Solving Statics Questions: Radio Tower, Fw, Fv, Fh

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In summary: I think it would be wiser to have a positive fixing at the foot of the mast to reduce the chances of the tower moving in the wind.
  • #1
eurekameh
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http://imageshack.us/photo/my-images/217/unledsdb.png/
So in the picture, we have a radio tower supported by a cable. W is the weight of the radio tower, Fw is the force of the wind, and T is the tension from the cable.

I want to clarify if Fv (a vertical force) is also the normal force on the tower from the floor.
I also want to ask why there is Fh, a horizontal force on the tower from the floor. Is this Fh a force of friction because the tower has an impending motion?
 
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  • #2
Fv is the vertical reaction of the ground on the antenna mast.

Fh is the horizontal reaction of the ground on the antenna mast.
It would be foolish to leave this to friction rather than a positive fixing at the foot of the mast.
If you think about it, it is the only horizontal force available to balance the horisontal component of the cable tension, if the wind does not blow.
I would hazard a guess that when Fh is calculated it turns out to be negative, since it appears (to me) to point in the wrong direction.

go well
 
  • #3
Studiot said:
Fv is the vertical reaction of the ground on the antenna mast.

Fh is the horizontal reaction of the ground on the antenna mast.
It would be foolish to leave this to friction rather than a positive fixing at the foot of the mast.
If you think about it, it is the only horizontal force available to balance the horisontal component of the cable tension, if the wind does not blow.
I would hazard a guess that when Fh is calculated it turns out to be negative, since it appears (to me) to point in the wrong direction.

go well

Fh is negative, so it does point in the opposite direction. But what I'm really asking is, what is causing this vertical / horizontal force on the tower / antenna mast(?) ?
Can we just say that it's a vertical / horizontal force on the tower from the floor? Or can we be more specific in saying that the vertical force is the normal force on the tower from the floor and that the horizontal force is a frictional force?
I'm not understanding "It would be foolish to leave this to friction rather than a positive fixing at the foot of the mast."
 
  • #4
I don't know you background but if you push (or pull) against something and it does not move it pushes back at you with equal force.

This return push is called the reaction or reactive force.

So when you stand on the ground you push down on the ground with your weight and the ground pushes back up on you with an equal and opposite reaction so you do not sink into the ground.

That is a vertical reaction.

Similarly if you push horizontally against a wall, it pushes back with an equal and opposite horizontal reaction force.

Since there are horizontal forces acting in your diagram, there must be a horizontal reaction.
This may be provided by friction or it may be because the base is spiked into the ground. It does not matter which, the force must be there or the base of the tower will move.

I simply observed that only a fool would rely on friction being strong enough in such a situation. There is no magic in this.
 
  • #5
If so, we can calculate the coefficient of friction between the tower and the floor using Fh and Fv.

To answer your first question, Fv is not necessarily the normal force on the tower from the floor. It could represent any vertical force acting on the tower, such as a force from a nearby building or structure.

As for Fh, it could indeed be a force of friction if the tower is experiencing impending motion. However, it could also represent any horizontal force acting on the tower, such as a force from a strong wind blowing in a horizontal direction.

In order to accurately solve these statics questions, we would need more information about the specific situation and the forces acting on the tower. We would also need to consider the geometry and materials of the tower and cable, as well as any external factors such as the wind speed and direction. By carefully analyzing all of these factors, we can determine the forces and tensions at play and ensure the stability and safety of the radio tower.
 

FAQ: Solving Statics Questions: Radio Tower, Fw, Fv, Fh

What is the purpose of solving statics questions?

Solving statics questions helps us understand the forces acting on an object in a state of equilibrium. It allows us to analyze and predict the behavior of structures and systems under various conditions.

What is a radio tower and why is it important in statics questions?

A radio tower is a tall structure used to support antennas for broadcasting and receiving radio signals. In statics questions, it is often used as an example to study the forces acting on a vertical structure in equilibrium.

How do we calculate Fw (weight force) in statics questions?

Fw, or weight force, is the force that acts on an object due to its mass and gravity. In statics questions, we can calculate Fw by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s²). This gives us the weight force in Newtons (N).

What are Fv and Fh in statics questions?

Fv and Fh represent the vertical and horizontal forces acting on an object in equilibrium, respectively. Fv is perpendicular to the surface the object is resting on, while Fh is parallel to the surface. These forces must be balanced for the object to remain in a state of equilibrium.

Can you provide an example of solving a statics question?

Sure, let's say we have a radio tower with a weight of 10,000 N and Fv and Fh forces of 8,000 N and 6,000 N, respectively. To solve for the remaining force, we can use the formula Fw = √(Fv² + Fh²). Plugging in the values, we get Fw = √(8,000² + 6,000²) = √(64,000,000 + 36,000,000) = √100,000,000 = 10,000 N. This means that the remaining force must be 10,000 N to keep the tower in equilibrium.

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