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

In summary: Fa=...Fa means the force applied. I think that's what I am looking for here.Fa means the force applied. I think that's what I am looking for here.Got it.
  • #1
DracoMalfoy
88
4

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?
 
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  • #2
DracoMalfoy said:

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
SammyS said:
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 that's what I am looking for here.
 
  • #4
DracoMalfoy said:
Fa means the force applied. I think that's what I am looking for here.

That's kind of what I thought.

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

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

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
SammyS said:
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
DracoMalfoy said:
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
SammyS said:
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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Likes SammyS

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

1. How do Newton's laws apply to objects on an incline?

Newton's laws can be applied to objects on an incline by considering the forces acting on the object. The component of the weight of the object parallel to the incline is countered by the force of static friction, while the component perpendicular to the incline is countered by the normal force. These forces interact to determine the acceleration of the object down the incline.

2. What is the role of static friction in Newton's laws?

Static friction is a force that acts to prevent two surfaces from sliding past each other when there is no relative motion between them. In the context of Newton's laws, static friction is important because it can counteract other forces, such as the weight of an object on an incline, and allow for an object to remain at rest or move at a constant velocity.

3. How does the weight of a kid on a sled affect the sled's motion?

The weight of a kid on a sled is a force that acts in the downward direction. According to Newton's second law, this force will cause the sled to accelerate in the same direction, unless there is an opposing force, such as static friction. The weight of the kid on the sled will also affect the normal force and the force of static friction acting on the sled.

4. Can an object on an incline ever have a net force of zero?

Yes, an object on an incline can have a net force of zero if it is at rest or moving at a constant velocity. In this case, the forces acting on the object are balanced, with the force of gravity being counteracted by the force of static friction. This is known as mechanical equilibrium and is described by Newton's first law.

5. How does the angle of an incline affect the acceleration of an object?

The angle of an incline can affect the acceleration of an object by changing the magnitude and direction of the forces acting on the object. As the angle of the incline increases, the component of the weight acting parallel to the incline also increases, resulting in a greater force of static friction. This can lead to a greater acceleration of the object down the incline.

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