Solve Statics Problem for Bone Rongeur Machine Force at E

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The discussion centers on calculating the force exerted on the bone at point E of a bone rongeur machine when two 25-lb forces are applied. The user initially attempts to use equilibrium equations from four free-body diagrams but struggles to derive the force at E. Another participant suggests that the symmetry of the machine allows for easier calculations using couple ratios, leading to the correct answer of 133.3 lb. The user later confirms they found the solution using couple ratios but still seeks clarification on applying equilibrium equations effectively. The thread highlights the importance of both methods in solving statics problems.
cyberdeathreaper
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Here's the problem:

"The bone rongeur shown [refer to attachment] is used in surgical procedures to cut small bones. Determine the magnitude of the forces exerted on the bone at E when two 25-lb forces are applied as shown."

I understand that this "machine" can be broken into 4 free-body diagrams, and then I can use the equilibrium equations on each one to supposedly find the answer. However, my equations don't readily give me a way to solve for the force at E. Any ideas?

Here's all the equilibrium equations I have come up with:
For the top left piece...
<br /> \sum F_x = 0 = D_x + B_x<br />
<br /> \sum F_y = 0 = F_E + D_y - B_y<br />
<br /> \sum M_D = 0 = -1.2 F_E - 1.6 B_y - 0.45 B_x<br />
For the top right piece...
<br /> \sum F_x = 0 = -B_x + A_x<br />
<br /> \sum F_y = 0 = B_y + A_y - 25<br />
<br /> \sum M_A = 0 = -110 -1.1 B_y + 0.45 B_x<br />

It should be obvious that the bottom pieces are symmetric with the top pieces, and similar in their equilibrium equations.

NOTE: The book indicates the answer is 133.3 lb.
 

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cyberdeathreaper said:
Here's the problem:

"The bone rongeur shown [refer to attachment] is used in surgical procedures to cut small bones. Determine the magnitude of the forces exerted on the bone at E when two 25-lb forces are applied as shown."

I understand that this "machine" can be broken into 4 free-body diagrams, and then I can use the equilibrium equations on each one to supposedly find the answer. However, my equations don't readily give me a way to solve for the force at E. Any ideas?

Here's all the equilibrium equations I have come up with:
For the top left piece...
<br /> \sum F_x = 0 = D_x + B_x<br />
<br /> \sum F_y = 0 = F_E + D_y - B_y<br />
<br /> \sum M_D = 0 = -1.2 F_E - 1.6 B_y - 0.45 B_x<br />
For the top right piece...
<br /> \sum F_x = 0 = -B_x + A_x<br />
<br /> \sum F_y = 0 = B_y + A_y - 25<br />
<br /> \sum M_A = 0 = -110 -1.1 B_y + 0.45 B_x<br />

It should be obvious that the bottom pieces are symmetric with the top pieces, and similar in their equilibrium equations.

NOTE: The book indicates the answer is 133.3 lb.

You are correct about the symmetry. For the upper and lower halves, each is just a coupled double lever. You can easily calculate the forces at B and C from the applied force and the distance ratios. Then do the same thing to find the force on each side of E.
 
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Can u post a detailed description on how u arrived at the above equations?
You are correct, the four pieces of the instrument would give rise to four free body diagrams. However the final equations that u have got certainly have some components missing. So if u post how u analysed the free body diagrams (if possible do post the free body diagrams u have considered), it would be easier to point out the mistake (if any) u have made or possibly point out what u missed.

-- AI
 
Okay, I figured out how to get the answer via the couple ratios.
(25)(4.4)/(1.1) = By
(By)(1.6)/(1.2) = FE = 133.3 lb

However, I'm still not understanding how I could arrive at the via the equilibrium equations... attached is my free body diagrams for the top pieces. Any help?
 

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