I need a math wiz, not car related

Jeanius · 04-08-2010, 11:51 AM

Ok folks, I need someone to help me with this problem, I get a diff answer than the book using the same formula for every other problem, and I dont know WTF!

Effort = Resistance x Resistance Distance / Effort Distance

Three people are trying to raise a 30-foot flagpole that measures 3 inches in diameter. The total weight of the flagpole is 95lbs. If one person served as the fulcrum by anchoring the end of the flagpole to the ground, and the other two people attempted to lift the flagpole at a point 12 feet from the fulcrum, how much resistance would they encounter?

Moriartii · 04-08-2010, 12:22 PM

So would that be
E=(95*30)/18
E=158.3 Repeating?

BigRigMike · 04-08-2010, 12:58 PM

Is the answer 118.75?

Jeanius · 04-08-2010, 01:14 PM

Quote:

Originally Posted by BigRigMike

Is the answer 118.75?

According to the book it is!

How did u get it?

BigRigMike · 04-08-2010, 01:21 PM

Resistance = 95lbs
Effort distance = 12 Ft

Resistance distance = 15ft - I assume this is where we differ because you should be using the center of gravity of the pole which is the middle or 15ft

Xanthos · 04-08-2010, 01:23 PM

Quote:

Originally Posted by BigRigMike

Resistance = 95lbs
Effort distance = 12 Ft

Resistance distance = 15ft - I assume this is where we differ because you should be using the center of gravity of the pole which is the middle or 15ft

This is the answer I got as well.
- X

Moriartii · 04-08-2010, 01:25 PM

But that assumes the CG is in the center of the pole which it is not since it is anchored at the end. So the RD would actually be 30 ft minus the fulcrum location of 12ft making it 18ft??

Cheers
K

ron10 · 04-08-2010, 01:27 PM

AnnaBear@SpartaChevy · 04-08-2010, 01:28 PM

Where to even start?!

Xanthos · 04-08-2010, 01:29 PM

Quote:

Originally Posted by Moriartii

But that assumes the CG is in the center of the pole which it is not since it is anchored at the end. So the RD would actually be 30 ft minus the fulcrum location of 12ft making it 18ft??

Cheers
K

But the fulcrum location is not at 12ft, its at 0ft because thats where the pole is anchored. Assuming that its a constant density, the center of gravity would be the halfway point of the pole.
- X

tradosaurus · 04-08-2010, 01:30 PM

Wouldn't the resistance vary (become easier as you lift the pole because the center of gravity changes as the angle of the pole approaches 90 degrees?

Moriartii · 04-08-2010, 01:31 PM

Ah I see what you are saying.
Cheers
K

Xanthos · 04-08-2010, 01:33 PM

Quote:

Originally Posted by tradosaurus

Wouldn't the resistance vary (become easier as you lift the pole because the center of gravity changes as the angle of the pole approaches 90 degrees?

Yes, but that could be said about any situation. This equation just solves for the force at a given point in time. In this case, immediately as you begin to lift.
- X

Jeanius · 04-08-2010, 01:38 PM

Quote:

Originally Posted by Moriartii

But that assumes the CG is in the center of the pole which it is not since it is anchored at the end. So the RD would actually be 30 ft minus the fulcrum location of 12ft making it 18ft??

Cheers
K

Thats what I thought too

thanks guys for the help, it makes sense now

04-08-2010, 11:51 AM	#1
Jeanius Drives: Blk/Blk 2018 Sierra SLT Join Date: Dec 2006 Location: Golden State Posts: 2,925	I need a math wiz, not car related Ok folks, I need someone to help me with this problem, I get a diff answer than the book using the same formula for every other problem, and I dont know WTF! Effort = Resistance x Resistance Distance / Effort Distance Three people are trying to raise a 30-foot flagpole that measures 3 inches in diameter. The total weight of the flagpole is 95lbs. If one person served as the fulcrum by anchoring the end of the flagpole to the ground, and the other two people attempted to lift the flagpole at a point 12 feet from the fulcrum, how much resistance would they encounter? __________________ SOLD: 2SS LS3 M6 - MGW - Dynatech LT's HF Cats - Magnaflow Street CB - ADM Race CAI - Pedders Xa Coilovers - Pfadt Sways TRADED: 2016 2SS M6

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04-08-2010, 12:22 PM	#2
Moriartii Drives: 11 IOM 2SS/RS M6 w/Hurst Join Date: Nov 2009 Location: Austin Texas Posts: 1,269	So would that be E=(95*30)/18 E=158.3 Repeating?

04-08-2010, 12:58 PM	#3
BigRigMike Drives: 2010 Ford Fusion Join Date: Aug 2008 Location: York, PA Posts: 374 View My Garage	Is the answer 118.75?

04-08-2010, 01:21 PM	#5
BigRigMike Drives: 2010 Ford Fusion Join Date: Aug 2008 Location: York, PA Posts: 374 View My Garage	Resistance = 95lbs Effort distance = 12 Ft Resistance distance = 15ft - I assume this is where we differ because you should be using the center of gravity of the pole which is the middle or 15ft

04-08-2010, 01:25 PM	#7
Moriartii Drives: 11 IOM 2SS/RS M6 w/Hurst Join Date: Nov 2009 Location: Austin Texas Posts: 1,269	But that assumes the CG is in the center of the pole which it is not since it is anchored at the end. So the RD would actually be 30 ft minus the fulcrum location of 12ft making it 18ft?? Cheers K

04-08-2010, 01:27 PM	#8
ron10 Drives: 2010 2SS IOM L99 Join Date: Jan 2010 Location: LOUISVILLE,KY.. Posts: 7,545

04-08-2010, 01:28 PM	#9
AnnaBear@SpartaChevy Drives: 2011 Camaro 1SS Manual Synergy Grn Join Date: Apr 2009 Location: Sparta Chevrolet, Sparta, MI Posts: 2,107	Where to even start?!

04-08-2010, 01:30 PM	#11
tradosaurus Account Suspended Drives: 2010 F150 Lariat, Candy Red Join Date: Aug 2009 Location: Texas Posts: 1,279	Wouldn't the resistance vary (become easier as you lift the pole because the center of gravity changes as the angle of the pole approaches 90 degrees?

04-08-2010, 01:31 PM	#12
Moriartii Drives: 11 IOM 2SS/RS M6 w/Hurst Join Date: Nov 2009 Location: Austin Texas Posts: 1,269	Ah I see what you are saying. Cheers K