• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Please post the problem on here
anonymous
  • anonymous
A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension and makes an angle of with the ceiling. Cable 2 has tension and makes an angle of with the ceiling.
anonymous
  • anonymous
Please draw a diagram and I will surely help you out

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1332704608666:dw|
anonymous
  • anonymous
I also need to know what they want you to find
anonymous
  • anonymous
???
anonymous
  • anonymous
find the expression fo t1 the tension is cable 1 does not depend on t2
anonymous
  • anonymous
The forces in the x and y-directions must equal zero since the object is not moving. \[\sum F_X = 0 \rightarrow -T_1 \cos(\theta_1) + T_2 \cos(\theta_2) = 0\]\[\sum F_Y = 0 \rightarrow T_1 \sin(\theta_1) + T_2 \sin(\theta_2) - mg = 0\]
anonymous
  • anonymous
how did you get these equations???
anonymous
  • anonymous
By summing the forces in the x and y-directions. Since T1 makes an angle with the horizontal, it has vertical and horizontal components. These are found using our basic trig relationships. Therefore, we get \[T_{1,x} = T_1 \cos(\theta_1) ~~{\rm and}~~ T_{1,y} = T_1 \sin(\theta_1) \] We can also observe that T_1 and T_2 act upwards in the vertical direction, while T1 acts to the left and T2 to the right in the horizontal direction, therefore we subtract them in the horizontal and add them in the vertical.
anonymous
  • anonymous
did u break it up into components
anonymous
  • anonymous
Yes.... \(T_1 \cos(\theta_1)\) is the x-component of \(T_1\), right?
anonymous
  • anonymous
yah csn u show me with the drawing i posted how u did it ( so i can see it)
anonymous
  • anonymous
You are familiar with the basic trig functions right? Is that where you are getting stumped?
anonymous
  • anonymous
|dw:1332708184548:dw|\[T_x = T \cos(\theta) ~~{\rm and}~~ T_y = T \sin(\theta)\] Do you understand how I obtained these?
anonymous
  • anonymous
|dw:1332708243165:dw|
anonymous
  • anonymous
is that how i break it up into components
anonymous
  • anonymous
|dw:1332708566454:dw|
anonymous
  • anonymous
i understand it better thank you. since t we were not told that the mass was moving we are assuming the net froce is zero
anonymous
  • anonymous
We can safely assume it is not moving. Because of this, yes, the forces must sum to zero.
anonymous
  • anonymous
and when u break up into components how do u know how to draw the y component and x component bc sometimes i get confused
anonymous
  • anonymous
The x-components will always be PARALLEL to the x-axis, and the y-components will always be PARALLEL to the y-axis. X and Y should be PERPENDICULAR.
anonymous
  • anonymous
|dw:1332709050421:dw|
anonymous
  • anonymous
is this correct how i broke it up
anonymous
  • anonymous
Exactly
anonymous
  • anonymous
jsut to make sure i got the hang of it |dw:1332709355994:dw|
anonymous
  • anonymous
Correct.

Looking for something else?

Not the answer you are looking for? Search for more explanations.