What is the magnitude of the upward force of friction

[kumar_23]

hi guys

can anyone help me with this?
a tree house has a vertical "fire pole" of smooth metal, designed for quick exits. a child of mass 35.7 kg slides down the pole with constant acceleration, starting from rest. the pole is 3.10 m high. the trip to the ground takes 2.00 s. what is the magnitude of the upward force of friction exerted by the pole on the child?

please help I am really confuzzeled

 

[cristo]

Since this is a homework question, please show some work before we can help you.

 

[kumar_23]

unknown = a d = v1(time) + 0.5(a)(time)2 ...time is squared
known
time= 2.00s a = 1.55m/s/s
distance = 3.10 m
v1= 0 m/s

 

[kumar_23]

this is what i have so far. what do i do after this? please help

 

[kumar_23]

can someone help me

 

[hage567]

Use Newton's second law to sum up the forces. Solve for the frictional force.

 

[kumar_23]

help

what do you mean?
what am i adding together?

 

[kumar_23]

i don't understand because right now i have the net force and acceleration.

 

[kumar_23]

can u please reply? thanks so much for the help =)

 

[hage567]

What forces are present? Can you tell me that? And their direction.

 

[kumar_23]

umm...there is net force and frictional force and that's it.

 

[hage567]

No, the net force is the difference between these two other forces. What is causing the kid to go down the pole in the first place?

 

[kumar_23]

sorry for the wait. i don't understand what you're asking?

 

[kumar_23]

"a tree house has a vertical "fire pole" of smooth metal"

 

[hage567]

I'm asking what forces are acting on the child as he goes down the pole. The is the frictional force, and there is one more. Can you tell me what it is? (What makes things go towards the Earth?)
Do you know how to draw free body diagrams?

 

[hage567]

OK, so you have gravity acting downwards, right? The frictional force acts which way?

 

[kumar_23]

Eureka! gravity is the force and yes, i know how to draw FBDs.

 

[kumar_23]

sorry about the repost.
frictional force is upwards away from the object

 

[hage567]

Yes, that's right. So what is your net force then?

 

[kumar_23]

frictional force minus gravity (not sure)

 

[hage567]

Yes, that's the right idea. So use that in Newton's second law, solving for the frictional force.

 

[hage567]

You don't know Newton's second law? Does F=ma look familiar?

 

[kumar_23]

sorry I'm a little slow at this because it's very new to me

 

[kumar_23]

yes but how is being used with this situation ?

 

[kumar_23]

doesn't f=ma have to do with acceleration and mass multplied together to give us fnet?

 

[kumar_23]

are you still there?

 

[hage567]

You said that "frictional force minus gravity" is your net force. I would change this to say "gravitational force - frictional force" (since I would call "down" the positive direction).
So on one side of your equation you will have "gravitational force - frictional force" and the other you will have "ma". Do you know how to calculate the gravitational force? You already know m and a. So isolate the frictional force (call it "f"), and solve for it.

 

[kumar_23]

It asks " what is the magnitude of the upward force of friction exerted by the pole on the child? " would that have direction and would it be in Newtons?

 

[cristo]

kumar, your replies seem to just be whatever you're thinking at that time. Consider the question before composing a reply.

It asks " what is the magnitude of the upward force of friction exerted by the pole on the child? " would that have direction and would it be in Newtons?

What do you think? Does it have a direction (or is the direction already specified in the question?). What is force measured in?

 

[hage567]

Yes, it would be in Newtons. Usually if a question just asks for the "magnitude" it does not care about the direction. In this case, it is stated right in the question that it is an upwards force, so I would assume you could just leave it at that.

 

[kumar_23]

would i have to islolate the f and use the upward direction and it would be in Newtons? i knew tha. sorry

 

[kumar_23]

thanks for you help. i really appreciated it. thanks very much

 

[hage567]

You're welcome.

 

[kumar_23]

can u just give me a hint on how to get started on another question as well? thanks so much

A 4.0*10 kg wind sled is gliding across a frozen lake with a constant velocity of 12 m/s (E) when a gust of wind from the southwest exerts a constant force of 1.0 * 10to the power of 2 on its sails for 3.0 s. with what velocity will the sled be moving after the wind has subsided?

can you just explain the question to me?

i won't take too much time but all i need is a hint.!=)

are you a physics teacher or some genius? because you explain very well.

 

[hage567]

What part are you having trouble with? Explain what you think it is saying.

 

[kumar_23]

is it asking for initial velocity? this is a tough one!

 

[hage567]

No, it has given you the initial velocity. What you need to find is the new velocity after the force of the wind has acted on the wind sled.

 

[kumar_23]

how do we figure that out? i think a free body diagram will be helpful in problem.
then, do we have to break the two directions into components?

 

[hage567]

Well, you know the force acting. What do you need to find the new velocity? Think Newton's second law.

 

[hage567]

I don't think you need to break it up into components.

 

[kumar_23]

do you use fnet= ma? or do i need to find a?

 

[hage567]

do you use fnet= ma? or do i need to find a?

I'm not too sure what you mean by that, but yes, this is the equation you need. Make sure you understand what the equation means.
One way to figure out what you're supposed to do with it is look at the three variables in that equation, which is the one you don't already know?

 

[kumar_23]

is it fnet i need to figure out ? do i use "constant velocity of 12 m/s (E) divided by 3.0 s. " to find acceleration ? then i figure out fnet by using fnet=ma after i figure out my acceleration.

****i'm not sure if this is how it's supposed to be done?****

 

[hage567]

READ the question! Write the information you have down in the proper form with the proper units so you can see what you have available to you.
The question states:

"a gust of wind from the southwest exerts a constant force of 1.0 * 10to the power of 2 on its sails"

This is your net force. So try to think of what you don't know but need to solve this problem. (Hint: you can find it by Newton's second law, it's the only thing you don't know in that equation)

 

[kumar_23]

sorry, i didnt notice the constant force part of the question
now, that tells me that i need to find acceleration using fnet = ma.
Do i then use the acceleration equation a = change in velocity /time. Do i then find final velocity by rearranging the equation?

 

[hage567]

that tells me that i need to find acceleration using fnet = ma. That's right!
Do i then use the acceleration equation a = change in velocity /time. Do i then find final velocity by rearranging the equation?

Yes, that should work.

 

[kumar_23]

so I am correct for the first time!

 

[kumar_23]

one more thing, is the initial velocity 0 m/s or 12 m/s in the equation?

 

[hage567]

What does the question say?

"A 4.0*10 kg wind sled is gliding across a frozen lake with a constant velocity of 12 m/s (E)"

Doesn't look like a zero to me! It says it's moving at a constant velocity, so you assume that's the initial velocity.

 

[kumar_23]

:shy: oh...the light bulb just went on!