# Force on bottles

1. Aug 5, 2012

### kapital

If we act with force on bottle in two different way, we get two results:

a) If push(with hand) bottle at upper side of it, it will fall down(go to horizontal position),but if we

b) push bottle with the same force at lower side, it will move forward.

Why is this happening? Could we calculate at which height is limit from where one occurrence come to other?

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2. Aug 5, 2012

### Danger

Welcome to PF, Kapital.
It's a matter of leverage. In a bottle like that, the centre of gravity is low. Hitting at or below that point will impart linear impulse. Hitting higher will impart torque that works around the CG. (If it was a very tall bottle, with the CG higher, hitting low would also tip the thing over, but in the opposite direction.)

3. Aug 6, 2012

### jbriggs444

A force will always impart a linear impulse. Unless it is applied directly toward the axis of rotation, it will also impart a rotational impulse.

Note that in the physics vernacular, the "axis of rotation" is an arbitrary choice. One chooses an axis of rotation that will make the analysis of a problem simpler. The center of gravity is not the only possible choice, but it is a reasonable one and makes the physics notion of "angular momentum" match up well with our everyday notion of rotation.

If the force in this case is applied as a sharp blow then a blow above the center of gravity will impart a forward rotational impulse. A blow below the center of gravity will impart a backward rotational impulse. In addition to this, the force will always impart a forward linear impulse.

Following the blow, the [now moving] bottle will experience torques from friction on the table and from the supporting force from the table.

Determining whether the resulting rotation is sufficient to take the bottle past the tipping point is a moderately complicated exercise that could involve the coefficient of static friction, the height and magnitude of the applied force, the width of the base of the bottle, the height of the center of gravity, and the bottle's mass and moment of inertia.

If the force were applied smoothly, steadily ramping up until something happens then the analysis is much simpler. Choose the far corner of the bottle's base as the axis of rotation. The higher the point of application of the force, the greater the torque it imparts.

There is a critical height which depends on the width of the bottles base and the coefficient of static friction. Push above this height and you get a tip. Push below this height and you get a slide. This height can be computed as:

h = 1/2 * b / μ

Where h is the height of the applied force, b is the width of the base and μ is the coefficient of static friction. [The height of the center of gravity does not matter in this situation]