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

In summary: Here's a check on your solution. Summation(Fy) = 0 gives Tab = 1741.7 N. Summation(Fx) = 0 gives Tac = 1517.3 N. Summation(Fz) = 0 gives Tad = 403.4 N.In summary, the problem involves a derrick boom that 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. The boom AB exerts a force A that is directed from B to A. By using statics and the given coordinates, the force in each of the two cables (AC and AD) can be determined.
  • #1
SnickerGTI
3
0

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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  • #2
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
 
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  • #3
A drawing of what?
 
  • #4
Anybody have a clue?
 
  • #5
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
 
  • #6
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.
 

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

1. What is net force?

Net force is the overall force acting on an object, taking into account all the individual forces acting on it. It is calculated by adding together all the forces in a given direction.

2. How is net force related to lifting a given mass?

When lifting a given mass, the net force is equal to the weight of the object being lifted. This is because the force of gravity acting on the object is the only force acting in the opposite direction to the lifting force.

3. What is the formula for calculating net force?

The formula for calculating net force is Fnet = ΣF, where Fnet is the net force and ΣF is the sum of all the individual forces acting on the object.

4. What is the unit of measurement for net force?

The unit of measurement for net force is Newtons (N). This is because force is defined as the product of mass and acceleration, and Newtons are the unit of measurement for both mass and acceleration.

5. How does the direction of net force affect the motion of an object?

The direction of net force determines the direction of motion of an object. If the net force is in the same direction as the object's motion, it will continue moving at a constant speed. If the net force is in the opposite direction, it will cause the object to decelerate or change direction.

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