The problem:

A $100\,\mathrm{kg}$ cubical box lies on the floor. A child pushes horizontally at the top edge. What is the magnitude of force to put the box on the verge of tipping over, given that there is sufficient friction between it and the floor to prevent sliding?

My questions:

At which point will gravity act? On center of mass? I am also confused about how the rotation equations should be applied to the box. Will the torque remain constant while box is being pushed?

  • $\begingroup$ "Will the torque remain constant while box is being pushed ?" Do you mean the torque due to gravity? $\endgroup$ – K. Kirilov Mar 31 '18 at 19:31
  • $\begingroup$ Actually, the force is essentially zero when the box is on the verge of tipping. $\endgroup$ – Hot Licks Mar 31 '18 at 19:53
  • $\begingroup$ Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – ACuriousMind Mar 31 '18 at 21:15

In fact you want to rotate the cube about the opposite bottom side. For this you have to counteract the moment of the gravitational force, indeed acting on the center of mass. Just make a drawing and you will figure it out.

The child may need some help, though!


Not the answer you're looking for? Browse other questions tagged or ask your own question.