SnickerGTI
Sep16-08, 03:53 PM
1. The problem statement, all variables and given/known data
"A derrick boom is guyed by cables AC and AD. A worker is lifting a 20kg block by pulling on a rope that passes through the pulley at A. Knowing that the boom AB exerts a force A that is directed from B to A, determine this force and the force in each of the two cables."
http://img213.imageshack.us/img213/4463/2673dsc0036ep9.jpg
ANSWER: (From back of Textbook)
F(AB) = 1742N
T(AC) = 1517N
T(AD) = 403N
2. Relevant equations
F(net) in every direction = 0 (Hence statics)
F(AB) = (TensionAB)(Unit Vector Lambda)
Lamda = AB(position vector from A to B)
----
|AB|(Magnitude of AB)
3. The attempt at a solution
Coordinates:
A = 0x + 6y + 0z
B = -6x + 0y -3z
C = -10.5x + 0y -8z
D = -6x + 0y -7z
E = +6x +1.5y +0z
All forces of all the ropes and the weight force must be equal to the one supporting object: the boom(Tab).
I drew a triangle with AE as the hypotenuse.
Tan(theta) = (opp/adj) = [(6-1.5)/6]
(theta) = arctan(4.5/6)
(theta) = 36.87degrees
[(4.5^2) + (6^2)]^(1/2) = Hyp
Hyp = 7.5
W = (m)(g) = 196N = Tae
.: (196N/7.5m) = (Taex/6m) = (Taey/4.5m)
Taex = (6m)(196/7.5m) = 156.8N
Taey = (4.5m)(196/7.5m) = 117.60N
Sum Fx = 0 = Tab(6/9) - Tac(10.5/14.5) - Tad(6/11) - 156.8N
Sum Fy = 0 = Tab(6/9) - Tac(6/14.5) - Tad(6/11) -(9.8*20kg) - 117.60N
Sum Fz = 0 = Tab(3/9) - Tac(8/14.5) - Tad(7/11)
From here I used simple substitution and elimination to solve for the variable Tab, Tac and Tad; but I keep getting numbers that are off by around 150N for each value. I have tried this problem so many times, can somebody check to see if my equations above are correct? I am assuming my problem lies there and not in my simultaneous equation solving skills.
"A derrick boom is guyed by cables AC and AD. A worker is lifting a 20kg block by pulling on a rope that passes through the pulley at A. Knowing that the boom AB exerts a force A that is directed from B to A, determine this force and the force in each of the two cables."
http://img213.imageshack.us/img213/4463/2673dsc0036ep9.jpg
ANSWER: (From back of Textbook)
F(AB) = 1742N
T(AC) = 1517N
T(AD) = 403N
2. Relevant equations
F(net) in every direction = 0 (Hence statics)
F(AB) = (TensionAB)(Unit Vector Lambda)
Lamda = AB(position vector from A to B)
----
|AB|(Magnitude of AB)
3. The attempt at a solution
Coordinates:
A = 0x + 6y + 0z
B = -6x + 0y -3z
C = -10.5x + 0y -8z
D = -6x + 0y -7z
E = +6x +1.5y +0z
All forces of all the ropes and the weight force must be equal to the one supporting object: the boom(Tab).
I drew a triangle with AE as the hypotenuse.
Tan(theta) = (opp/adj) = [(6-1.5)/6]
(theta) = arctan(4.5/6)
(theta) = 36.87degrees
[(4.5^2) + (6^2)]^(1/2) = Hyp
Hyp = 7.5
W = (m)(g) = 196N = Tae
.: (196N/7.5m) = (Taex/6m) = (Taey/4.5m)
Taex = (6m)(196/7.5m) = 156.8N
Taey = (4.5m)(196/7.5m) = 117.60N
Sum Fx = 0 = Tab(6/9) - Tac(10.5/14.5) - Tad(6/11) - 156.8N
Sum Fy = 0 = Tab(6/9) - Tac(6/14.5) - Tad(6/11) -(9.8*20kg) - 117.60N
Sum Fz = 0 = Tab(3/9) - Tac(8/14.5) - Tad(7/11)
From here I used simple substitution and elimination to solve for the variable Tab, Tac and Tad; but I keep getting numbers that are off by around 150N for each value. I have tried this problem so many times, can somebody check to see if my equations above are correct? I am assuming my problem lies there and not in my simultaneous equation solving skills.