Magnitude of Kinetic Frictional Force on Fireman Sliding Down Pole

[lalalah]

The alarm at a fire station rings and a 96.0-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.55 m). Just before landing, his speed is 1.79 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?



Relevant equations:
I'm not sure, but I believe this has something to do with Newton's second law of motion? So this may be helpful {\SigmaFnet }/mass = acceleration , although I couldn't see a way to use it myself

I mainly used: y = (v²_y - v²_oy)/(2a_y), where a_y can be substituted with a_y= (-f_k)/mass

The Attempt at a Solution

Sooo, my messy and essentially incorrect attempt:

y = -3.55 m
v₀ = 0 m/s
v_oy = 1.79 m/s
mass = 96.0 kg

using the equation for the y-displacement, I attempted to find f_k:

-3.55 m = (1.79² - 0 m/s) / (2(-f_k)/96kg)

multiplying both sides by -(2/96) fk ----> .07396 fk = 1.79^2

fk = 43.3 N


if someone could point me in the right direction i would be terribly grateful!

 

[Doc Al]

Solve this in two stages. First find the acceleration, using the kinematic formula that you stated. Then apply Newton's 2nd law to find the unknown friction force. What forces act on the fireman?

 

[lalalah]

i found the acceleration to be -.04512, but I'm confused right now about \Sigma<br /> Fy and -fk.<br /> <br /> if \Sigma<br /> Fy = ma_y and a_y = -fk/mass, then are these two force values the same?

 

[Doc Al]

i found the acceleration to be -.04512,

Recheck that calculation. (Pay attention to the decimal point.)

When using \Sigma F = m a, \Sigma F represents the net force on the man. What forces act on the man?

 

[lalalah]

he is being acted on by a friction force from the pole... which is slowing down his acceleration

 

[lalalah]

oh sorry! i meant to type -0.4512 m/s^2!

in that case, would it be correct of me to say that the net force and not the kinetic frictional force is 43.3 N ?

 

[lalalah]

forces which act on the man:
gravity, kinetic frictional force

 

[Doc Al]

he is being acted on by a friction force from the pole... which is slowing down his acceleration

Good. That's one of the forces acting on him. What's the other?

oh sorry! i meant to type -0.4512 m/s^2!

Good. (I assume you made it negative to indicate that the acceleration acts down.)

in that case, would it be correct of me to say that the net force and not the kinetic frictional force is 43.3 N ?

Correct.

 

[Doc Al]

forces which act on the man:
gravity, kinetic frictional force

Good!

Now write an equation. Add up the forces to find the net force (direction counts!) and set it equal to "ma".

 

[lalalah]

So, Fnet = -W + f_k = ma = 43.3 N?

= -(mg) + f_k = ma = 43.3 N
= -(96 kg)((9.8 m/s) + f_k = 43.3 N
= f_k = 940.8 N

does this look right? it doesn't seem right to me...

 

[Doc Al]

So, Fnet = -W + f_k = ma = 43.3 N?

Careful with signs. Which way does the net force point?

 

[lalalah]

downward.
so, would it be
Fnet = -W + f_k = ma = -43.3 N?

so... f_k = (96 kg)(9.8 m/s) -43.3 N

and f_k = 897.5 N ?

i hope its right :x

 

[Doc Al]

Looks good.

 

[lalalah]

hoorayy~! and yet i would never have expected the number to come out so big... thank you for your help!