How to Find the Force at the End of a Toggle Clamp Arm?

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Homework Help Overview

The discussion revolves around determining the force at the end of a toggle clamp arm, specifically in the context of a swivel type pneumatic cylinder setup. The original poster mentions a force exerted by the cylinder and provides a diagram for reference.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the force exerted by the cylinder and its components relative to gravity. There are attempts to balance moments around the pivot and questions about the correctness of the calculated force at the end of the arm.

Discussion Status

The discussion is active with participants providing insights on the complexities involved in the problem. Some guidance has been offered regarding the need for clarity in the diagram and the importance of using correct units. However, there is no explicit consensus on the calculations presented.

Contextual Notes

There are mentions of unresolved components of force and the necessity for clearer diagrams and unit specifications. The original poster's calculations and assumptions are under scrutiny, particularly regarding the use of mass versus force units.

mathuria1986
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i want to use a swivel type pneumatic cylinder to make a toggle clamp with a swinging arm along a pivot. the line diagram is shown in the figure can anyone help me finding the force at the end of arm.
force by cylinder on one end of arm= 300kg
 

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Hello mathuria1986,

This could be quite complex to solve, or simple depending on how this system is orientated with respect to the local gravity.

The 300Kg mass will have a force F=ma along this local gravity. You have drawn it such that there is an angle 23' to the x-axis (call it the x-axis such that the vertical is y). Is this is the direction of local gravity you can find the component acting perpendicular to the pivot then use
\textrm{moment}=F_{perp} d
where d is the distance to the pivot, then you simple have that the force ? is the moment divided by the length 10.0 (units?).

However you have an unresolved component that will form a net force on the whole system, if the system is not under a net force this will require another force to acting in the opposing direction.
 
i tried to solve it this way. i took moment along pivot. and balanced the force on the other end of arm ehich comes to 235 kg. is it correct?/

the file is attached here
 

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Hi,
when you say a force of 235Kg this does not make any sense. A Kg is a Mass, so under the acceleration of gravity it will have a force 235g=2350N in the negative vertical direction.

Since your diagrams do not provide this direction I can't tell you if you are making a mistake.

I do not wish to sound patronizing but it will help you if you simplify your diagram as much as possible and add units to the lengths.

Then for the force's either draw forces and label them F1 F2 etc (which should have units of N) or draw circles for the mass and draw the acceleration (gravity).
 

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