A father spins his child on a cart

[Cc518]

Homework Statement

The father stands at the summit of a conical
hill as he spins his 20 kg child around on a 5.0 kg cart with a
2.0-m-long rope. The sides of the hill are inclined at 20 degrees. He
keeps the rope parallel to the ground, and friction is negligible.
What rope tension will allow the cart to spin with the same
14 rpm it had in the example?

Homework Equations

F=mrw^2

The Attempt at a Solution

I drew the fbd diagram, and I got two equation :1. Fny +Ty=mg and 2. Tx-Fnx=mrw^2
where Fnx and Fny is the x, y-component of the normal force, Tx and Tx is the x, y-component of the Tension.
But I don't know how to solve for Fn

Can anyone help me?
The link to the image: https://cloud.smartdraw.com/share.aspx/?pubDocShare=3584F28114431EA1F27BA700C6061213131

 

[haruspex]

You have two unknowns, Fn and T. You want to find T, not Fn. use one equation to eliminate Fn from the other.

 

[Cc518]

Thank you for reply!
I got tention as 179.2n, but the answer I got from other sources is 185n.
The fbd they used is different than mine though.

https://cloud.smartdraw.com/share.aspx/?pubDocShare=3584F28114431EA1F27BA700C6061213131
They said Tention= mrw^2-mgx
Which one is correct?

 

[haruspex]

Thank you for reply!
I got tention as 179.2n, but the answer I got from other sources is 185n.
The fbd they used is different than mine though.

https://cloud.smartdraw.com/share.aspx/?pubDocShare=3584F28114431EA1F27BA700C6061213131
They said Tention= mrw^2-mgx
Which one is correct?

I get 178.7N.
Did you mean they get mrω²+mg_x?
That overlooks that the rotation is in the horizontal plane, so need to take the upslope component of mrω² to align it with the tension. And r here, of course, is the radius of the rotation, not the length of the rope.

 

[Cc518]

I get 178.7N.
Did you mean they get mrω²+mg_x?
That overlooks that the rotation is in the horizontal plane, so need to take the upslope component of mrω² to align it with the tension. And r here, of course, is the radius of the rotation, not the length of the rope.

Sorry, I meant mrω2+mgx actually.
So is the rotational plane horizontal or is it parallel with the incline of the slope?
In other words, does the Tx act as centripetal force or does T act as the centripetal force?
Thank you :)

 

[haruspex]

is the rotational plane horizontal or is it parallel with the incline of the slope?

What do you think?

 

[Cc518]

What do you think?

I think it should be horizontal because if the plane is parallel to the incline of the slope, the object, in this case, the cart, will move up the slope, which is not what happened.

 

[haruspex]

I think it should be horizontal because if the plane is parallel to the incline of the slope, the object, in this case, the cart, will move up the slope, which is not what happened.

Right.

 

[Cc518]

Right.

Thank you so much for your help! I really appreciated it!

 

[haruspex]

does the Tx act as centripetal force

It contributes to the centripetal force. The normal force also affects it.

 

[Cc518]

It contributes to the centripetal force. The normal force also affects it.

Oh, yeah. Thank you for the reminder :)