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

  • #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.
 
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