<p>For one of the questions in the Barron's book, it takes about the frictional force of an object on a plane as the plane is rotated from zero to fifty degrees.</p>
<p>We are asked to determine how the frictional force changes during this time.</p>
<p>I reasoned that F is proportional to the N = mg cos theta, which of course is decreasing. Therefore, I thought that the frictional force is steadily decreasing. I also considered the extreme case when theta = 90 degrees; there is no frictional force. </p>
<p>However, the Barrons explanation read: "As long as there is no sliding, the force of friction is just equal and opposite to the component of the weight PARALLEL to the plane..." </p>
<p>Hmm, I always thought it was proportional to the normal force, which is PERPENDICULAR to the plane. </p>
<p>if u look at mgsintheta, and if the coeff of static friction is equal to mgsintheta, then it wouldn't move, if the prob didn't say anything about the box moving down.</p>
<p>i would have thought the same thing that if the box were rotated, then the normal force would decrease, but only when it's moving. hope my exp is ok.</p>
<p>I forgot to realize that static friction is LESS THAN OR EQUAL TO the coefficient of friction x normal force. So yes, it would equal mg sin theta, which obviously increases. Got it, thanks.</p>
<p>Not too bad. I got 760 on the Barrons test. I took Physics B like three years ago, so I had to review all the B topics (they weren't covered in Physics C), which i just finished today. Hmm, I guess i still have to get a bit better on ray diagrams and a bit better at modern physics. </p>