Solve Crane Cable Tension & Moment Around Point A | Test Question

• pooface
In summary: Tsin10. then the torque due to that is -Tsin10*12 (minus sign indicates counteclockwise). The component parallel to the arm is Tcos10... but this creates 0 torque...So to get tension:10kN*6cos30 + 25kN*12cos30 - Tsin10*12 = 0
pooface

Homework Statement

The image is a crane.
For the diagram given, find the tension in the cable and find the moment around point A.
The length of the crane the dark bar is 12 m.

http://img249.imageshack.us/img249/679/exqjm6.jpg

sigma m = F *D

The Attempt at a Solution

I had this question on a test today and am very curious to know the answer.
if someone strong in physics please solve it.

I got 51 something Kilo Newton for the reaction at A.
For the tension i got 46.08 Kilo Newton i believe.

thank you.

EDIT: In my reaction/moment calculation I included the force of the cable. that being said i forgot to account for the y direction of the cable. should the cable have been accounted for?

Last edited by a moderator:
Here's what I get:

let T be the force of tension..

let B be the force in the bar...

Bcos30 - Tcos20 = 0

B = 1.085T

Bsin30 - 35 - Tsin20 = 0

plug in B = 1.085T

I get T = 174.6N

Then the torque about A:

10kN*6cos30 + 35kN*12cos30 - Tsin20*12

= 10kN*6cos30 + 35kN*12cos30 - 174.6sin20*12
= -300.9 kNm

so the reaction has to be +300.9 kNm ie 300.9 kNm clockwise

I was a little thrown off by the problem... because I initially just tried to get T by making the torque about A to 0... then when I saw the question asking for the moment about A I changed...

Not very confident with how i did this problem.

If someone as intelligent as you are not confident then I am sure my class took a strong beating during this test.

Btw the weight hanging is 25KN i should have wrote it neater.

Yup totally blew the last page of this test. I even got the tension wrong..which was the easier question.

He gave us 50 minutes to solve 4 questions. By the time i got to the 4th had very little time left.

but listen in your moment calculation why didnt you account for the x component of the tensin in the string?

pooface said:
If someone as intelligent as you are not confident then I am sure my class took a strong beating during this test.

Thanks I appreciate that. But in this question, I don't feel confident at all because I changed the way I solved the problem after reading the second part of the question. That's never a good sign. ;)

The assumption I'm making here that there's a reaction force acting at A along the axis of the beam... along with a reaction moment at A. I don't feel good about this assumption that the net reactional force acts along the axis...

Btw the weight hanging is 25KN i should have wrote it neater.

yeah, I used 25kN... The 35kN... is the 25kN + the 10kN of the beam...

Yup totally blew the last page of this test. I even got the tension wrong..which was the easier question.

He gave us 50 minutes to solve 4 questions. By the time i got to the 4th had very little time left.

but listen in your moment calculation why didnt you account for the x component of the tensin in the string?

I was using the component of tension perpendicular to the beam... ie Tsin10. then the torque due to that is -Tsin10*12 (minus sign indicates counteclockwise). The component parallel to the beam is Tcos10... but this creates 0 torque...

You can also divide into the x (horizontal) and y(vertical)-components... magnitude of x-component of tension would be Tcos20. torque due to that is -Tcos20*(12sin30). The torque due to the y-component of tension is Tsin20*12cos30... net torque would be Tsin20*12cos30 - Tcos20*(12sin30) = Tsin(20-30) = Tsin(-10) = -Tsin10. comes out to the same answer...

I would have blown this question on a test too.

Go back and look at the moment about A and ask yourself "Is the boom arm rotating about A?" If not, what does that say about the moments?

TVP45 said:
Go back and look at the moment about A and ask yourself "Is the boom arm rotating about A?" If not, what does that say about the moments?

Oh... I was making this whole problem way more complicated than it was... for some reason I was fixated on the answer to the second part not being 0.

so to get tension:

10kN*6cos30 + 25kN*12cos30 - Tsin10*12 = 0

T = 149.6 kN ?

Last edited:
do you mean T cos 20? because the triangle of the given 10 degrees is not a right triangle.

149.6N doesn't SEEM right because how could the tension be so low compared to the other forces?

Last edited:
pooface said:
do you mean T cos 20? because the triangle of the given 10 degrees is not a right triangle.

149.6N doesn't SEEM right because how could the tension be so low compared to the other forces?

I blundered... it should be 149.6 kN.

I'm taking the component of tension perpendicular to the arm... that's Tsin10

Thanks ok I understand what you did there.

Can someone clarify for what TVP45 meant about the boom bar please?

Why should the moment/reaction be 0 again?

Last edited:
Since the boom bar is not rotating about A, the moments about that point must sum to 0. That is a necessary condition for static equilibrium.

That is a fairly simple equation to solve.

1. What is the purpose of solving for crane cable tension and moment around point A?

The purpose of solving for crane cable tension and moment around point A is to ensure the stability and safety of the crane and its load. By knowing the tension and moment at this critical point, engineers can determine if the crane is capable of lifting the intended load without tipping over.

2. How are crane cable tension and moment around point A calculated?

Crane cable tension and moment around point A are calculated using the principles of statics and the equations of equilibrium. The forces acting on the crane and its load, including the weight of the load, the weight of the crane, and the tension in the cables, are all taken into account to determine the tension and moment at point A.

3. What factors can affect the tension and moment around point A?

The tension and moment around point A can be affected by several factors, including the weight and size of the load, the length and angle of the crane's boom, the position of the crane's center of gravity, and any external forces such as wind or uneven ground.

4. Why is it important to accurately solve for crane cable tension and moment around point A?

Accurate calculation of crane cable tension and moment around point A is crucial for the safety of the crane, its operators, and anyone in the surrounding area. If the tension and moment are not properly calculated, the crane could potentially tip over, causing serious damage and injury.

5. What are some methods for testing crane cable tension and moment around point A?

There are a few different methods for testing crane cable tension and moment around point A, including using load cells to measure the tension in the cables, performing a stability analysis using computer software, and conducting physical tests with a full-scale crane and load. The method chosen will depend on the specific needs and resources of the project.

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