# Tension and Cliffs/Pulley systems

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#### Jepstein30

##### Full Member

Having trouble visualizing tension as it relates to cliffs or pulley systems with masses on both sides.

Specifically, EK Physics 1001 questions 241 and 244.

241) 50 kg woman holding a 50 kg mass from the end of a rope (pulley over the cliff to make it a straight drop). Question asks for the tension in the rope.

I assumed that the mass in in equilibrium so the forces up = forces down. Gravity is pulling down at 500N so the tension should also be 500N?

The answer is 250N but I can't really follow the explanation. How/Why does the mass of the woman matter?

244) Mass of m on the left side of a pulley, mass of 2m on the right side of the pulley. What is the tension in the rope in terms of mg?

No clue how to approach this one, I completely sucked at pulleys in physics and ended up just using shortcuts like mechanical advantage = # of ropes supporting pulley.

Basically, I don't know how to approach problems with masses on both sides of the pulley whether the mass be on a surface (first problem) or hanging in the air (second problem)?

Anyone want to explain the answers and/or point me towards a resource I can use to understand this type of problem??

Thanks!

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
Having trouble visualizing tension as it relates to cliffs or pulley systems with masses on both sides.

Specifically, EK Physics 1001 questions 241 and 244.

241) 50 kg woman holding a 50 kg mass from the end of a rope (pulley over the cliff to make it a straight drop). Question asks for the tension in the rope.

I assumed that the mass in in equilibrium so the forces up = forces down. Gravity is pulling down at 500N so the tension should also be 500N?

The answer is 250N but I can't really follow the explanation. How/Why does the mass of the woman matter?

244) Mass of m on the left side of a pulley, mass of 2m on the right side of the pulley. What is the tension in the rope in terms of mg?

No clue how to approach this one, I completely sucked at pulleys in physics and ended up just using shortcuts like mechanical advantage = # of ropes supporting pulley.

Basically, I don't know how to approach problems with masses on both sides of the pulley whether the mass be on a surface (first problem) or hanging in the air (second problem)?

Anyone want to explain the answers and/or point me towards a resource I can use to understand this type of problem??

Thanks!

The way to approach this is to set up two equations, one for the net force acting on each mass.

In regards to the first question,

the equation for the mass is Fnet = Fg (force of gravity acting on mass) - T (tension in the rope)
the equation for the woman is Fnet = T (because only T is the acting force assuming no friction)

Set both equations equal to each other: T = Fg - T, this will become 2T = mg = 500N, so T = 250N

In regards to the second question,

for the mass on the right side of the pulley, Fnet = 2mg - T
for the mass on the left side of the pulley, Fnet = T - mg

Setting both equations equal to each other: 2mg - T = T - mg, solving for T gives you T = 3/2mg

If you're not provided with a diagram, make sure to draw one to have a better understanding of how forces work together to produce Fnet.

#### Jepstein30

##### Full Member
The way to approach this is to set up two equations, one for the net force acting on each mass.

In regards to the first question,

the equation for the mass is Fnet = Fg (force of gravity acting on mass) - T (tension in the rope)
the equation for the woman is Fnet = T (because only T is the acting force assuming no friction)

Set both equations equal to each other: T = Fg - T, this will become 2T = mg = 500N, so T = 250N

In regards to the second question,

for the mass on the right side of the pulley, Fnet = 2mg - T
for the mass on the left side of the pulley, Fnet = T - mg

Setting both equations equal to each other: 2mg - T = T - mg, solving for T gives you T = 3/2mg

If you're not provided with a diagram, make sure to draw one to have a better understanding of how forces work together to produce Fnet.

Ah awesome!

That makes so much sense and is much easier than I was making it out to be. Going to look at both of these problems again and hopefully find some more material of this sort..

#### Jepstein30

##### Full Member
Wait, for the second problem the answer is 1.3mg. Sorry I didn't provide that up there.

The answer key involves 'ma' .

Guessing we do that here but not the problem before because the other one we can assume was in equilibrium (i.e. holding the mass, mass isn't moving) while this one is just the picture (i.e. there's nothing explicitly saying its in equilibrium?)

So it really should be:

Left weight: T - mg = ma
Right weight: 2mg - T = ma2
Since the weight on the right is 2x the one on the left, it will accelerate due to gravity 2x as much?

2T - 2mg = 2ma
2mg-T = 2ma
2T - 2mg = 2mg - T
3T= 4mg
T = 1.3mg

EDIT: math wrong.. I get 1.3 but can someone confirm the use of 'ma' in this problem and how i just made the acceleration of the second weight 2x the first? little shaky on that still.

ALSO: EK 1001 has a follow-up question on this pulley problem

245. In the question above, as the mass becomes infinitely large, the tension in the rope becomes:
A) 0 N
B) 500 N
C) 1000 N
D) infinite

I think it'll become infinite.. but the answer key says otherwise (the explanation also talks about a man though so it's clear that it has no clue what it's talking about)

Last edited:

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
Wait, for the second problem the answer is 1.3mg. Sorry I didn't provide that up there.

The answer key involves 'ma' .

Guessing we do that here but not the problem before because the other one we can assume was in equilibrium (i.e. holding the mass, mass isn't moving) while this one is just the picture (i.e. there's nothing explicitly saying its in equilibrium?)

So it really should be:

Left weight: T - mg = ma
Right weight: 2mg - T = ma2
Since the weight on the right is 2x the one on the left, it will accelerate due to gravity 2x as much?

2T - 2mg = 2ma
2mg-T = 2ma
2T - 2mg = 2mg - T
3T= 4mg
T = 1.3mg

EDIT: math wrong.. I get 1.3 but can someone confirm the use of 'ma' in this problem and how i just made the acceleration of the second weight 2x the first? little shaky on that still.

ALSO: EK 1001 has a follow-up question on this pulley problem

245. In the question above, as the mass becomes infinitely large, the tension in the rope becomes:
A) 0 N
B) 500 N
C) 1000 N
D) infinite

I think it'll become infinite.. but the answer key says otherwise (the explanation also talks about a man though so it's clear that it has no clue what it's talking about)

you are right. Sorry I made a mistake in my previous post when I assumed that both Fnet's are equal to each other, when in fact, one is equal to twice that of the other. Fnet is ma, so for the heavier mass, Fnet is equal to 2ma. However, in the previous question the mass of the woman was equal to the mass of the object and that made their Fnet equal

#### Jepstein30

##### Full Member
you are right. Sorry I made a mistake in my previous post when I assumed that both Fnet's are equal to each other, when in fact, one is equal to twice that of the other. Fnet is ma, so for the heavier mass, Fnet is equal to 2ma.

No problem, makes total sense to me now! Except in the first problem, it doesn't even matter what the women's mass is.. no? Because the only acceleration possible is to the left off the cliff due to T. A follow-up question increases the mass of the woman to 100 kg and has the same answer.

Any thoughts on the second question I posted that EK has the wrong answer key for? Since we know T = 1.3mg .. as m increases to infinite, so will T?

But given that gravity is the force doing the accelerating here.. will the acceleration not be able to top 9.8 m/s^2 ? A similar question asked about the acceleration and I correctly chose that as the mass increases towards infinite, the acceleration becomes 10 m/s^2. Not so sure about the force though since technically there is no upper limit on the force: f = mg

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
No problem, makes total sense to me now! Except in the first problem, it doesn't even matter what the women's mass is.. no? Because the only acceleration possible is to the left off the cliff due to T. A follow-up question increases the mass of the woman to 100 kg and has the same answer.

Any thoughts on the second question I posted that EK has the wrong answer key for? Since we know T = 1.3mg .. as m increases to infinite, so will T?

But given that gravity is the force doing the accelerating here.. will the acceleration not be able to top 9.8 m/s^2 ? A similar question asked about the acceleration and I correctly chose that as the mass increases towards infinite, the acceleration becomes 10 m/s^2. Not so sure about the force though since technically there is no upper limit on the force: f = mg

In regards to the woman's mass, the only force acting on her in the x direction is T. Her weight only acts in the y direction. If friction is to be considered, then yes, the more massive she is greater the opposing force to T friction will be.

In regards to the second question, is the answer zero?

#### Jepstein30

##### Full Member
In regards to the woman's mass, the only force acting on her in the x direction is T. Her weight only acts in the y direction. If friction is to be considered, then yes, the more massive she is greater the opposing force to T friction will be.

In regards to the second question, is the answer zero?

Right, so if no friction, the woman can weigh 1 kg or 1000 kg, doesn't change the T. With friction, the weight comes into play because N = mg and f= Nu. So more mass = more force opposing T.

No clue as to the 2nd question since EK's answer key is clearly mistaken. It says C but references a man from an earlier problem in the explanation.

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
Right, so if no friction, the woman can weigh 1 kg or 1000 kg, doesn't change the T. With friction, the weight comes into play because N = mg and f= Nu. So more mass = more force opposing T.

No clue as to the 2nd question since EK's answer key is clearly mistaken. It says C but references a man from an earlier problem in the explanation.

My first hunch would be infinite as well. But when I thought of it, I'm leaning now toward zero more. If m is increased to infinity, then 2m will be infinity as well. The 2 factor won't matter any more. This will cause the system to be at equilibrium and the net force will be zero. I don't understand how the answer could be 1000N when the question did not specify the mass.