Dynamics Problem

[Tetra]

A sign for a restaurant has a mass of 82 kg. It is held out from the wall by a light horizontal steel rid which supports no weight and a wire at 40 degrees to the horizontal. (This will form a right triangle). Find the tension in the wire and the compressions in the steel wall.

 

For the tension in the string, I know that once we find the vertical force of the sign (803.6 N [down]), then we can use 82sin40 to find the tension in the string. But what about the wire? Is the force just the force of the object itself (so 803.6N [down])?

 

Also, in a question where an object is hanging from the center of a rod or string or something, will the tension in either side of the string be identical? Or is there some way for them to be different?

Edited July 11, 2010 by Tetra

[cypress]

I assume you mean wire not string and rod not the wire. Not sure where the 82sin40 came from the tension in the wire is a force not a mass. but sin40 does figure into it.

 

No the load in the rod will not be equal to the sign.

 

Also watch your significant digits as you go. Gravity is 9.807 m/s² not 9.8 and as you continue with the problem you may introduce error by rounding off early.

 

So You seem to be on your way to finding the tension in the wire but your formula needs a little work. Next note that the 40 degree angle creates a horizontal load in the rod. Do you see why? The question tells you the load is a compressive load into the building, do you see that? If so what would be the formula that would result in a static balance of loads?

Edited July 12, 2010 by cypress

[ewmon]

You should really learn the logic behind formulating the equations that involve trig functions (and you can get that knowledge from your text, notes, classmates, teacher, etc).

 

Other than learning this logic academically, you can also develop a "gut feeling" for the trig with thought experiments using extreme angles such as 0° and 90° to the horizontal. Imagine the forces acting through the wire and rod in both cases. You don't need to imagine a specific amount of weight to use or to write any equations for these imaginary cases, but making diagrams would help you visualize it. These cases are especially helpful because the sine and cosine of these angles are either 1 or 0 (ie, "all" or "nothing"). So, first you'll want to develop "word answers" for the forces in the wire and rod, then you'll develop equations from these word answers.

 

Developing a gut feeling about such problems is also helpful for tests where the teacher might give a problem never encountered by the students before, but must be worked out through their understanding of the subject matter.

 

Just a friendly PS — If my memory serves me properly, this is a statics problem, not dynamics.

Edited July 12, 2010 by ewmon

[Tetra]

Late reply, but...

 

To cypress

 

Oh, your right...I forgot that 82 was a mass. So, if I found what the force of gravity is, then I can form a right-angle triangle. Knowing that the angle is 40 degrees, and the opposite is 803.6 (the force of gravity) I can use 803.6sin40 to find the hypotenuse, or the tension force.

 

(I would upload an image but...it needs to be on a URL?)

 

And for the rod, since we know (or calculated) the tension force, then we can draw a right triangle again and solve for that horizontal force...right?

 

But I don't get it....if the object is in equilibrium, shouldn't the forces pulling on it equal 0 in the end? Like, if you break it into horizontal/vertical components, it's not 0...or then would that mean it's not right?

 

To ewmon:

 

Oh, I know how to formulate trig functions (at least..I'm pretty sure I did it right). But could you explain this "gut feeling" thing? I can draw everything out, but I don't get the part about "developing word answers" and whatnot.

 

And this is supposedly physics...the dynamics unit in physics.

[cypress]

Late reply, but...

 

To cypress

 

Oh, your right...I forgot that 82 was a mass. So, if I found what the force of gravity is, then I can form a right-angle triangle. Knowing that the angle is 40 degrees, and the opposite is 803.6 (the force of gravity) I can use 803.6sin40 to find the hypotenuse, or the tension force.

 

(I would upload an image but...it needs to be on a URL?)

 

You could use imageshack, but don't bother I think I understand. Now check your formula again. You got the force due to gravity OK and it is vertical thus the use of sin but remind yourself the definition of sin. Is it not opposite over hypotenuse? If so, what does that imply about your formula? It may help to think about the impact of a wire (or rod) holding a force but at an angle. As the vertical angle decreases from 90 to 0 what happens to the required force? Does it increase or decrease?

 

And for the rod, since we know (or calculated) the tension force, then we can draw a right triangle again and solve for that horizontal force...right?

 

Yes, good!

 

But I don't get it....if the object is in equilibrium, shouldn't the forces pulling on it equal 0 in the end? Like, if you break it into horizontal/vertical components, it's not 0...or then would that mean it's not right?

 

When it is static (in equilibrium) the forces are not zero but are balance so only the net forces (not each force in isolation) acting on each and every item is zero. In the case of the sign, gravity is exerting a downward force and the wire is exerting an equal upward force. Now since the wire is not vertical, the wire, in the process of pulling tension on the sign also pulls outward on the building and thus something must push back on the wire to keep it from collapsing inward toward the building. Thus the rod pushes back putting it into compression and the building pushes on the rod with an equal force to keep it static.

 

So how about those formulas?

[Cap'n Refsmmat]

(I would upload an image but...it needs to be on a URL?)

Under Additional Options in the posting screen is an Attach Files section. Hit Manage Attachments.

 

Of course, with the upgrade tomorrow morning, it'll be completely different...

[ewmon]

To ewmon: could you explain this “gut feeling” thing?

For example, when the angle of the wire to the horizontal, α, is 90°, the wire bears “all” the sign’s weight, and the rod has “no” force applied to it (and, in fact, you can remove it without any consequence). The sin(90°)=1, and cos(90°)=0. So, the tensile force in the wire involves the sin(α) and the compressive force in the rod involves cos(α).

 

When α=0°, then an infinite (ie, ∞) tensile force through the wire would hold up the sign, and the rod also bears “all” that tensile force. The sin(0°)=0, and cos(0°)=1. It's important to remember that the tensile force in the wire is the hypotenuse, and from there, use the definitions of the sine and cosine.

 

It may be easier to use another method starting with the point where the sign, wire and rod meet. Draw the tensile force in the wire from the point diagonally upward toward the wall, draw the compressive force through the rod horizontally from the point outward (ie, in the direction away from the wall), and draw the gravitational force downward from the point. Think of what forces must counterbalance other forces in order for this system to be in equilibrium.

[Tetra]

To Cap'n Refsmmat:

 

Ok, I see what I did wrong now! It was a silly mistake, I drew the triangle but didn't realize that it was the hypotenuse that I was supposed to be looking for. So, for the tension in the wire, it's supposed to be force of 'gravity / sin40'. Similarly, the tension in the rod is the horizontal component of the tension in the wire, so that formula is 'Force of tension cos49'.

 

Thanks for all your help, today and with other stuff!

 

To ewmon:

 

Hmmmmm I think I get your gut feeling thing. It's like visualizing the actual sin and cos waves, where it goes up and down at certain degrees. So that way we kind of get an idea of the magnitude of the actual value we're getting (small, huge, and the size relative to other objects in the situation.)

(P.S. And I didn't realize before that compression meant and outwards force..thanks for clearing that up!