- #1
gnits
- 137
- 46
- Homework Statement
- To find the forces acting in a system in equilibrium
- Relevant Equations
- moments
Could I please ask where I have gone wrong with my reasoning in the following question:
The answers in given in the book are:
(1/2)W tan(Ө)
W vertically
(1/2)W tan(Ө) horizontally
Here is my diagram:
Considering the system as a whole:
(In the text below "Ya" and "Xa" are the forces at the hinge at A)
Resolving vertically gives:
Ya + R = 2W + ω (call this "equation 1")
Resolving horizontally gives:
F = Xa (this agrees with the book answer)
Taking moments about C gives:
W * a sin(Ө) + W * 3a * sin(Ө) = Ya * 4a * sin(Ө)
which gives:
Ya = W (this agrees with the book answer)
And so from "equation 1" we have that R = W + ω (call this "equation 2")
Now, considering the ring alone and taking moments about B gives:
R * 2a * sin(Ө) = ω * 2a * sin(Ө) + F * 2a * cos(Ө)
which gives:
R * tan(Ө) = ω * tan(Ө) + F
So using "equation 2" gives:
F = W * tan(Ө)
But book answer is F = (1/2) * W * tan(Ө)
Where have I reasoned wrongly?
Thanks for any help.
The answers in given in the book are:
(1/2)W tan(Ө)
W vertically
(1/2)W tan(Ө) horizontally
Here is my diagram:
Considering the system as a whole:
(In the text below "Ya" and "Xa" are the forces at the hinge at A)
Resolving vertically gives:
Ya + R = 2W + ω (call this "equation 1")
Resolving horizontally gives:
F = Xa (this agrees with the book answer)
Taking moments about C gives:
W * a sin(Ө) + W * 3a * sin(Ө) = Ya * 4a * sin(Ө)
which gives:
Ya = W (this agrees with the book answer)
And so from "equation 1" we have that R = W + ω (call this "equation 2")
Now, considering the ring alone and taking moments about B gives:
R * 2a * sin(Ө) = ω * 2a * sin(Ө) + F * 2a * cos(Ө)
which gives:
R * tan(Ө) = ω * tan(Ө) + F
So using "equation 2" gives:
F = W * tan(Ө)
But book answer is F = (1/2) * W * tan(Ө)
Where have I reasoned wrongly?
Thanks for any help.