When does an object rotates and translate at the same time? I have a ruler on the table. 
If I push it with my finger on its end, it will rotate. 
but if I push it hard enough, it will rotate and slide slightly. 
Which force cause this sliding motion? and how to calculate it? If my ruler was floating in space, does it still slide when I push it hard on its end?

 A: The fact that it rotates on one end without sliding probably has something to do with the shape of the ruler and how it is in contact with the table.  Presumably, the middle part of the ruler is in good contact, and the frictional force on the ruler at that point provides a pivot point allowing it to rotate.  If you push it hard enough, you are likely overcoming the static frictional force and the ruler will then slide as well as rotate.
The two concepts at play here are linear momentum and angular momentum.  Because you are pushing the ruler away from its center, you are applying a torque to the ruler.  This torque supplies angular momentum to the ruler causing it to rotate.  When applying a force to an object, you are also supplying some amount of linear momentum if the net force is not zero.  When the ruler doesn't slide, the frictional force supplied at the pivot point is canceling out the force you supply, resulting in zero net force, but not canceling the net torque on the ruler.
In the situation with the ruler in space, there would be no other forces acting on the ruler other than the force of you pushing it.  Therefore you would be supplying both a net force and a net torque, meaning the ruler would go away from you while also rotating.
To calculate the linear acceleration, you can use Newton's laws, which simply states that
$F_{net} = ma$
In this case the net force would be the vector sum of the frictional force on the ruler and the applied force.  Here, the frictional force will oppose the force you apply, so, $F_{net} = F_{app} - F_{fric}$.  If the ruler doesn't move, $F_{net} = 0$.  The frictional force can be found from $F_{fric} = \mu mg$, where $\mu$ in the case of the sliding ruler is the kinetic coefficient of friction.
The torque on the ruler is given by
$\tau = r\times F$
where $r$ is the distance from the center of mass that you are applying the force.  If you apply the force perpendicular to the ruler, the torque is simply $\tau = rF$
