Stuck on these boom problems

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In summary: It's trying to use the cosine of the angle between the cable and the horizontal. I have no idea what this is supposed to do!
  • #1
smith5029
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A wrecking ball (weight = 5400 N) is supported by a boom, which may be assumed to be uniform and has a weight of 3800 N. As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32°, and the angle between the boom and the horizontal is 48°.

what i have done which i guess is wrong is.
L*5400 - Lsin(32)Ft + 1/2Lsin(48)(3800)=0
and then i got rid of the L's
and got 5400 - sin(32)Ft + 1/2sin(48)3800=0

but this was wrong, if you could just explain to me how to do this it would really help a lot thanks.
 
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  • #2
you have posted only the background info, not the actual problem (whatever it is)! Also...I don't see where "L" is defined...so I have no idea what that is either.
 
  • #3
it wants to know the force of the cables tension, the length isn't defined but we can get it out right?
 
  • #4
Okay. So the problem is asking you to find the force due to tension in the support cable. When I said that L wasn't defined, I didn't mean that the problem didn't specify the length, I meant that you introduced the symbol "L" without stating what it was. In fact, it is still unclear what it is. Remember that we can't see the picture, so it's hard to figure out what's going on.

Your equation seems to be trying to sum the torques around the point where the boom is attached to the tractor. So your terms seem to be

L*5400 = the torque on the boom (around that point) due to the weight of the wrecking ball (so, I'm inferring that you meant that L is the length of the boom). I don't think this term is correct, because you want to multiply the force by the moment arm, or lever arm, which is the perpendicular distance from the point of application of the force to the pivot point. That is not just the distance along the lever. The force is downward, so this perp. distance is the horizontal distance.

L/2*sin(48)*3800 seems to be the torque on the boom due to its own weight. It is in the same direction as the torque above. It has an error too. We want the moment arm, or the horizontal distance in this case. Is that given by Lsin(48)?

L*sin(32)*Ft

Now this one is very confusing. It seems that you're trying to calculate the counter torque provided by the support cable, (ie the torque that must prevent the boom from actually falling). But if so...then why did you use L? Isn't L the length of the boom? It's the force in the cable we're talking about now. And even if you had used the right length, can you see again why sine is not correct?
 
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1. What are the "boom problems" in question?

The term "boom problems" refers to a type of engineering problem that involves issues with a crane's boom, which is the long, horizontal arm used for lifting and moving heavy objects. These problems can range from structural issues to mechanical failures.

2. Why are boom problems important to address?

Boom problems can pose serious safety risks, both to the workers operating the crane and to those in the surrounding area. They can also result in costly damage to equipment and project delays.

3. What are some common causes of boom problems?

Some common causes of boom problems include improper use or maintenance of the crane, overloading the crane, structural weaknesses, and mechanical failures.

4. How can boom problems be prevented?

To prevent boom problems, regular maintenance and inspections of the crane should be conducted. Proper training and adherence to safety protocols are also crucial. It is also important to follow the manufacturer's recommended load limits and avoid overloading the crane.

5. What steps should be taken if a boom problem occurs?

If a boom problem occurs, the first step is to ensure the safety of all personnel in the area. The crane should then be shut down and the problem should be assessed by a qualified engineer or technician. Depending on the severity of the issue, repairs or replacements may need to be made before the crane can be safely used again.

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