Solving a Crane Jib Problem with Tension and Angles

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steven10137
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Homework Statement


A crane with jib AB, 7m long. Centre of gravity 3m from A, inclined at 30 degrees to the vertical.
The horizontal tie is attached at C, 5m from A.
If the mass of the jib is 200kg and the load at B is 600kg, what is the tension in the tie.

http://img139.imageshack.us/img139/2818/cranezu9.jpg

Homework Equations


[tex] \sum {M_{CW} } = \sum {M_{ACW} } [/tex]

The Attempt at a Solution


OK, well if we take the pivot about the point A, we have;
[tex] \begin{array}{l}<br /> \sum {M_{CW} } = \sum {M_{ACW} } \\ <br /> \left( 3 \right)\left( {200 \times 9.8} \right) + \left( 7 \right)\left( {600 \times 9.8} \right) = \left( 5 \right)\left( {T\sin 30} \right) \\ <br /> \Rightarrow T = \frac{{5880 + 41160}}{{2.5}} = 18816N \\ <br /> \end{array}[/tex]

This tension seems very high for the system.

Could someone please explain the principle behind the Tension and angle components of the crane, because I don't think I fully understand it.

Thanks in advance
Steven.
 
Last edited by a moderator:
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steven10137 said:

Homework Statement


A crane with jib AB, 7m long. Centre of gravity 3m from A, inclined at 30 degrees to the vertical.
The horizontal tie is attached at C, 5m from A.
If the mass of the jib is 200kg and the load at B is 600kg, what is the tension in the tie.

http://img139.imageshack.us/img139/2818/cranezu9.jpg

Homework Equations


[tex] \sum {M_{CW} } = \sum {M_{ACW} } [/tex]

The Attempt at a Solution


OK, well if we take the pivot about the point A, we have;
[tex] \begin{array}{l}<br /> \sum {M_{CW} } = \sum {M_{ACW} } \\ <br /> \left( 3 \right)\left( {200 \times 9.8} \right) + \left( 7 \right)\left( {600 \times 9.8} \right) = \left( 5 \right)\left( {T\sin 30} \right) \\ <br /> \Rightarrow T = \frac{{5880 + 41160}}{{2.5}} = 18816N \\ <br /> \end{array}[/tex]

This tension seems very high for the system.

Could someone please explain the principle behind the Tension and angle components of the crane, because I don't think I fully understand it.

Thanks in advance
Steven.
You are not evaluating the moments correctly. The moment of a force about a point is equal to the force times the perpendicular distance from the line of action of the force to the point.
 
Last edited by a moderator:
Draw a free body diagram. That should help you see what you need to do to solve this. Its actually just a one step problem.
 
thanks for your reply Jay

"The moment of a force about a point is equal to the force times the perpendicular distance from the line of action of the force to the point."

So for taking the moments about the point A:
The centre of mass for the jib has moment; (3sin30)(200x9.8)

Following this procedure;
(3sin30)(200x9.8) + (7sin30)(600x9.8) = 5T
therefore T = 4704N
 
steven10137 said:
thanks for your reply Jay

"The moment of a force about a point is equal to the force times the perpendicular distance from the line of action of the force to the point."

So for taking the moments about the point A:
The centre of mass for the jib has moment; (3sin30)(200x9.8)

Following this procedure;
(3sin30)(200x9.8) + (7sin30)(600x9.8) = 5T
therefore T = 4704N
Almost there. But the perpendicular distance from the tension force tothe point is (5)sin60, right?
 
ya but that's not perpendicular.. that's horizontal, its COS not sin
 
ya exactly cos30 or sin60.. either works
 
OK so this perpencidular distance we are looking for the is the distance 'x' in the diagram yea?

http://img165.imageshack.us/img165/5121/craneforcepp2.jpg

cos30 = x / 5
the distance x is the 5cos30

Using this;
(3sin30)(200x9.8) + (7sin30)(600x9.8) = 5cos30T
therefore T = 5.43x10^3 N
 
Last edited by a moderator:
dots.jpg


The perpendicular distance, for each of the forces, will be the shortest distance from the green dot to each dotted line.

Its the distance to the line of action of the force, which is infinite.
 
steven10137 said:
OK so this perpencidular distance we are looking for the is the distance 'x' in the diagram yea?

http://img165.imageshack.us/img165/5121/craneforcepp2.jpg

cos30 = x / 5
the distance x is the 5cos30

Using this;
(3sin30)(200x9.8) + (7sin30)(600x9.8) = 5cos30T
therefore T = 5.43x10^3 N
Correct.
 
Last edited by a moderator:
excellent :)
thanks for your help Vidatu and PhanthomJay