Really need help with a question given only mg and coefficient of the friction of the two walls. Cannot find a equation that leads to actual number. Coefficient is 0.48, mg is 212N
Hi, Theresa. So, the short answer is that you don't, and there are plenty of physics teachers out there who ONLY use the sine function (which comes out of the full vector treatment using the cross product). When you do that, you let the angles between the vectors run from 0 to 2Pi, and negative signs come out of that as a matter of course. I've found that it's usually less confusing to forego the vector treatment and deal with magnitudes, assigning positives or negatives based on an inspection of what sort of rotation a particular torque will produce. I use the geometry of the system to my advantage, peeling off distances using either sine or cosine depending on which side of a triangle I'm trying to find. So, I get both functions because the forces can be oriented at any direction, and I'm starting off with a set angle given by the problem statement. I run through the process quickly starting at 6:15 in the video.
I've watched several ladder problem videos today, and you've made the best. Thanks Stephen!
My pleasure. Thanks for your comment!
Definitely the best video on this set of problems. Thanks!!
Really need help with a question given only mg and coefficient of the friction of the two walls. Cannot find a equation that leads to actual number. Coefficient is 0.48, mg is 212N
thanks very much this was very helpful !
Great video! Thank you :)
Not a problem, thanks for commenting.
When figuring out the torque equations for each force, why do we have to use sin for N2 and cos for gravitational torque?
Hi, Theresa. So, the short answer is that you don't, and there are plenty of physics teachers out there who ONLY use the sine function (which comes out of the full vector treatment using the cross product). When you do that, you let the angles between the vectors run from 0 to 2Pi, and negative signs come out of that as a matter of course. I've found that it's usually less confusing to forego the vector treatment and deal with magnitudes, assigning positives or negatives based on an inspection of what sort of rotation a particular torque will produce. I use the geometry of the system to my advantage, peeling off distances using either sine or cosine depending on which side of a triangle I'm trying to find. So, I get both functions because the forces can be oriented at any direction, and I'm starting off with a set angle given by the problem statement. I run through the process quickly starting at 6:15 in the video.
good video
Thanks!
War eagle
Awesome