Wind drag problem with a ball hanging on a rope

[crememars]

Homework Statement

A 1.2 kg ball is hanging from the end of a rope. The rope hangs at an angle of 25 degrees from the vertical when a 15.0m/s horizontal wind is blowing. If the wind’s force on the rope is negligible, what drag force does the wind exert on the ball?

Relevant Equations

the answer is 5.84 N

I drew a FBD but I feel like it's wrong because there are too many missing values. I tried this:

Fy = 0
Tcosθ - Fg = 0
Tcosθ = mg
T = (1.2)(9.8) / cos(25)
T = 12.98 N

Fx = ma
Fwind - Tsinθ - Fdrag = ma
Fwind - (12.98)(sin25) - Fdrag = ma

I don't know how to find these missing values. I feel like I'm approaching the situation wrong? I saw one solution solve it like this:

g(tanθ) = a
(9.8)(tan25) = 4.57

Fx = ma
Fx = (1.2)(4.57)
Fx = 5.48 N

but this doesn't make any sense to me.

any help would be appreciated, thank you !

 

[topsquark]

Homework Statement:: A 1.2 kg ball is hanging from the end of a rope. The rope hangs at an angle of 25 degrees from the vertical when a 15.0m/s horizontal wind is blowing. If the wind’s force on the rope is negligible, what drag force does the wind exert on the ball?
Relevant Equations:: the answer is 5.84 N

View attachment 315675
I drew a FBD but I feel like it's wrong because there are too many missing values. I tried this:

Fy = 0
Tcosθ - Fg = 0
Tcosθ = mg
T = (1.2)(9.8) / cos(25)
T = 12.98 N

Fx = ma
Fwind - Tsinθ - Fdrag = ma
Fwind - (12.98)(sin25) - Fdrag = ma

I don't know how to find these missing values. I feel like I'm approaching the situation wrong? I saw one solution solve it like this:

g(tanθ) = a
(9.8)(tan25) = 4.57

Fx = ma
Fx = (1.2)(4.57)
Fx = 5.48 N

but this doesn't make any sense to me.

any help would be appreciated, thank you !

You have too many forces on your diagram. The force from the wind is the force of the drag. So you only have the force from the wind acting to the right.

-Dan

 

[crememars]

You have too many forces on your diagram. The force from the wind is the force of the drag. So you only have the force from the wind acting to the right.

-Dan

I think I understood, thank you! so the drag would be equal to the x component of the tension.

 

[topsquark]

I think I understood, thank you! so the drag would be equal to the x component of the tension.

Yes.

-Dan

 

[Cassie97]

You have too many forces on your diagram. The force from the wind is the force of the drag. So you only have the force from the wind acting to the right.

-Dan

Would that not make it 4.97 N if it is equal to the x-component of the tension in the string?

 

[nasu]

No.
But if you show how you get this value, someone may figure out the error.

 

[David Lewis]

It helps to draw your diagram approximately to scale. Then you can tell if the answer you came up with is in the ballpark (sanity check). If you draw the correct arrangement of the 1.3kg force vector on the ball, the drag vector you are solving for, and the 25-degree angle in the right place, that will be everything you need.