STATICS: Chapter 2, net force of lifting a given mass.

AI Thread Summary
The discussion focuses on solving a statics problem involving a derrick boom supporting a 20kg block. The user attempts to calculate the forces exerted by the boom and cables, using equations based on equilibrium conditions. They report discrepancies in their results, consistently off by about 150N from the textbook answers. Feedback from other users points out potential sign errors in their force summation equations, suggesting corrections to improve accuracy. The conversation emphasizes the importance of careful calculation and verification in statics problems.
SnickerGTI
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Homework Statement



"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


Homework 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)

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.
 
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SnickerGTI said:

Homework Statement



"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


Homework 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)

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.

How about a drawing, I can't visualize what you are describing.

CS
 
Last edited by a moderator:
A drawing of what?
 
Anybody have a clue?
 
SnickerGTI said:
A drawing of what?

A drawing of the problem. If you have inserted one already, I'm not able to see it for some reason (most likely my company is blocking the link behind the scenes).

CS
 
SnickerGTI: Nice work. You just made a couple of sign errors. Notice the fourth term in your summation(Fx) equation should be 156.9, not -156.8 N. (Also, generally always carry four or five significant digits for all quantities throughout your calculations, then round the final answer to three, or maybe four, significant digits.) Secondly, the third term in your summation(Fz) equation should be Tad*(7/11), not -Tad*(7/11). Give it one more try.
 
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