Simple Weight Problem

  • #1

Homework Statement



An air-conditioner unit is fastened to a roof that slopes upward at an angle of 36.0˚ . In order that the unit not slide down the roof, the component of the unit's weight parallel to the roof cannot exceed 520N. What is the maximum allowed weight of the unit?


Homework Equations





The Attempt at a Solution



My attempt at the solution is an attached file. We haven't learned about force yet but this problem is in the homework. My answer seems wrong because I don't think you can add two different weights unless they are in the same direction. And my answer seems too large. So how do I find the weight of the AC unit? Do I use some kind of Pythagorean identity? Thanks for all the help.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

  • AC unit problem.pdf
    56 KB · Views: 407

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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In your solution, you have W1 which is a vector parallel to the roof and W2 which is a vector perpendicular to the roof. The problem says that the "component of the unit's weight parallel to the roof" cannot be larger than 520 N. That's just W1.
 
  • #3
Sorry I'm not following. I thought w1 is perpendicular to the ground. How can w1 be parallel to the roof when it intersects the roof? I think I'm missing something big.
 
  • #4
gneill
Mentor
20,934
2,877
Sorry I'm not following. I thought w1 is perpendicular to the ground. How can w1 be parallel to the roof when it intersects the roof? I think I'm missing something big.

No, you're okay. Your diagram shows w1 as the weight of the air conditioner. It is vertical. Your w2 is the normal force produced by that weight on the surface of the roof, and your "520" labelled vector is the downslope force produced by the weight.

So you've done the correct calculation: w1 = 520N/sin(36°)

That's the maximum weight of the unit.
 
  • #5
Alright thanks a lot. Sorry one more question: When I calculate 520/sin(36˚ ) I get 884.6768407. So I shouldn't round up (right?) because the max is that number so rounding to 885 would be over max weight? Well I submitted my answer as 884 because of the reason I just stated and it counted my answer as correct, but said the true answer is 885 and to use the number 885 if I need it for future calculations. Can you explain why it is 885 and not 884? Thanks
 
  • #6
gneill
Mentor
20,934
2,877
The rounding of results should be based upon the significant figures available in the given information, not on the particulars of the "application" of the calculation.

In real life, each given value should have attached to it an error estimate (or precision estimate). So, for example, the maximum downslope force might be specified as 520N +/- 15N. Proper rounding and carrying forward of the errors through the calculation would result in a value for the allowed weight with a tolerance that an engineer could interpret according to application.
 
  • #7
Oh ok gotcha. Thanks again
 

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