- #1
quietrain
- 648
- 2
if i take moments at point O, does it mean that
LTcos(a) = (L/2) mg
and hence T = 32?
but answer is 16.
what am i missing out here? thanks!
Solving LTcos(a)= (L/2)mg: Find T=16
- Thread starter quietrain
- Start date
-
- Tags
- Moments
Click For Summary
Homework Help Overview
The discussion revolves around a physics problem involving tension in a system, specifically analyzing the equation LTcos(a) = (L/2)mg. Participants are attempting to understand the relationship between tension and the forces acting on the system, while also addressing the geometry involved in the setup.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking
Approaches and Questions Raised
- Participants discuss taking moments about point O and question the implications of the normal force from the wall. There are attempts to resolve forces into components and clarify the geometry involved in the problem. Some participants express confusion about the correct application of the moment equation and the role of different forces.
Discussion Status
The discussion is active, with participants providing insights and sketches to aid understanding. There is recognition of mistakes in the initial approach, and some participants suggest that getting the geometry correct is crucial. While there are differing interpretations of the problem, guidance has been offered regarding the application of moments and the importance of considering all components of tension.
Contextual Notes
Participants note the potential for infractions due to posting in the wrong section, indicating a concern about adhering to forum rules. There is also mention of the need to resolve forces correctly and the implications of the geometry on the calculations.
if i take moments at point O, does it mean that
LTcos(a) = (L/2) mg
and hence T = 32?
but answer is 16.
what am i missing out here? thanks!
- #2
Drakkith
Mentor
- 23,205
- 7,687
You might want to try this in the Homework section. They should be able to help you there!
- #3
- #4
quietrain
- 648
- 2
oh shucks... i forgot to put in homework ... they are going to infract me again :(
anyway, i think i know why i got it wrong, the wall has a normal force that balance the x component of the tension
but i have a question, where is the normal force of the wall? at P or O? but since i take moments at O, will that come into play?
anyway, i think i know why i got it wrong, the wall has a normal force that balance the x component of the tension
but i have a question, where is the normal force of the wall? at P or O? but since i take moments at O, will that come into play?
- #5
quietrain
- 648
- 2
Studiot said:
erm i assume you resolve the weight to match the tension?
i was trying to solve by resolving in x and y components
- #6
Studiot
- 5,440
- 11
No I didn't resolve anything.
Your original idea of moments about O was good - you just fell down in the implementation.
I hoped my sketch would help since it was the geometry you got wrong, but I wanted to leave something for you to do.
Remember the moment is the perpendicular distance from O to the line of action. This is why I drew in some perpendiculars.
Your original idea of moments about O was good - you just fell down in the implementation.
I hoped my sketch would help since it was the geometry you got wrong, but I wanted to leave something for you to do.
Remember the moment is the perpendicular distance from O to the line of action. This is why I drew in some perpendiculars.
- #7
quietrain
- 648
- 2
Studiot said:
er, sry, ignore the sketching on the original diagram, those aren't mine :(
but anyway, i know where i went wrong already
LT = L/2 mg
in component form
summation of torques at O ,
x comp + y comp = weight comp
Tsin(alpha)* Lcos(alpha) + Tcos(alpha)* Lsin(alpha) = (L/2)mg sin(alpha)
T=16.171my mistake : i forgot that the summation of moments is the whole tension and not only the y-component of the tension. did you meant that? and thank you for drawing, i appreciate it
ok thanks everyone!
- #8
Studiot
- 5,440
- 11
Just for the record, my amended version of your original equation leads to
Lsin(2a)T = 1/2mgLsin(a)
T = 1/2mg sin(a) / sin(2a)
using my sketch and not resolving into components.
go well
I am sorry but my Latex doesn't seem to work any longer.
Lsin(2a)T = 1/2mgLsin(a)
T = 1/2mg sin(a) / sin(2a)
using my sketch and not resolving into components.
go well
I am sorry but my Latex doesn't seem to work any longer.
Similar threads
- · Replies 11 ·
- · Replies 95 ·
- · Replies 2 ·
- · Replies 2 ·
- · Replies 28 ·
- · Replies 1 ·
- · Replies 2 ·