A question about torque

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JasonRox

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There is an example in the book that goes like this:

F1---------------A----------B--------F2

F1 = Vertical Pole Number 1
F2 = Vertical Pole Number 2
A = Centre of Gravity of Horizontal Pole
B = Extra Weight

The numbers aren't necessary.

Because B is farther away from pole F1, I'm assuming the force on F1 is greater than the force on F2.

The book says otherwise.

Who is right?

Note: The book has been wrong more than once.
 
N

NetSquirrel

Hi, Jason.

If this is set up like a high jump bar, then you'd expect there to be more force acting on F2, because the weight of B is not split equally between F1 and F2. (E.g. if F1 and F2 were your outstretched arms, you would expect F2 to be holding most of the weight.)

However, there would obviously be a greater moment acting (clockwise) about F1 than (anticlockwise) about F2. The book might not have made the distinction between the two.
 

JasonRox

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On the other hand, if we cut the bar at B and split the weight of B, greater force will be place on F1, right?

Visual

B/2=C
F1--------------------------C
F2-------C
 

russ_watters

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You are closer on your second try, JasonRox. The issue with the first example is that torque isn't relevant there. Assume for a moment that the beam is supported from underneath by pins. The only thing the beam can do to the pins is push down. In your second example, pins wouldn't work - the beam has to be supported rigidly (perhaps with an angled brace) to handle the torque.
 

Doc Al

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Maybe this mental image will help: Imagine two men holding each end of a heavy plank. If it's uniform and held level, they exert equal forces to hold up their ends. Now place an additional weight on top of the plank. If that weight was placed in the middle, the two men would share the load. But if it were placed closer to man #2, which man would have to lift harder?
 

JasonRox

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Doc Al, that's what I was thinking about the other day. It's all good now.

Thanks, guys.
 

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