DOC PREVIEW
CU-Boulder PHYS 1110 - Lecture 11

This preview shows page 1-2 out of 7 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Physics 1110 Fall 2004L11 - 1Lecture 11 17 September 2004 Announcements• No Reading Assignment for Mon, Sept 20.• Exam 1 is Tuesday evening, Sept 21, 7:30-9pm. The examcovers the material in Chapter 1 through 4.4 of the text,Mastering Physics HW sets 1-3 and the concept tests. Theexam will be part long answer (roughly 1/4) and multiple choice.• Review – The Monday Sept 20th lecture will be a review. Inaddition there will be an informal Q&A session in room DuaneG-1B20 from 5-7pm on Monday evening.• Review efficiently; focus on the topics that give you difficulty.Try to make up exam questions for each topic.Equilibrium Problem 2 (continued)Here is the statement of the problem. Last time we analyzed theforces and developed the FBD shown below.Example 2: Consider a block at rest on an inclined surface (a ramp)with an incline angle of 15°. The block has a weight of 55 N. What arethe forces on the block from the ramp?Physics 1110 Fall 2004L11 - 2Now that we’ve done the analysis, the next step is to define acoordinate system and get the components, so we can write outNewton’s 2nd law for this problem. We will need an x-y coordinatesystem, but we have a choice to make; should we use the commonhorizontal-x, vertical-y choice, or a tilted coordinate system, with they-axis perpendicular to the ramp surface, and the x-axis parallel tothe ramp surface. I will choose the tilted system based on thefollowing idea: if I choose the tilted system, then finding thecomponents of 2 of the 3 forces (the normal and friction forces) istrivial, and I only have to use trigonometry for the weight vector. If Iuse the horizontal/vertical choice, then finding the components of theweight vector is easy, but now the normal and friction forces bothhave 2 components, requiring the trigonometric analysis.So here is my FBD again, but now with the tilted coordinate system:We wish to write r F net= 0 using the variables from this problem. Thisgives r F net()x= nx+ fx+ wx= 0r F net()y= ny+ fy+ wy= 0.Now we find the individual components. By our choice of coordinatesystem, nx=0 and fy=0. To find the weight components, we need touse trigonometry as shown below:Physics 1110 Fall 2004L11 - 3From this figure we can see that wx= +w sin  and wy= -w cos . Sonow we insert this into our 2nd law equations: r F net()x= 0 + fx+ w sin= 0r F net()y= ny+ 0  w cos= 0.We can immediately solve both equations (separately), and find nowthe final answers for the friction and normal forces:fx= w sin= (55 N)sin15°=14.2 Nny= w cos= (55 N)cos15°=53.1 N .Now that we have answers, we must go think about whether theymake sense. First, note that relative to our coordinate systems, thesigns make sense. Second, notice that if we took the angle  to thevalue zero, we would get back the solution of example 1 in theprevious lecture. Note also that the vector arithmetic is critical; thefriction and normal force magnitudes do NOT add up to the simpleweight of 55 N.Physics 1110 Fall 2004L11 - 4FrictionIn example 2, if we had increased the ramp angle more and more,eventually the block would start to slide down the ramp. This is inagreement with our everyday experience that if we push on someobject, initially it remains at rest until we push hard enough to make itslide. While the object is being pushed on, but remaining at rest, wecall the friction force static; when the object begins to slide, we callthe friction force kinetic. The size of the static force depends on thesize of the force it is resisting, while the kinetic friction force dependson the magnitude of the force of the surface on the sliding object andon the surfaces themselves, e.g., whether they are smooth or rough.On a microscopic level, friction is a very complicated interactionbetween the atoms on the surfaces of the objects sliding against oneanother. A simple but fairly accurate model for this complexinteraction is that the magnitude of kinetic friction f is equal to µkr n where r n is the normal force. The direction is always parallel to thesurface and opposing the direction of motion. The quantity µk is calledthe coefficient of kinetic friction and is a unitless number, whichdescribes just how weak or strong the friction forces are.As we saw in example 2 above, we find the static friction by simplysolving the equations of equilibrium. However, we just discussed thatstatic friction eventually is not strong enough to hold the object atrest, and we get kinetic friction as the object slides. This maximummagnitude for static friction also depends on the magnitude of thenormal force and on the details of the surface. So we can write fmaximum static=µsr n ,where the quantity µs is the coefficient of static friction. Note well thatthe formula above just gives the maximum value that the static frictioncan take; usually it will be something different. The only time you areactually likely to have this particular value for the static friction iswhen the problem says something like “pushing the object hardenough just to make it slip.” Then you know you are at the maximumvalue of the static friction.There is another type of friction called rolling friction that acts verymuch like kinetic friction in slowing something down, although at thePhysics 1110 Fall 2004L11 - 5atomic level it is relatively different because the rolling motion is quitedifferent. We will return to this when we study rolling objects.Application of Newton’s laws: Accelerating ObjectsNow we turn to cases where we do not have equilibrium, so that thereis a non-zero net force causing acceleration. In order to demonstratethis, we’ll follow up example 2, with example 3 where now the block issliding down the ramp.Example 3: Consider a block sliding down an inclined surface (aramp) with an incline angle of 35° and a coefficient of kinetic frictionof 0.4. The block has a weight of 55 N. What are the forces on theblock from the ramp and what is its acceleration?In fact, the only difference between the analysis of this problem andthat of example 2 is that since the friction is kinetic, we know aformula for its value. The figures above are identical, and if I use thesame tilted coordinate system, then the components are the same aswell. So we can jump right to applying Newton’s 2nd law in its fullform: r a =r F netm.First we write out the equations for the individual components: ax=r F net()xm=nx+ fx+ wxmay=r F net()ym=ny+ fy+ wym.Now we insert


View Full Document

CU-Boulder PHYS 1110 - Lecture 11

Documents in this Course
Load more
Download Lecture 11
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 11 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 11 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?