Static Forces of a Trebuchet

  • #1

Homework Statement


The goal is to calculate the static forces in the throwing arm of the trebuchet when it's pinned as shown in the image. The counterweight is equal to 1435 newtons. The mass of the arm is equal to 114 newtons. Dimensions are shown in inches. The pin is in double shear since it passes through three plates.

Homework Equations


∑M, ∑Fx, ∑Fy

The Attempt at a Solution


∑MAxle = (1435 N)(11.12") - (114 N)(9.09") + (FYPin)(49.59") - (FXPin)(33.23") = 0

∑Fy = -1435 - 114 + FYAxle + FYPin = 0

∑Fx = -FXAxle + FXPin = 0

So this is where I'm stuck. If FXAxle = FXPin then I'm left with 2 equations and 3 unknown values. And there's no moments about the axle correct? - since it rotates freely? Any advice would be greatly appreciated.

https://www.photobox.co.uk/my/photo/full?photo_id=501397650911
 
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Answers and Replies

  • #2
gneill
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Your image is behind a login wall, hence cannot be viewed. We discourage remotely located images for reasons such as this. Please upload your image to our server using the UPLOAD button at the bottom right of the edit window.
 
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  • #3
Your image is behind a login wall, hence cannot be viewed. We discourage remotely located images for reasons such as this. Please upload your image to our server using the UPLOAD button at the bottom right of the edit window.
Thanks for the heads up. I've attached the pdf file. I've also added one more equation:

ΣMPin = (1435 N)(60.71") - (FYAxle)(49.59") + (FXAxle)(33.23") + (114 N)(40.5") = 0.

So this gives me four equations and four unknowns. This will allow me to solve for the X and Y forces at the pin and at the axle correct?
 

Attachments

  • #4
gneill
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So this gives me four equations and four unknowns. This will allow me to solve for the X and Y forces at the pin and at the axle correct?
I would think so, yes, so long as you've got all the signs and distances correct for all your moments. For example, in your first equation how did you decide on the signs and distances for your FYpin and FXpin?
 
  • #6
Note the lines of action of the Fpin forces:
View attachment 235160
I chose the sign convention to be positive in the counter-clockwise direction. Here's my work. Other than direction of the forces on the x direction, I think it's correct. Would you mind looking it over?
 

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  • #7
gneill
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I chose the sign convention to be positive in the counter-clockwise direction. Here's my work. Other than direction of the forces on the x direction, I think it's correct. Would you mind looking it over?
I can barely make out the diagram and writing in the image, so I'm afraid I won't be spending time on it. I will comment on the work you've typed in though.

In your first equation:
∑MAxle = (1435 N)(11.12") - (114 N)(9.09") + (FYPin)(49.59") - (FXPin)(33.23") = 0
Where does the coefficient 49.59" on the FYPin term come from? It looks like it might be the span of the arm between the pin and the axle. That means that you consider the Y-component of the pin force to be acting perpendicularly to the arm. That contradicts your other usage of FYPin where you consider it to act vertically.
For your third equation:
∑Fx = -FXAxle + FXPin = 0
I don't understand the negative sign on FXAxle, probably because I didn't find any labelled forces on your first image. You might find it simpler to just assume generic positive values for the force components and then let the math sort out the signs.
upload_2018-12-5_10-17-18.png


I also question your coefficient of 60.71" on the counterweight term:
ΣMPin = (1435 N)(60.71") - (FYAxle)(49.59") + (FXAxle)(33.23") + (114 N)(40.5") = 0.
 

Attachments

  • #8
I can barely make out the diagram and writing in the image, so I'm afraid I won't be spending time on it. I will comment on the work you've typed in though.

In your first equation:

Where does the coefficient 49.59" on the FYPin term come from? It looks like it might be the span of the arm between the pin and the axle. That means that you consider the Y-component of the pin force to be acting perpendicularly to the arm. That contradicts your other usage of FYPin where you consider it to act vertically.
For your third equation:

I don't understand the negative sign on FXAxle, probably because I didn't find any labelled forces on your first image. You might find it simpler to just assume generic positive values for the force components and then let the math sort out the signs.
View attachment 235255

I also question your coefficient of 60.71" on the counterweight term:
Hi, sorry my replies are few and far between. Dead week is aptly named. I've learned that I can treat the force at the pin as just one resultant force, perpendicular to the throwing arm. So I have three equations and three unknowns. I found the angle by taking the arctan(vertical distance from pin to axle/horizontal distance from pin to axle) = arctan(28.52/40.50) = 31.15°. I then found the complementary angle 90° - 31.15° = 54.85°. I also used the original units of lbs since those were the original units and are a bit easier for me to understand. The three equations are:

ΣFx = Fxaxle - Fpin*cos(54.85) = 0
ΣFy = Fyaxle - Fpin*sin(54.85) - 322.5 lbs - 25.74 lbs
+CCWΣMpin = (322.5 lbs)(52.62") - (Fyaxle)(40.50") - (Fxaxle)(28.52") + (25.74 lbs)(31.41") = 0

Solving, I get:
403.57 lbs for Fyaxle
67.68 lbs for Fpin
38.96 lbs for Fxaxle

I believe these are the correct values.
 
  • #9
gneill
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In your moment equation, where does the 52.62" distance on your counterweight term come from?
 
  • #10
In your moment equation, where does the 52.62" distance on your counterweight term come from?
A typo on my end, 11.12 + 40.50 = 51.62. I used the correct number in my work.
 
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