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Physics: Balancing a beam

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shub

New Member
Can anyone help me with finding the formulas for this problem? I am also confused, because I don't see the 800 N weight on the diagram. I am not sure if this a goof by the instructor or something I am not understanding. Thanks in advance to anyone that can help.

"A 900 N weight and an 800 N weight act in the directions shown on a 5 meter long uniform rigid beam that has a mass of 41 Kg. (a) How much force and in what direction at point A is needed to balance the beam? (b) How much force is exerted by the support at B?"

8479-balance.JPG


p.s. I am not flooding the forum with problems. I may ask for one more for a total of 3, but I am trying to figure that one out on my own at the moment.
 

dknguyen

Well-Known Member
Most Helpful Member
I too see no 800N weight.

(a) You just calculate the torque that each for produces around fulcrum B (using right hand role, + would be CW and -would be CCW) and add them up. Then choose the force at A so the torque produced by A cancels the other sout.

(b) We are not talking about torques here. We are talking about net forces (ie. translation, not rotation). So you just add them up.
 
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shub

New Member
Okay. So I am not just stupid then. Perhaps we could work under the assumption that the 800 is supposed to be where the 75 N is?
 

dknguyen

Well-Known Member
Most Helpful Member
Okay. So I am not just stupid then. Perhaps we could work under the assumption that the 800 is supposed to be where the 75 N is?

It's your choice whether to follow the diagram or the text.
 

RCinFLA

Well-Known Member
Looks straight forward to me. I see no issue between diagram and text.

Start with just the beam by itself. You have uniformly distributed weight.

5 meters = 41 kg of weight so there is 41kg * 0.8m/5m on left, 41kg * 4.2m/5m on right.

6.56 kg of beam on left, 34.44 kg on right.

Now imagine beam with no weight. Equivalent to real is 6.56 kg at 0.4 m to left and 34.44 kg at 2.1m to right, or 3.28 kg at 0.8 m to left and 17.22 kg at point A for leverage around point B.

Now add the additional weights at their leverage point on the weightless imaginary beam. Move and recalculate the equivalent weights based on their leverage from the point 'B' to balance at point 'B'

The weight on B is just the sum of up's (negative number) and down's (positive number) of real weights/force (don't forget 41kg down force for beam). No leverage rating here. 1 newton = 9.807 kg
 
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Russ Hensel

New Member
I do not see a conflict between the diagram and text either, but am unclear on weather or not you are to include gravity, I would guess yes, else why tell you the mass. Force and weight are almost always in N not Kg ( mass ). It just says the weights act on the bean, not how, they could not be placed on it as they would both point down. So strings and pulleys may be involved but not show. The fact that the forces are from weights also suggest a gravitational field. I would give a C to the author of the question.
 
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