mathuria1986
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How to Find the Force at the End of a Toggle Clamp Arm?
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gash789
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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
[itex]\textrm{moment}=F_{perp} d[/itex]
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.
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
[itex]\textrm{moment}=F_{perp} d[/itex]
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.
mathuria1986
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gash789
- 57
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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).
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).