What is the Torque on Point A?

[OneCosmicGuy]

Homework Statement

I only need help on one part of this question - if you look at the pict in the link below, I ultimately need to determine the torque on point A.
Length F and D are equal
LM (Large Mass) is twice that of the Small Mass (SM) or point G & E respectively.

I know I need to calculate the downward force but all the force is applied at point C which, being angular, is actually two forces, one vertical and one horizontal. It seems like the arraignment and size difference between the two masses might shift the horizontal component of the force?

Homework Equations

The Attempt at a Solution

http://members.cox.net/cosmicatom3/CC-Torque.jpg"
Thank you all in advance - I do appreciate it!

 

[korkscrew]

I'm still a bit unclear on the set up. Are you given any angles between the lever arms and the force vectors? I'm assuming this diagram is drawn with gravitational force downward? I believe you can just add the masses and calculate the torque with the lever arm B and the angle between K and B, if you've got it...

 

[tiny-tim]

Welcome to PF!

Hi OneCosmicGuy! Welcome to PF!

I only need help on one part of this question - if you look at the pict in the link below, I ultimately need to determine the torque on point A.

I know I need to calculate the downward force but all the force is applied at point C which, being angular, is actually two forces, one vertical and one horizontal. It seems like the arraignment and size difference between the two masses might shift the horizontal component of the force?

For the torque about the point A,

you only need the external forces …

so don't worry about the forces at C …

they're only internal.

(and the horizontal components will cancel out anyway)

 

[OneCosmicGuy]

... Are you given any angles between the lever arms and the force vectors? I'm assuming this diagram is drawn with gravitational force downward? ...

Yes, gravity is downward and the angle the major angle CAGround would be 45 in this example. Thank you for your reply :-)

 

[OneCosmicGuy]

Hi OneCosmicGuy! Welcome to PF!


For the torque about the point A,

you only need the external forces …

so don't worry about the forces at C …

they're only internal.

(and the horizontal components will cancel out anyway)

Thank you for the reply and the welcome! I spent considerable time browsing through a lot of threads before posting - you all have quite a nice group here and I'm sure many people appreciate your help. Just for the purposes of full disclosure (always been a fan of honesty), I'm not actually in an academic class at this point - this is small piece of a much bigger project/concept I'm working on at home (does that still count as homework? haha) and trying to validate the math before I actually spend money and time attempting to build something that might not even work.

As for internal and external... that kind of makes sense, but I'm not exactly sure how to apply that I guess... I mean, if segment F was say, four times longer and D stayed the same, wouldn't that affect the overall torque being applied to A? Or is that what you mean by external? The way I see it, we have two downward forces at work, one for each mass. But the downward force is applied through point C, correct? Although you said to ignore C because that's internal... so apparently I'm not understanding how the forces act in this situation... would you mind expanding on your above answer? (and thank you in advance)

 

[tiny-tim]

… The way I see it, we have two downward forces at work, one for each mass. But the downward force is applied through point C, correct? …

I assume you're worrying about how to construct A so as to bear the whole torque …

in that case, yes you're right, the forces at G and E are applied through C (that's the only way they can get to A!) …

but it doesn't affect the calculations.

Imagine that GC is a separate rod, glued to ACE at C …

then the forces on GC are the weight J, the weight of the rod itself, and the force exerted on GC from ACE …

and there's also forces on ACE …

including (Newton's third law) an equal-and-opposite reaction force from GC on ACE …

these two forces are the internal forces I'm talking about …

in a sense, you do have to take them into account, but they always cancel out.

 

[OneCosmicGuy]

I assume you're worrying about how to construct A so as to bear the whole torque …

Partially but I'm mostly concerned about a balance situation (against something not shown in the diagram). Think teeter-toter or wheel type of situation - I need to know if it's going to rotate or balance.

Imagine that GC is a separate rod, glued to ACE at C …

then the forces on GC are the weight J, the weight of the rod itself, and the force exerted on GC from ACE …

(Q1) So what I'm hearing is the distance of G and E from C really doesn't matter when looking at the torque experienced at A??

(Q2) If not, then does the distance of G and E matter with respect to A? I kind of thought it would since torque is force x distance (hence the red and blue dotted lines - which would then be broken down into horizontal and vertical components)

(Q3) Just for the sake of my understanding, if I were to alter GE and... say it's perpendicular to CA... the overall weight is unchanged, but the distance of the large mass and the small mass to point A has now changed - not by much but it has changed so wouldn't that alter the torque experienced at A?

If it would be easier to illustrate, I can assign numbers to the parts. Maybe seeing how the numbers are used would make that light bulb reach 100 watts? LOL

Adam <--- feels some what "dim-witted" right now LOL

 

[tiny-tim]

100 watts!

Hi Adam!

… teeter-toter …

oooh … i had to look that up on google … turns out it's what we in England call a see-saw!

(Q1) So what I'm hearing is the distance of G and E from C really doesn't matter when looking at the torque experienced at A??

That's right.

(Q2) If not, then does the distance of G and E matter with respect to A? I kind of thought it would since torque is force x distance (hence the red and blue dotted lines - which would then be broken down into horizontal and vertical components)

Also right …

except I'm not sure you've got the "x" right …

torque = force "cross" distance

(or torque = force times perpendicular distance = force times lever arm) …

so the red and blue dotted lines don't matter, but the horizontal distances from A to J and L do.

(Q3) Just for the sake of my understanding, if I were to alter GE and... say it's perpendicular to CA... the overall weight is unchanged, but the distance of the large mass and the small mass to point A has now changed - not by much but it has changed so wouldn't that alter the torque experienced at A?

If it would be easier to illustrate, I can assign numbers to the parts. Maybe seeing how the numbers are used would make that light bulb reach 100 watts? LOL

As in (Q2), it's the horizontal distances from A to J and L which matter.

(and no, the letters are fine … good diagram, btw! )

™

 

[OneCosmicGuy]

Okay, you probably feel like this is a zombie thread... you think it's dead and it comes back to haunt you LOL... so its the distance from A to J and L that matter... I'm assuming you simply add the two together to find the total force acting on A?

Hope your holidays were great!

Adam

 

[tiny-tim]

happy new year!

Okay, you probably feel like this is a zombie thread... you think it's dead and it comes back to haunt you LOL... so its the distance from A to J and L that matter... I'm assuming you simply add the two together to find the total force acting on A?

Hi Adam!

yes, the total torque at A is the weight of G times AJ plus the weight of E times AL

(torque is a vector, so of course it obeys the laws of vector addition, but in this case the vectors are parallel, so yes you simply add the magnitudes … but if GE was twisted out-of-plane, you would have to add the torques properly as vectors)

 

[OneCosmicGuy]

It's that zombie post coming back to life again haha

I had a feeling the horizontal positioning of GE made it easy so let's change it up a bit just to stretch my brain a bit more.

I understand that we now have to deal with force vectors of which there are two in different directions who's sum(?) = net force J and L ??

J = CG & GA ??
L = CE & EA ??

Those aren't right angles to each other which means... I've got it wrong??

Hope your 2009 is going slower then mine - I can't believe January is gone already.

Thanks in advance :-) This is fun actually :-)

Adam

 

[tiny-tim]

Hi Adam!

… J = CG & GA ??
L = CE & EA ?? …

It's still only the horizontal distances from A to J and L which matter.

This is fun actually :-)

Physics is always fun!
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