Firefighter sliding down a pole

[petrichortea]

Homework Statement

A firefighter whose weight is 812 N is sliding down a vertical pole,
her speed increasing at the rate of 1.8 m/s2. Gravity and friction are the two significant forces acting on her. What is the magnitude of the frictional force?

Relevant Equations

w=mg
a=F-net/m
F-net=F-fk

First, I solved for the firefighters mass and got 82.9kg.
Then, I plugged in the mass and the 1.8m/s2 acceleration into a=Fnet/m
This is where I get confused. I think I'm supposed to use Fnet = F-fk to solve for fk (and that would be the answer?) but I don't know what to input for F.

 

[Lnewqban]

Welcome, @petrichortea !
What force is pulling the man downwards?
What force is resisting that downwards movement?

 

[petrichortea]

Welcome, @petrichortea !
What force is pulling the man downwards?
What force is resisting that downwards movement?

Gravity is pulling down.
N is resisting, I think. Which is normal force, according to my class notes

 

[hmmm27]

Gravity is pulling down.
N is resisting, I think. Which is normal force, according to my class notes

Irrelevant to the question, which outright gives you the downwards force, the net downwards acceleration, and asks for the upwards force (which, in this case is provided by friction, but could just as well be rocket boots or something).

 

[erobz]

Gravity is pulling down.
N is resisting, I think. Which is normal force, according to my class notes

Not quite on the second part.

 

[kuruman]

Gravity and friction are the two significant forces acting on her. What is the magnitude of the frictional force?

This is where I get confused. I think I'm supposed to use Fnet = F-fk to solve for fk (and that would be the answer?) but I don't know what to input for F.

The net force is the vector sum of all the forces acting on the firefighter.
You are told that gravity and friction are the two significant forces acting on her.
If Fnet = F - fk and fk is friction, what could F possibly be?

 

[Lnewqban]

Gravity is pulling down.
N is resisting, I think. Which is normal force, according to my class notes

Friction against the pole is what is preventing gravity to accelerate the firefighter to slide downwards in free fall (acceleration of 9.81 m/s2).
Why is the downwards acceleration the given 1.8 m/s2?

Normal force and coefficient of friction of hands and legs against the pole create that friction force that slows the fall down.
In this problem, the question is the magnitude of that friction force.

What happens if that force becomes bigger and bigger until being as big as the weight of the person?

 

[haruspex]

N is resisting, I think. Which is normal force

A normal force exerted by one body on another is defined to be the contact force at right angles ("normal ") to their contact surface. The pole is vertical, so the normal force it exerts on the firefighter is horizontal. It cannot resist a vertical force such as gravity.

The frictional force of contact is parallel to the surfaces, so in this case is vertical and will resist gravity.

 

[Orodruin]

Here are the two big hints:

Gravity and friction are the two significant forces acting on her.

So keep that in mind.

Then, I plugged in the mass and the 1.8m/s2 acceleration into a=Fnet/m

Ok, so you have the net force Fnet.

This is where I get confused. I think I'm supposed to use Fnet = F-fk to solve for fk (and that would be the answer?) but I don't know what to input for F.

Fnet is the total force and fk friction. What force other than friction has been mentioned as relevant? (See above.)

What force is pulling the man downwards?

Woman:

… down a vertical pole,
her speed increasing at the rate of 1.8 m/s2.

 

[petrichortea]

So, based on your (all) help. I think I've figured out that F is her weight - but in Newtons not kg. I came up with 662.78 N as an answer. That's 67.6 kg of frictional force acting on her 82.8 kg weight. That's a lot more friction than my gut feeling was expecting but that's what the numbers say so now I'm waiting to find out!
Thank you!

 

[erobz]

What force is pulling the man downwards?

Woman:

… down a vertical pole,
her speed increasing at the rate of 1.8 m/s2.

Call the Fed's! Lnewqban accidentally misgendered the imaginary firefighter in the back story!

 

[Lnewqban]

So, based on your (all) help. I think I've figured out that F is her weight - but in Newtons not kg. I came up with 662.78 N as an answer. That's 67.6 kg of frictional force acting on her 82.8 kg weight. That's a lot more friction than my gut feeling was expecting but that's what the numbers say so now I'm waiting to find out!
Thank you!

Units for force are never kilograms, which is reserved for mass.
You can have a mass, but you will not have a force associated to it, until you see some acceleration.
Gravity is a constantly present acceleration, reason for which it is easy to confuse mass with force in daily life.

The problem is that the firefighter needs to reduce her landing velocity as much as possible for a safe landing.
Commercial elevators don’t accelerate passengers above 1 m/s.s.

 

[petrichortea]

Units for force are never kilograms, which is reserved for mass.
You can have a mass, but you will not have a force associated to it, until you see some acceleration.

I think that's where my confusion comes in - I can imaging Fahrenheit and Celsius, kilograms and pounds...but Newtons a foreign concept so it's hard to "see" it in my minds eye.

 

[haruspex]

I can [imagine] … kilograms and pounds...but Newtons a foreign concept

Or maybe it's the other way about. When you imagine a kg, you are probably thinking about its weight in your hand, but that's not a kg - it's a force of 9.81N. To have a feel for a mass of 1kg you need to think in terms of inertia, of the force needed to accelerate it at some rate across a smooth flat surface. For a given rate of acceleration that would be the same on the moon as on Earth.

 

[Orodruin]

Or maybe it's the other way about. When you imagine a kg, you are probably thinking about its weight in your hand, but that's not a kg - it's a force of 9.81N. To have a feel for a mass of 1kg you need to think in terms of inertia, of the force needed to accelerate it at some rate across a smooth flat surface. For a given rate of acceleration that would be the same on the moon as on Earth.

In other words, what is being imagined is likely the units of kgf (kilogram-force or kilopond) and lbf (pound-force), which are units of weight.

 

[Lnewqban]

I think that's where my confusion comes in - I can imaging Fahrenheit and Celsius, kilograms and pounds...but Newtons a foreign concept so it's hard to "see" it in my minds eye.

Wait until you deal with Pascals (pressure unit = N/m2)!

It is a common confusion in many regions of the world.
I understand you, as I received an education based on the old technical units of kilogram-force, for which the weight of one cubic meter of pure cold water was 1 ton or 1000 kgf.

 

[PeroK]

Homework Statement:: A firefighter whose weight is 812 N is sliding down a vertical pole,
her speed increasing at the rate of 1.8 m/s2. Gravity and friction are the two significant forces acting on her. What is the magnitude of the frictional force?
Relevant Equations:: w=mg
a=F-net/m
F-net=F-fk

First, I solved for the firefighters mass and got 82.9kg.
Then, I plugged in the mass and the 1.8m/s2 acceleration into a=Fnet/m
This is where I get confused. I think I'm supposed to use Fnet = F-fk to solve for fk (and that would be the answer?) but I don't know what to input for F.

I would look at it like this. $812N$ is her weight, which is the force of gravity acting on her. We know that the acceleration of gravity is $g = 9.8 \ m/s^2$. That would be her acceleration in free fall, under the full gravitational force of $812N$.

Now let's try various frictional forces, $f$, and calculate the net acceleration:

$f = 0, \ a = g = 9.8 \ m/s^2$

$f = 81N = 0.1g, \ a = 0.9g = 8.82 \ m/s$

$f = 162N = 0.2g, \ a = 0.8g = 7.84 \ m/s^2$

And, hopefully, you can see what's happening. In order to reduce the acceleration to $1.8 \ m/s^2$, which is about $0.18 g$, we need a frictional force of over 80% of her weight. That must be about $660 N$.