chapter12



   [|Go to applet here]   Use this to check your answers    Recall that the static friction force will **only be as large as it needs to be**. Say you push a box of weight 120N with a force 50N. If the coefficient of static friction for the surface is .5, then the frictional force will be 50N, even though it can be a maximum of 60N, because the static friction force will **only be as large as it needs to be**.

Think about something tipping. It rotates around a fixed point without undergoing any translational motion. So, within the context of this applet, this means that there can’t be a net horizontal force. If there was a net horizontal force, the box would simply slide in the direction of this force.

There are two forces causing torque in this situation: the applied force, applied at the upper left corner of the box, and the normal force, applied at the middle of the bottom of the box. With, respect to torque, all that needs to happen in order for the box to tip is the torque due to the applied force has to be greater than the torque due to normal force.


 * CATEGORY || SCORE (1-4) || POINTS (0-20) ||
 * Content || 4 || 20 ||
 * Organization || 4 || 20 ||
 * Accuracy || 4 || 20 ||
 * Appearance || 4 || 20 ||
 * Participation || 4 || 20 ||
 * TOTAL || 100 ||