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

  • Thread starter Thread starter mathuria1986
  • Start date Start date
  • Tags Tags
    Forces Pivot
AI Thread Summary
To determine the force at the end of a toggle clamp arm using a swivel type pneumatic cylinder, one must consider the orientation of the system relative to local gravity. The force exerted by the cylinder is 300 kg, which translates to a force of 2940 N when accounting for gravity. Calculating the moment involves finding the perpendicular force component acting on the pivot and using the formula moment = F_perp * d, where d is the distance to the pivot. There is confusion regarding the mention of a 235 kg force, as this should be expressed in Newtons for clarity. Simplifying the diagram and clearly labeling forces and units will aid in accurately solving the problem.
mathuria1986
Messages
3
Reaction score
0
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
 

Attachments

  • pivot.jpg
    pivot.jpg
    11.8 KB · Views: 482
Physics news on Phys.org
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
 

Attachments

  • COMP-Model.jpg
    COMP-Model.jpg
    13.9 KB · Views: 477
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).
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top