piper210_355 Posted October 9, 2008 Share Posted October 9, 2008 Okay so this is probably a really easy problem, but my teacher literally doesn't teach us any thing, so can someone please help me. The coefficient of static friction between a 2.5kg box and a wooden board is .45 and the coefficient of kinetic friction between the box and the board is .28. The box is placed on the board, which is slowly lifted at one end until the box starts to slide down the board, at which point the angle theta of the board is held constant. Determine the angle theta of the board and the acceleration of the box as it slides down the board. Link to comment Share on other sites More sharing options...
timo Posted October 9, 2008 Share Posted October 9, 2008 So? What's your ideas on how to tackle the problem? Can you calculate theta or at least have an idea how it could be done or what it would depend on? If you knew theta could you calculate the acceleration? Could you calculate the acceleration in the absence of friction? ...? Link to comment Share on other sites More sharing options...
Klaynos Posted October 9, 2008 Share Posted October 9, 2008 OK, so, what can you tell me about the relationship between frictional forces and coefficients for friction? the relationship between force along a slope and angle of the slope? Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 I know that acceleration equals final velocity minus initial velocity over time. other than that i don't know what to do. I have no idea how to find theta or what to do with the friction. Like i said my teacher hasn't taught me anything. I just need an explaination of how to do it. Link to comment Share on other sites More sharing options...
timo Posted October 9, 2008 Share Posted October 9, 2008 You'll have to look at the forces (the acceleration will follow from them via a=F/m later) for this problem. Link to comment Share on other sites More sharing options...
big314mp Posted October 9, 2008 Share Posted October 9, 2008 The static force of friction is equal to the normal force (the force perpendicular to the surface) times the coefficient of friction. Hint: Break your forces into vectors. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 I think i know how to go about finding the acceleration now, but how do I find what the angle is? I just need to know what equations to use and i think i'll be good from there. The static force of friction is equal to the normal force (the force perpendicular to the surface) times the coefficient of friction. Hint: Break your forces into vectors. don't i need to know what the angle is to find the normal force? Link to comment Share on other sites More sharing options...
Klaynos Posted October 9, 2008 Share Posted October 9, 2008 I think i know how to go about finding the acceleration now, but how do I find what the angle is? I just need to know what equations to use and i think i'll be good from there. don't i need to know what the angle is to find the normal force? Yes... What is true just before the object starts to move? There's a very special situation that happens. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 Yes... What is true just before the object starts to move? There's a very special situation that happens. I know that the initial velocity is 0. I know that in this case the normal force would be found by dividing the force of friction by the coeffiecent of friction. Is that what you mean? Are the static and kinetic friction the force of friction? And if so how do i find the coeffiecent of friction? *I apologieze if this is way off, but I'm just gathering facts from my book and stuff i'm finding on the internet and trying to make sense of it all. Link to comment Share on other sites More sharing options...
Klaynos Posted October 9, 2008 Share Posted October 9, 2008 Think about gravity as well... At the point of moving there is something special about the gravitation and friction forces... *I apologieze if this is way off, but I'm just gathering facts from my book and stuff i'm finding on the internet and trying to make sense of it all. There's no need to apologise, I'm trying to help you work it threw yourself, I'm sure you can, you're far more likely to learn this way as well... Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 Well i know that gravity is pushing down on the box and the board is pushing up on it. I'll take a guess and say that when the box starts to move the force of gravity is greater then the force of the board, causing the box to move down. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 i'm thinking that maybe i can some how use the friction that is given to find the angle. is that right? Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 I think you have it right. Basically the force of friction is exceeded by the force of gravity. Now, right at the instant the box begins to move, the force of gravity parallel to the board (remember that one component of the gravitational force will be your normal force, the other will be your acceleration force) will equal the force of static friction. Draw a picture with all of the forces broken into components. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 i have a picture drawn, but how does this help me find theta? Link to comment Share on other sites More sharing options...
Klaynos Posted October 12, 2008 Share Posted October 12, 2008 I think you've got this in ideas but havn't realised the consequences... At the point of moving the force of gravity is equal to the force of friction (well infinitesimally larger), in parallel with the ramp. So you can write them down as = Now you also know what the normal force is (clue it's a component of gravity), so if you combine that with the fact there equal and write down the equations you should then be able to solve for theta. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 ok so in this case gravity=static friction I'm thinking to find force normal i use F(s)=F(normal)u(s) F(normal)=F(s)/u(s) F(normal)= 9.81/.45 F(normal)= 21.8 am i ok so far? Link to comment Share on other sites More sharing options...
Klaynos Posted October 12, 2008 Share Posted October 12, 2008 What's F(s)? And no probably not, F_normal = R = some componant of gravity so it should have some theta term in. You need to split gravity into parallel to the ramp and perpendicular with it. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 so F(gravity)=F(normal)+F(parallel) ? Or should it be something else...probably since you said theta had to be in there. so is R another way to denote normal force? I'm sorry this is taking so long I'm scrambling all over the place trying to find some stuff that can help me with these physics problems, since the teacher didn't Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 I drew a pic to help explain. I omitted some of the forces for the sake of clarity. Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 The unlabeled blue line is just the green line shifted over, to make it easier to see the right triangle. Think about what u equals, and then think about what tan(theta) equals, in terms of the forces given in the picture. The angle between the unlabeled blue line, and the red F(gravity) line is also theta, btw. Link to comment Share on other sites More sharing options...
Klaynos Posted October 12, 2008 Share Posted October 12, 2008 Yeah R is just another way of writeing "normal force" because it's "special" so has it's own letter Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 ok so i looked at the graphic and in terms of tan though tan(theta)=F(x)/F(normal) Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 F(x) is equal to what at the moment when the box begins to move? Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 would it be the static friction since in this case the force of gravity = static friction Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 F(x) = F(static friction) So substitute this into the expression you had before (the tan(theta) one). Link to comment Share on other sites More sharing options...
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