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Inclined plane FNET

  • #1

Homework Statement


A sight seen on many bunny hills across ontatrio is young skiers pushing on ski poles and gliding down a slope untill they come to a rest . Obeserving from a distance , you note a young person (approximatly 25kg) pushing off with the ski poles to give herself an intial velocity of 3.5m/s. If the inclination of the hills is 5 degress and hte coefficient of kinetic friction for the skis on dry snow is 0.20 calculate.

A) the time taken for the skier to come to a stop



Homework Equations



solving for t
a=v/t
solving for a
a=Fnet/m


The Attempt at a Solution








vi-3.5m/s m=25kg mu=0.20 angle=5degress

I'm assuming that the skier without velocity would be stationary on the inclination.
This is my first problem, is she accelerating down the ramp or moving at a constant velocity

Fg =mg=25kg(9.8m/s^2)=245N
Fn=Fg-Fy=mg-mgsin5=25kg(9.8m/s^2)-25kg(9.8m/s^2)sine5=224N

Fg=245N
Fn=224N


Ff=muFn=0.20mgcos5=49N

49 newtons on the yaxis

This is where I'm confused
I have another equation that goes like this

Fa=Fgsine5 + muFn
=masine5+mumascos5
=ma(sine +mucos5)
-25kg(9.8m/s^2)(sine5+0.20(cos5))
=70N

I'm assuming this equation accounts for acceleration, subtracting the Ff.

Fnet=x+y=21+70
=91N

a=91/25=3.64m/s^2
t=3.5m/s/3.64m/s^2=0.9s

This is wrong


oh and can someoen give me a more detailed explaantion what Fnormal is. I"m assuming it is the x component Fg subtracted from Fg.
 
Last edited:

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
Welcome to PF.
Fn=Fg-Fy=mg-mgsin5=25kg(9.8m/s^2)-25kg(9.8m/s^2)sine5=224N
Not quite right.

Your Fnormal is the normal component of Fg which is m*g*Cos5 which makes the retarding force from friction μ*m*g*Cos5. Then you have the component down the incline of m*g*Sin5

So your acceleration is g*sin5 - μ*g*Cos5.

V = a*t so ...

t = 3.5/(g*sin5 - μ*g*Cos5)
 
  • #3
ohhhhhh

Now I theaoratically understand, We are breaking the slope into x and y compents.

y= masine5
x=macos5

friciton lies in the x axis, we are subtracting the force from the y axis
ohh so it's a negative because thier is no accelleration.

t = 3.5/(g*sin5 - μ*g*Cos5)
Does this include mass

Thanks,
Do you have any good sites ti do more of these types of examples, generally working with x and y components?
 
  • #4
t = 3.5/(g*sin5 - μ*g*Cos5)


This piece
(g*sin5 - μ*g*Cos5)

I'm suppose to do algebrai expression

mg(sin5-μCos5)

If I aplly ur method, a= -1 meaning t=3.5s

If I do algbrai method it's 29 t-0.1
I guess what I'm saying is that the numebers do not seem right, but I give u the bebefit of the doubt..
 
Last edited:
  • #5
LowlyPion
Homework Helper
3,090
4
t = 3.5/(g*sin5 - μ*g*Cos5)


This piece
(g*sin5 - μ*g*Cos5)

I'm suppose to do algebrai expression

mg(sin5-μCos5)

If I aplly ur method, a= -1 meaning t=3.5s

If I do algbrai method it's 29 t-0.1
I guess what I'm saying is that the numebers do not seem right, but I give u the bebefit of the doubt..
Well fwiw I get a = -1.1

I don't understand your saying "29 t-0.1". Where does that come from?

The acceleration you should note in this case is independent of mass.
 
  • #6
Oh thank you , for answering
ok
i pluged in algbraic expression

Fa=mgsine5-o.20mgcos5
=mg(sine5-0.20cos5)
=29

t=v/a=3.5/29=0.12


The acceleration you should note in this case is independent of mass. I'll let that one sink in
oh so if the object was accelerating, it would be dependant on mass, got it.

well, hmm, i'm going to go watch a youtube vid on skiing ... so i know the dynamics better
thanks again
 
  • #7
LowlyPion
Homework Helper
3,090
4
Oh thank you , for answering
ok
i pluged in algbraic expression

Fa=mgsine5-o.20mgcos5
=mg(sine5-0.20cos5)
=29

t=v/a=3.5/29=0.12
The acceleration is independent of mass because the force affecting the acceleration is also proportional to mass. The mass cancels out completely.

Your expression (3.5/29) is incorrect.

You have equated force in Newtons to acceleration. Now that I see that 29 is your force, then you must still divide the 29 by the 25, which makes your result 1/25 of what it should be.
 
  • #8
so
t=3.5/(25/29) =4s


wait hmm

25kg/29N =0.86m/s^2

t=v/a
t=3.5/0.86=4s

This sounds more realistic


DAMN UR OFFLINE!
 
  • #9
LowlyPion
Homework Helper
3,090
4
so
t=3.5/(25/29) =4s
wait hmm

25kg/29N =0.86m/s^2

t=v/a
t=3.5/0.86=4s
This sounds more realistic
DAMN UR OFFLINE!
No. Your a = F/m

a = 29/25.

But I would disagree with your calculation for 29N. I get 27.5N

That yields 27.5/25 = 1.1

Oh. wait 1.1 .

So t = 3.5/1.1, as I suggested a few posts ago.
 
  • #10
oh right I completly forgot about a=Fnet/m

since she their is no acceleration, both forces on the y axis cancel eacthother out. Fa account for friction in the equation leaving Fnet=27.3.

a=Fnet/m=27/25=1.1m/s^2

t=v/a=3.5/1.1=3.1s

phew ^.^
thanks(I feel like I'm walking in circles when I do formulas that cancel eachother out)
 

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