal to the force caused by the mass which creates the tension in the string.


so why doesn't angle affect this?
Original post by thefatone
so why doesn't angle affect this?


Because as I said the string it will accelerate by the same amount along all of it, the acceleration won't differ from the angel. The tension will be along the string as well and the tension when it is the held makes acceleration = 0 so force on both sides is equal along the string. The tension on the right is equal to the mass's force from gravity and tension on the left needs to cancel the mass's force so it is equal to it and the tension on the right.
Reply 6
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
Because as I said the string it will accelerate by the same amount along all of it, the acceleration won't differ from the angel. The tension will be along the string as well and the tension when it is the held makes acceleration = 0 so force on both sides is equal along the string. The tension on the right is equal to the mass's force from gravity and tension on the left needs to cancel the mass's force so it is equal to it and the tension on the right.


right i see....
(let's say the force is 20 N)
so if i split things up into horizontal and vertical components and used pythagoras will this still get me tension is equal on both sides?

20cos⁡302+20sin⁡302\sqrt{{20\cos 30}^2 + {20\sin 30}^2}?
Original post by thefatone
right i see....
(let's say the force is 20 N)
so if i split things up into horizontal and vertical components and used pythagoras will this still get me tension is equal on both sides?

20cos⁡302+20sin⁡302\sqrt{{20\cos 30}^2 + {20\sin 30}^2}?


No there is nothing about vertical and horizontal forces. The only force on the string is along it.
Reply 8
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
No there is nothing about vertical and horizontal forces. The only force on the string is along it.


in that case forget this thread then.

\thread

xD
Original post by thefatone
in that case forget this thread then.

\thread

xD


The dashed dot is where the hand is holding.

(edited 7 years ago)
Reply 10
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
The dashed dot is where the hand is holding.



but it's an angle so that should make a difference

i'm getting a U in physics. I might aswell turn in and screw the exam and just ask for a U
Original post by thefatone
but it's an angle so that should make a difference

i'm getting a U in physics. I might aswell turn in and screw the exam and just ask for a U


Send an email to your teacher or something. I can't really say it in any other way to show how it doesn't follow horizontal and vertical components.
Reply 12
Avatar for thefatone
thefatone
OP
6
Original post by Vikingninja
Send an email to your teacher or something. I can't really say it in any other way to show how it doesn't follow horizontal and vertical components.


yea nvm just leave it ^-^ you tried your best

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you