Introductory Static Friction on an Incline Problem

Introductory Static Friction on an Incline Problem


– Good morning. If we slowly raise this end of the board we can solve for the
coefficient of static friction between the book and the incline. – ♫ Flipping Physics ♫ (book scrapes) – Bobby, what are our known values? – Well, it looks like
the book starts to slide when the incline angle is 15 degrees. I think that’s it. – [Mr. P.] And what
exactly are we solving for? – The coefficient of static
friction equals question mark. – The moment we are interested in in this demonstration is
the moment right before the book starts to slide. And the angle of the incline at that moment is 15 degrees. (book scrapes) – Billy, please give
us a free body diagram of all the forces acting on the book right before the book starts to slide. – The force of gravity is down, the board pushes up on the book, which is the force normal
which is perpendicular to the incline and up. The force of static friction is parallel to the incline and opposes
the motion of the book. Without friction the book
would slide down the incline so the force of static friction is up and parallel to the incline. – [Mr. P.] And now could you please break the force of gravity into its components. – Okay. Instead of the force of gravity we can now substitute in its components the force of gravity parallel, which is parallel to and down the incline, and the force of gravity perpendicular which is perpendicular
to the incline and down. – Bo, could you please do the next step? – Next we need to sum the forces. Let’s do the parallel direction. The net force and the parallel direction equals force of gravity parallel, which is positive because
it is to the right and force of static friction, which is negative
because it’s to the left. And this also equals
mass times acceleration in the parallel direction, but I don’t know what to do now. – Bo, that’s fine. Class, what is the
acceleration of the book in the parallel direction before the book starts to move? – [All Students] Zero. – Therefore, before
the book starts to move the force of gravity parallel and the force of static friction are equal in magnitude. So notice as the incline angle increases the force of gravity down the incline, what we call the force of
gravity parallel, increases. Therefore, the force of static friction which adjusts in
magnitude to keep the book from sliding relative to
the incline, also increases. The force of static friction continues to increase until the magnitude of the force of gravity parallel is just barely above the magnitude of the maximum force of static friction. At which point the friction switches from static to kinetic, decreases in value and the book begins to accelerate down the incline. Therefore, we use the maximum force of static friction in our equation because the force of static friction is at its maximum value when the incline is at 15 degrees. And because it is static friction we know the book isn’t moving yet, which means the
acceleration in the parallel direction is equal to zero. – Bobby, why don’t you substitute
in some equations, please. – Force of gravity parallel is mass times gravity– – [Bo] Acceleration due to gravity. – Mass times acceleration due to gravity times sin theta which equals force of static friction maximum. Which is the coefficient
of static friction times force normal. We need force normal. – [Bo] Force normal
equals force of gravity. – [Billy] No it doesn’t! – [Bobby] No. – [Bo] Yes it does. Force normal equals
mass times acceleration due to gravity. I’ve got it in my notes, somewhere. (paper crinkles) – Force normal does not
equal the force of gravity. Many students think it
does, however it does not. Now I will admit that often it does; however, you have to every time draw a free body diagram
and sum the forces in order to solve for the force normal. – Bo, please solve for the force normal. – The net force in the
perpendicular direction equals force normal minus force of gravity perpendicular. Which equals mass times
acceleration perpendicular. Acceleration in all directions is zero, so force normal equals force
of gravity perpendicular. Which equals mass times
acceleration due to gravity times cosine theta. Pretty much equals force of gravity it’s just force of gravity
perpendicular instead. It doesn’t really matter. – Ooo. Billy, please continue. – We can substitute for force normal mass times acceleration due to gravity times cosine theta, into the equation we had from the net force
and the parallel direction. We now have mass times
acceleration due to gravity times sine theta equals the coefficient of static friction times
mass times acceleration due to gravity, times cosine theta. Everybody brought mass to the party! – [Bobby] Actually, everybody brought the force of gravity to the party. (upbeat, funky digital music) ♫ Everybody brought mass ♫ – And we can divide
both sides by the cosine of the incline angle and the
sine of theta over cosine theta equals tangent theta. Therefore, the coefficient
of static friction between the book and the incline is equal to the tangent
of the incline angle right before the book starts to slide. Pretty cool, hey? – [Billy] That is pretty cool. – [Bobby] Yeah.
– [Bo] I guess. – We can plug in 15 degrees
for our incline angle and we get that the
coefficient of static friction between the book and the incline is 0.267949 or 0.27 with two significant digits. Thank you very much for
learning with me today. (chuckles) I enjoyed learning with you.

32 comments / Add your comment below

  1. You need to somehow attract more subscribers. These videos definitely need to be discovered by students. The more people know about these, the more people subscribe. And this stuff is remarkable. Better than any physics tutorials out there. What's your motivation behind making these?

    By the way, try contacting PBS and see if you guys can collab. Might help your channel out.

  2. Very nice and enjoying way of learning… Bro u make me visualise the force in real world, those force arrows gives clear insight whats happening around. 😊

  3. Bro i just wanna say.. I got one my type of physics understanding guy , its you 😎.. I love discussing quantum Mechanics and wants to go in every detail to unify macro level physics to micro.

  4. But since the books begins to slide when the angle is 15°, then aren't you solving for the minimum coefficient of kinetic friction to move the book?

  5. I certainly enjoy your videos. I have a question based on this video. If a block is at rest on the incline plane at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for fs(max) = UsMgcos (minimum angle needed for block to slide)?

  6. that's good but u didn;t tell how the inclination angle is equal to the sinO or cosO that you're substituting ….

Leave a Reply

Your email address will not be published. Required fields are marked *