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Why is tension in a rope the same at every point?
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So, I had this lecture where it was explained that if the pulley is friction-less, then the tension on any point of the rope is going to be same.
I can understand the friction-less part, as pulley is not applying any resistive forces that change the magnitude of the tension force.
But even then, how can the force of tension be same at any point on the rope?
newtonian-mechanics forces mass free-body-diagram string
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
asked Sep 1 at 5:13
Daksh Miglani
1285
add a comment |Â
up vote
3
down vote
favorite
So, I had this lecture where it was explained that if the pulley is friction-less, then the tension on any point of the rope is going to be same.
I can understand the friction-less part, as pulley is not applying any resistive forces that change the magnitude of the tension force.
But even then, how can the force of tension be same at any point on the rope?
newtonian-mechanics forces mass free-body-diagram string
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
asked Sep 1 at 5:13
Daksh Miglani
1285
Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So, I had this lecture where it was explained that if the pulley is friction-less, then the tension on any point of the rope is going to be same.
I can understand the friction-less part, as pulley is not applying any resistive forces that change the magnitude of the tension force.
But even then, how can the force of tension be same at any point on the rope?
newtonian-mechanics forces mass free-body-diagram string
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
asked Sep 1 at 5:13
Daksh Miglani
1285
So, I had this lecture where it was explained that if the pulley is friction-less, then the tension on any point of the rope is going to be same.
I can understand the friction-less part, as pulley is not applying any resistive forces that change the magnitude of the tension force.
But even then, how can the force of tension be same at any point on the rope?
newtonian-mechanics forces mass free-body-diagram string
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
asked Sep 1 at 5:13
Daksh Miglani
1285
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
edited Sep 1 at 6:05
Qmechanic♦
96.7k121631021
96.7k121631021
asked Sep 1 at 5:13
Daksh Miglani
1285
asked Sep 1 at 5:13
Daksh Miglani
1285
asked Sep 1 at 5:13
Daksh Miglani
1285
1285
Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15
add a comment |Â
Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15
Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15
Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
If the tension changed throughout the rope, there would be a piece of the rope experiencing different tension forces on its ends, and hence experiencing a net force.
Newton's second law says that $F = m a$, and the acceleration of the rope is the same as the acceleration of the blocks. Since the rope is light, that means the net force on each piece of the rope has to be very small. That means the change in the tension must be very small. Usually the rope is so light compared to the blocks that we can neglect the change in tension along it entirely, so the tension is the same at every point.
answered Sep 1 at 5:29
knzhou
33.6k897169
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If the tension changed throughout the rope, there would be a piece of the rope experiencing different tension forces on its ends, and hence experiencing a net force.
Newton's second law says that $F = m a$, and the acceleration of the rope is the same as the acceleration of the blocks. Since the rope is light, that means the net force on each piece of the rope has to be very small. That means the change in the tension must be very small. Usually the rope is so light compared to the blocks that we can neglect the change in tension along it entirely, so the tension is the same at every point.
answered Sep 1 at 5:29
knzhou
33.6k897169
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
add a comment |Â
up vote
4
down vote
accepted
If the tension changed throughout the rope, there would be a piece of the rope experiencing different tension forces on its ends, and hence experiencing a net force.
Newton's second law says that $F = m a$, and the acceleration of the rope is the same as the acceleration of the blocks. Since the rope is light, that means the net force on each piece of the rope has to be very small. That means the change in the tension must be very small. Usually the rope is so light compared to the blocks that we can neglect the change in tension along it entirely, so the tension is the same at every point.
answered Sep 1 at 5:29
knzhou
33.6k897169
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If the tension changed throughout the rope, there would be a piece of the rope experiencing different tension forces on its ends, and hence experiencing a net force.
Newton's second law says that $F = m a$, and the acceleration of the rope is the same as the acceleration of the blocks. Since the rope is light, that means the net force on each piece of the rope has to be very small. That means the change in the tension must be very small. Usually the rope is so light compared to the blocks that we can neglect the change in tension along it entirely, so the tension is the same at every point.
answered Sep 1 at 5:29
knzhou
33.6k897169
If the tension changed throughout the rope, there would be a piece of the rope experiencing different tension forces on its ends, and hence experiencing a net force.
Newton's second law says that $F = m a$, and the acceleration of the rope is the same as the acceleration of the blocks. Since the rope is light, that means the net force on each piece of the rope has to be very small. That means the change in the tension must be very small. Usually the rope is so light compared to the blocks that we can neglect the change in tension along it entirely, so the tension is the same at every point.
answered Sep 1 at 5:29
knzhou
33.6k897169
answered Sep 1 at 5:29
knzhou
33.6k897169
answered Sep 1 at 5:29
knzhou
33.6k897169
answered Sep 1 at 5:29
knzhou
33.6k897169
33.6k897169
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
add a comment |Â
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
so we treat the rope like a system?
– Daksh Miglani
Sep 1 at 5:32
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
@DakshMiglani Yes, I am applying $F=ma$ to the system of the rope.
– knzhou
Sep 1 at 5:33
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
alright, it makes sense thanks :)
– Daksh Miglani
Sep 1 at 5:34
add a comment |Â
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Related: physics.stackexchange.com/q/156413/2451 and links therein.
– Qmechanic♦
Sep 1 at 6:15