- #1
quietrain
- 655
- 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
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In summary, the conversation discusses the calculation of tension using moments at point O and the mistake made in the implementation. After resolving the weight and tension components, it was realized that the summation of moments should include the whole tension, not just the y-component. A sketch was provided to help understand the geometry and the final equation for tension was determined to be T = 1/2mg sin(a) / sin(2a).
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
- 23,093
- 7,499
You might want to try this in the Homework section. They should be able to help you there!
- #3
- #4
quietrain
- 655
- 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
- 655
- 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
- 9
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
- 655
- 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
- 9
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.
FAQ: Solving LTcos(a)= (L/2)mg: Find T=16
What is LTcos(a) and how does it relate to the equation?
In this equation, LTcos(a) represents the tension force in a system with a mass hanging from a string or cable at an angle a. It is related to the equation because it is equal to the force of gravity (mg) multiplied by the length of the string (L) and the cosine of the angle (a).
What are the units of measurement for the variables in this equation?
The units for tension (T) are typically Newtons (N), the units for mass (m) are kilograms (kg), and the units for length (L) are meters (m). The gravitational constant (g) has units of meters per second squared (m/s^2).
How do I solve this equation for T?
To solve for T, you will need to use basic algebraic principles. First, move the term with T to one side of the equation and all the other terms to the other side. Then, divide both sides by the remaining coefficient of T. Finally, use a calculator to find the value of T.
What is the significance of the value 16 in this equation?
The value 16 represents the acceleration due to gravity on Earth, which is approximately 9.8 meters per second squared (m/s^2). It is multiplied by the length of the string (L) to determine the force of gravity (mg) and is used to solve for the tension force (T).
Can this equation be used in real-world scenarios?
Yes, this equation can be used in real-world scenarios where a mass is hanging from a string or cable at an angle. It is commonly used in physics and engineering to calculate the tension force in a system. However, it is important to note that this equation assumes ideal conditions and may not be accurate in all real-world situations.