Can you balance an object on a point if its centre of mass lies outside? If yes, how?
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$\begingroup$ take open box for example $\endgroup$– UmaxoOct 23, 2019 at 15:54
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1$\begingroup$ Like a horseshoe? Why couldn't you balance it? $\endgroup$– user234190Oct 23, 2019 at 15:59
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3$\begingroup$ What do you mean by balance? Hang a horseshoe on a nail, or concentricity like a ring, or total concentricity of a hollow sphere? $\endgroup$– Adrian HowardOct 23, 2019 at 17:01
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$\begingroup$ Like a ring, when its centre of mass lies in space. Can we balance the ring on a point on its periphery? $\endgroup$– PlatoOct 25, 2019 at 10:23
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3 Answers
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Provided the centre of gravity is directly under or over the point of support it will be balanced.
It is safer if the centre of gravity is below the support. It is unstable otherwise.
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Yes but only if the Vector ,originating from the center of gravity of the body in question and having the direction of weight , coincidences with a surface point of the body such that the normal Vector of the surface at the said point is also collinear with weight and in same direction. (If it coincidences at multiple such surface points, it can be balanced by supporting at each of the points)
For the body to be balanced, net moment of force needs to be zero in addition with net force being zero.
Let's assume that the body is balanced by supporting at a point A. Hence normal force exerted by the pivot on the body at point A is exactly equal in magnitude and opposite in direction to that of Weight of the body (mg). But this condition is not sufficient, net moment must also be zero. But we know two equal and opposite forces constitute a non zero couple, unless they are having same line of action.
The reaction at pivot A is in the direction of normal to the surface of the object at that point (assuming frictionless). The weight is acting from the center of mass. Hence to make their line of action same, the condition stated in the beginning of the answer must be satisfied.
Note : Normal to the surface must not only be collinear with weight but also in the same direction because with know normal force is exerted in a direction opposite to the normal vector of a surface and normal force is always greater than zero.
answered Oct 23, 2019 at 19:06
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You can balance it but there needs to be something physical connecting the location of the center of mass to the rest of the structure so that the object can be supported at its center of mass. The mass of that something needs to be negligible compared to the mass of the object to the extent that it does not materially alter the location of the center of mass of the object.
Check out the picture of the forks in a cork at the bottom of following link. Note the addition of the toothpick.
http://dev.physicslab.org/Document.aspx?doctype=3&filename=RotaryMotion_CenterMass.xml
Hope this helps.
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$\begingroup$ You can also see this effect with tight-rope walkers if they have a heavy pole and they hold that pole low enough. $\endgroup$ Oct 23, 2019 at 18:02
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$\begingroup$ As mentioned in the question comments, a horseshoe can be balanced without necessarily having anything physically connected to it's center of mass. $\endgroup$– Time4TeaOct 23, 2019 at 18:31
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$\begingroup$ @Time4Tea The OP is talking about the situation where the center of mass falls outside the object. The center of mass of a horseshoe oriented in a vertical plane falls within the horseshoe. But the COM of a horseshoe oriented in a horizontal plane does not. My answer pertains only to the latter. $\endgroup$– Bob DOct 23, 2019 at 19:00
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$\begingroup$ Sorry, I don't follow your sentence: "But the COM of a horseshoe oriented in a horizontal plane does not." In what way does it not? The COM always lies within the envelope of the object. $\endgroup$– Time4TeaOct 23, 2019 at 19:03
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$\begingroup$ @Time4Tea I'm not talking about the "envelope" of the object. I'm talking about the COM not being located on the physical object itself. Where would you put your finger on a horseshoe to balance it when oriented in a horizontal plane? Did you look at the two fork example at the bottom of the link I provided? How would you balance the forks if the toothpick wasn't provided? $\endgroup$– Bob DOct 23, 2019 at 19:14