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

[DracoMalfoy]

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?

 

[SammyS]

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.

 

[DracoMalfoy]

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.

 

[SammyS]

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?

 

[DracoMalfoy]

That's kind of what I thought.

Do you see why you got it wrong?

No :| I am trying to figure it out

 

[SammyS]

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?

 

[DracoMalfoy]

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.

 

[SammyS]

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.

 

[DracoMalfoy]

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