Applying Newtons Laws: Incline, static friction, kid on sled

  • #1

Homework Statement


[/B]
A child sits on a sled (50kg) on a snowy hill with an incline of 5 degrees. The coefficient of static friction is 0.05. What downward force must a parent apply parallel to the normal force, to prevent the child from sliding down the hill? (Answer: 366N)


Homework Equations



Ffs= MsFn[/B]


The Attempt at a Solution


[/B]
I drew out the problem on an x-y axis.

Ffs(-x direction)
Fn(+y direction)
Fa(+x direction)
Fg(-y direction, x direction)
Fgx(-y, +x direction)
Fgy(-y direction)

Looks like this ^ when mapped out. I then separated equations based on both directions

Fg=Ma
Fg= 50kg⋅9.8m/s^2
Fg= 490N

x direction
-Ffs+Fgx+fa=Ma,x
-Ffs+Fg(sinθ)= 0m/s^2

y direction

Fn-Fgy=Ma,y
Fn-Fg(cosθ)= 0m/s^2
Fn= Fg(cosθ)

Solving
  • Fn= 490N(cos5)
Fn= 488N

  • Ffs= .05(488N)
Ffs=24.4N

  • -Ffs+fg(sinθ)+fa= 0m/s^2
-24.4N+490N(sin5)+fa=0


Fa= 67N ...... What did i do wrong here?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


[/B]
A child sits on a sled (50kg) on a snowy hill with an incline of 5 degrees. The coefficient of static friction is 0.05. What downward force must a parent apply parallel to the normal force, to prevent the child from sliding down the hill? (Answer: 366N)

Homework Equations


[/B]
Ffs= MsFn

The Attempt at a Solution


[/B]
I drew out the problem on an x-y axis.

Ffs(-x direction)
Fn(+y direction)
Fa(+x direction)
Fg(-y direction, x direction)
Fgx(-y, +x direction)
Fgy(-y direction)

Looks like this ^ when mapped out. I then separated equations based on both directions

Fg=Ma
Fg= 50kg⋅9.8m/s^2
Fg= 490N

x direction
-Ffs+Fgx+fa=Ma,x
-Ffs+Fg(sinθ)= 0m/s^2

y direction

Fn-Fgy=Ma,y
Fn-Fg(cosθ)= 0m/s^2
Fn= Fg(cosθ)

Solving
  • Fn= 490N(cos5)
Fn= 488N
  • Ffs= .05(488N)
Ffs=24.4N
  • -Ffs+fg(sinθ)+fa= 0m/s^2
-24.4N+490N(sin5)+fa=0

Fa= 67N ...... What did i do wrong here?
It would help a lot if you would describe the variables.

I guess the following:
Ffs : The friction (static) force.
Fn : The normal force
Fa : Not sure here. Maybe the force supplied by the parent ?
Fg : The force of gravity.​

Reading the problem statement again, it says " What downward force must a parent apply parallel to the normal force, ... "

It also appears that the x-axis is parallel to the surface of the hill and is inclined at 5° from the horizontal, and the the y-axis is perpendicular to that and upward at an angle of 5° from the vertical.

If I am right regarding the above things, then you have the direction of the force supplied by the parent wrong.
 
  • #3
It would help a lot if you would describe the variables.

I guess the following:
Ffs : The friction (static) force.
Fn : The normal force
Fa : Not sure here. Maybe the force supplied by the parent ?
Fg : The force of gravity.​

Reading the problem statement again, it says " What downward force must a parent apply parallel to the normal force, ... "

It also appears that the x-axis is parallel to the surface of the hill and is inclined at 5° from the horizontal, and the the y-axis is perpendicular to that and upward at an angle of 5° from the vertical.

If I am right regarding the above things, then you have the direction of the force supplied by the parent wrong.
Fa means the force applied. I think thats what im looking for here.
 
  • #4
SammyS
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Fa means the force applied. I think thats what im looking for here.
That's kind of what I thought.

Do you see why you got it wrong?
 
  • #5
That's kind of what I thought.

Do you see why you got it wrong?
No :| im trying to figure it out
 
  • #6
SammyS
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As I said in Post #2, my first post :
...

If I am right regarding the above things, then you have the direction of the force supplied by the parent wrong.
( I was referring to Fa there. )

What does the problem statement say about the direction of the applied force?
 
  • #7
As I said in Post #2, my first post :

( I was referring to Fa there. )

What does the problem statement say about the direction of the applied force?
Preventing the kid from slinding down the hill. so the force is applied downward... oh i see.
 
  • #8
SammyS
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Preventing the kid from slinding down the hill. so the force is applied downward... oh i see.
Yes, in the (negative) y direction, directly opposite the direction of the normal force.
 
  • #9
Yes, in the (negative) y direction, directly opposite the direction of the normal force.
got it.

x-direction
-Ffs+Fgx=Ma,x
-Ffs+Fg(sinθ)=0m/s^2
-Ffs= Fg(sinθ)
-Ffs= 490sinθ
Ffs= 42.7N

  • Ffs=MsFn
Fn=42.7/.05
Fn= 854N

y-direction
Fn-Fgy-Fa=Ma,y
Fn-Fg(cosθ)-fa=0m/s^2
Fn= Fg(cosθ)+Fa
854N=490cos5+fa
Fa=854N-488N
Fa=366N
 
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