How Much Weight Can the Air-Conditioning Unit on a Sloped Roof Hold?

In summary: I think the latter is more likely. In that case, the 425N would be -ve and the component would be +ve. But I don't see any problem with using the 425N as the maximum magnitude, and then the component would be -ve. The question does not explicitly state which is the correct sign convention, but I think it's reasonable to assume that the author would use the more common sign convention.I think there's a mistake in the solution, as well. I get 425N/sin 55 = 518.3N, which is the same as the solution. The solution is correct as far as it goes, but then it goes on to say that the answer
  • #1
dust66
3
0

Homework Statement


An air-conditioning unit is fastened to a roof that slopes at an angle of 35° above the horizontal (Fig. 1.33). Its weight is a force on the air conditioner that is directed vertically downward. In order that the unit not crush the roof tiles, the component of the unit’s weight per- pendicular to the roof cannot exceed 425 N. (One Newton, or 1 N, is the SI unit of force. It is equal to 0.2248 lb.) (a) What is the maxi- mum allowed weight of the unit? (b) If the fasteners fail, the unit slides 1.50 m along the roof before it comes to a halt against a ledge. How much work does the weight force do on the unit during its slide if the unit has the weight calculated in part (a)

Homework Equations


Fy=Fsinθ
W=FS

The Attempt at a Solution


I created x / y axes with x being on the slope of the roof
then try to find F as I know the Fy and the angle so:
Fy=Fsinθ or F=Fy/sinθ
F=425N / sin125 = 425N / 0.82 = 518.3 N

and then for (b)
First find Fx = Fcosθ = 518.3N x cos55 = 518.3N x 0.574 = 297.5N
then
W=FS = 297.5N x 1.50m = 446.25Nm

Does this sound about right??
the results in the answers are: (a)5.2 x 10^2 N
(b) 4.5 x 10^2 Nm

which look similar but i think I am missing one zero somewhere?
link to the page of the book:
http://s9.postimg.org/nhklria4f/Screen_Shot_2016_01_13_at_16_02_15.png

Thanks in advance! Hope its ok to post things like this :)
 
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  • #2
Don't your answers agree with the textbook answers, or am I missing something?
 
  • #3
oh crap they do :)
lol
sorry!
its been a long day i was counting 10^2 as 1000 o0)

time for a brake i think :)
 
  • #4
dust66 said:
oh crap they do :)
lol
sorry!
its been a long day i was counting 10^2 as 1000 o0)

time for a brake i think :)

You mean, like one of these?
​
Or, one of these?
Outside-Brain-Break-Tools.jpg

​
 
Last edited by a moderator:
  • #5
love the break joke !
i always spell it wrong :)
 
  • #6
I know this is an old thread and I'm really sorry for resurrecting it, but this is quite related to my question:
I did the same thing as OP on the x/y axis. Using basic geometry, I got that the angle between the x-axis and the vector is 55°. Since it goes from x to y, it's either -25° or 325°. Thus, I fail to understand how OP got 125° there. Can someone explain it to me?
 
  • #7
I'm not quite sure if people are still going to visit this very old thread, but there are two major flaws in the above given solution.

For the First, I'd like to state that the angle (that's been measured as 125° here) must be measured from the +ve x-axis to the vector. And when this would be done flawlessly, the angle corresponding to the given problem condition would come out to be 135° and not 125°

Secondly, and I'm not very sure of this myself, the component perpendicular to the roof is given as 425N. Although, the Weight force is acting downwards and lies in the third quadrant of the plane, and hence the given component cannot be +ve in any case Or by any chance whatsoever. Consequently, I suggest considering the given component as -425N in place of 425N.

Lastly, I'd suggest using these values to calculate the desired variable and I'm quite sure one would get the correct solution and agree with me. Thank you. All the best.
 
  • #8
Marq2345 said:
the angle corresponding to the given problem condition would come out to be 135° and not 125°
I get 125. Please explain how you arrive at 135.
Marq2345 said:
I suggest considering the given component as -425N in place of 425N.
Whether it is positive or negative depends on your choice of sign convention.
The question only makes sense if we take the 425N either as being the maximum magnitude or as measured down into the roof.
 

1. What is a vector component problem?

A vector component problem is a type of physics problem that involves breaking down a vector into its horizontal and vertical components to solve for unknown quantities, such as magnitude or direction.

2. How do you find the horizontal and vertical components of a vector?

To find the horizontal and vertical components of a vector, you can use trigonometric functions such as sine and cosine. The horizontal component is found by multiplying the magnitude of the vector by the cosine of the angle, and the vertical component is found by multiplying the magnitude by the sine of the angle.

3. What is the difference between a scalar and a vector?

A scalar is a physical quantity that has only magnitude, while a vector has both magnitude and direction. Examples of scalars include temperature and mass, while examples of vectors include velocity and force.

4. How do you add or subtract vectors?

To add or subtract vectors, you can use the head-to-tail method or the parallelogram method. In the head-to-tail method, you place the tail of one vector at the head of the other vector and draw a new vector from the tail of the first vector to the head of the second vector. In the parallelogram method, you draw both vectors from the same starting point and then complete the parallelogram to find the resulting vector.

5. Can vectors have negative components?

Yes, vectors can have negative components. The direction of a vector component can be positive or negative depending on the orientation of the vector in relation to a chosen coordinate system. For example, a vector pointing to the left would have a negative horizontal component in a standard Cartesian coordinate system.

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