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In a horizontal surface, a block (cube) is sliding due to a sudden push. When the block slides, there is frictional force which is acting on the block.

Frictional force will have a torque around the center of mass, so why does the block not rotate/roll around (a horizontal axis through) the center of mass instead of sliding?

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/79656 (admittedly this doesn't have much of an answer either) $\endgroup$ – nivag Jun 5 '14 at 15:09
  • $\begingroup$ What makes you think the frictional force will have a torque about the center of mass? $\endgroup$ – Pranav Hosangadi Jun 5 '14 at 15:10
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    $\begingroup$ @PranavHosangadi: the friction is on the bottom of the block, the CM is presumably halfway up, so there is a torque. $\endgroup$ – Ross Millikan Jun 5 '14 at 15:11
  • $\begingroup$ @RossMillikan Ah, I didn't think of it in that sense. I thought OP was asking about rotation about the vertical axis. $\endgroup$ – Pranav Hosangadi Jun 5 '14 at 15:13
  • $\begingroup$ The torque is cancelled by the torque of the force normal to the plane if the object does not rotate. Often, problems in mechanics assume only displacement of the object, not rotation. $\endgroup$ – auxsvr Jun 5 '14 at 15:18
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First consider the initial push. If the friction is high enough and the push is high enough on the block it will roll instead of slide. Generally static friction is greater than dynamic friction, so if it starts sliding it will continue.

If the friction gets great enough, the block can roll. Say you have a box sliding on ice that comes to the edge of the ice and starts sliding on dirt. The friction will increase considerably. You compare the torque the friction applies to the box against the torque required to lift the back edge against gravity. If it is greater, the box will start to roll.

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The image below shows a rough diagram of the forces on a box just as it would begin to rotate.

enter image description here

In this case we can consider the normal force to the surface as focused on the leading corner about which the box would rotate.

This results in 2 torques on the box. The first from friction which is causing the box to rotate $$\frac{L}{2}Fr$$

and the second due to the normal force which is in the opposite direction. $$\frac{H}{2}mg$$.

If the first is bigger the box will at least begin to rotate. Note it may not actually flip as the ratio can change as the box decelerates and begin to rotate. This will be true if $$LFr > Hmg$$ using the standard friction law we get $$L\mu mg > Hmg $$ $$H > L\mu$$

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The normal force of the surface on the block supplies a counter-balancing torque. With regard to these torques, the system is static, that is, the block is not rotating. The normal force is able to provide whatever torque is needed to prevent rotation ... to a point. In some cases if you put a block on a plank and push the block it will slide, but if the coefficient of static friction is high enough, it will tumble instead. (Imagine that you are pushing with your finger toward the center of mass of the block so that your finger provides no torque around that point.)

If it helps, recall that the normal force is able to provide whatever magnitude of force is needed to prevent the block from accelerating downward (through the table, as it were).

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