Solve Statics Problem for Bone Rongeur Machine Force at E

[cyberdeathreaper]

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...
$$\sum F_x = 0 = D_x + B_x$$
$$\sum F_y = 0 = F_E + D_y - B_y$$
$$\sum M_D = 0 = -1.2 F_E - 1.6 B_y - 0.45 B_x$$
For the top right piece...
$$\sum F_x = 0 = -B_x + A_x$$
$$\sum F_y = 0 = B_y + A_y - 25$$
$$\sum M_A = 0 = -110 -1.1 B_y + 0.45 B_x$$

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.

 

[OlderDan]

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...
$$\sum F_x = 0 = D_x + B_x$$
$$\sum F_y = 0 = F_E + D_y - B_y$$
$$\sum M_D = 0 = -1.2 F_E - 1.6 B_y - 0.45 B_x$$
For the top right piece...
$$\sum F_x = 0 = -B_x + A_x$$
$$\sum F_y = 0 = B_y + A_y - 25$$
$$\sum M_A = 0 = -110 -1.1 B_y + 0.45 B_x$$

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.

 

[TenaliRaman]

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

 

[cyberdeathreaper]

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?