Friction Definition and 998 Threads

The answer key claims the answer is -0.9 J. By my math it should be -0.501 J, is that incorrect?

I tried to use the energy route but the acceleration was not a plausible one. It was just way too big.

I am trying to obtain the expression for the potential transmission of torque using friction. I could derive the formula assuming constant pressure between planar surfaces. To have it in LATEX so it is easier to read, this is the expression for the torque transmission using friction...

Good evening, all! Some quick background: I am working on a robot for a competition that strictly limits entry weights to 3lb. Robot speed & acceleration are highly desirable in this competition, so I have been focusing on ways to maximize my entry's performance. Drive motors for these...

So in (a) the answer is just mgsintheta. When doing (b), I got 8.44x10^3N to be the NET force the tractor must exert. Does the tractor not need to overcome its own friction as well? If so, shouldn't the equation be F_tractor = F_exerted - F_friction, and the question is asking for F_exerted...

Easier case: Elevator is at rest. We need to prevent box from free fall so friction should be bigger than "mg".(And they can be equal) When we push with force F we know that the maximum static friction is ##u_sF##. "mg" should be smaller than ##u_sF## or should be equal to it so the minimum...

I used kinetic friction and did mgμ_k=mv^2/r. However, the solution is mgμ_s=mv^2/r. I am confused on why we consider static friction and not kinetic friction, thanks!

My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same. I cannot find an answer online so can anyone help verify this? Thank you.

I hold my identification card on a low-friction surface by one of its edges. I slightly lean it, and it starts to fall. Before it falls over, I place my finger against the card, and this prevents it from falling all the way over. Then, I withdraw my finger without pushing or pulling the card and...

The thing with this exercise is that I don't think that the question makes sense at all (or, at least, is incomplete). First of all, we don't know if the mass moves with any of those tensions, therefore I cannot know which coefficient apply. Second of all, even if we suppose that the mass is...

Hi, I had those exercises and want to know if they're correct. Also, feedback/tips would be great from you, professionals. $$A$$ 1. Let's consider the oscillator with a friction parameter... \begin{equation} m \ddot{x}+\alpha \dot{x}=-\kappa x \end{equation} but with \begin{equation}...

We have a following setup (see below). A plastic rod is placed inside a plastic ring. A wire is wrapped around a plastic rod from which it goes to the outer surface of the plastic ring. The friction coefficient between the wire and the plastic ring is about 0.1. We have several different sizes...

So, ignore the -0.72, I was just trying to see if I had a sign error (I then remembered magnitude is absolute value) but basically: Since the weight is 12.0N, theta is 53.1, and the coefficient of kinetic friction is 0.100, I just plugged those values into the equations above...

I don't undertand the equation. It is Newtons's second law of motion, so it decribes a force that acts on a single disc relative to the ground. So when the force is proportional to velocity, shouldn't it be ##-bv##? Because the dics's velocity is ##v## relative to the ground. Relative to the...

The picture is here: I understand that friction I need to overcome is ##5mg\mu##, but why do I need to accelerate the slab on acceleration ##g\mu##?

First i show the sketch of the setup: My first attempt was just to balance out the forces on the box. On the sketch below i have shown the situation where the spring is stretched distance L. In this situation we get the equations: Which when solved leads to All good. I then looked at the...

For this, I don't understand why they don't have a negative sign as the torque to the friction should be negative. To my understanding, I think the equation 5.27 should be ##I\frac{d \omega}{dt} = -F_{friction}R## from the right hand rule assuming out of the page is positive. Noting that ##f_k...

Hi, my question is about electricity, hydrostatic pressure, potential energy and friction. Using an electrically powered compressor (such as compressors used to fill diving bottles) if I fill a tank of volume 10 cubic meters to a pressure of 100psi, 1. what is the potential energy stored in that...

Simple question. Let's say a solid cylinder has an initial speed ##v_o## and it's rotating on infinitely hard ground without air resistance. The cylinder will come to a stop eventually. There are two sources of friction. Since the wheel/cylinder is deformed at the contact patch, there is some...

(mentor note: moved from Classical Physics forum hence no template) Hello, I am having trouble with this question: Imagine in real life there was a coefficient of kinetic friction of 0.4 between the plastic wheels of the cart and the wooden ramp. If there is only friction on the flat part of...

Consider the standard pendulum with a weightless rod of length b and a mass point m and mg is applied. In the hinge there is a torque of viscous friction which is proportional ##\omega^2##. Now release the pendulum from the horizontal position. What biggest height does the point m attain after...

Let's say that the mass of the objest is suddenly bigger, so when I want to maintain the constant movement, my force must increase as well. But will the velocity have the same magnitude? I think that the velocity will be smaller, so if I want to have the same initial velocity, I must apply an...

Homework Statement: A door opened at 75 degrees from the latch remains at rest. When moved to a position 60 degrees from the latch, it slowly closes on its own. Is there another possible cause besides a sloping floor? Relevant Equations: \tau = r x F; v = \omega x r; a_T = \alpha x r, where...

We have 2 objects, m1 and m[SUPlB]2[/SUB] Friction is present between the two objects but not between m1 and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of...

Topography of both the object and the surface. Mass/inertia. Moisture, but that can probably fall under topography. I suppose atmospheric pressure, maybe. Or wind. Magnetism. Any others?

Question picture: My solution: Where: S is the lineforce Ff is the force as a result of friction a is the resulting acceleration F is the acting force The answear is supposed to be a=(F-2mg(mu))/(m+M) Any idea what i could have missed? Thanks for your help on beforehand!

The question was this: My calculations show that the answer should be equal to work done on crate to make it reach the same velocity which is equal to 216 J but the answer given is 432 J It is believed that extra energy is needed to overcome friction but friction is an internal force and...

My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt: Let block 1 be the vertically moving block and let block 2 be the horizontally moving one. Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0...

The answer is (D), but I don't understand why. Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts: 1) motion on snowy surface Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction...

So basically I need to find the coefficient of friction given the listed information. What bothers me is that I am getting two different accelerations for two different approaches. When I calculate acceleration using Fg=mgsin60 I do it this way: Fg=mgsin60 -> ma=mgsin60 ->a=gsin60 -> a=8.66. But...

So I have difficulty understanding this https://physics.stackexchange.com/questions/331173/why-is-friction-only-on-the-back-wheel With constant speed cycling uphill the front wheel has no friction. But if that would be the case how does the wheel even spin?

First of all, the pulling force is 300N cos(30) = 260 N At this point, I try to find the friction force Fn = mg = 20kg * 9.81 m/s^2 = 196.2 N Then, Ff = μ * Fn = 0.5 * 196.2 N = 98.1 N So after canceling the horizontal forces, 260N - 98.1N = 161.9N And the acceleration will be 161.9N / 20kg =...

There is a well-known Abraham-Lorentz equation describing radiative friction. Suppose a particle moves in an electromagnetic field. ma(t)=q(E+vxB) + m(tau)a’(t) By solving this equation numerically, I get non-physical solutions(runaway solutions) Although, it would seem that an electron in an...

We want to figure out how much work friction does on a block as it slides down an inclined plane with a rough surface. we find the force due to gravity that pulls the block down the ramp, that's found by M * g * sin(θ), The normal force on the block is given by M * g * cos(θ). The force of...

So the only problem I am having is determining the direction of static friction. I did the same problem but while they were going in a vertical circular motion instead, where the static friction force was in the direction of centripetal force (pointing to the center of the circle). Would it be...

The block starts to slide if friction can no longer hold the block. F=u*n and F=(m1+m2)a so: (m1+m2)a=uN=>am1+am2=uN=>am2=(uN)/(am1) So:am2=(uN)/(am1) is the force. The answer is F=(u*m1g(m1+m2))/m2 I do not see how the acceleration terms are canceled. Is my answer equivalent to this?

Are we allowed to use free body diagrams in dynamic systems where there is non-negligible friction in say the rotational pivot between two linkages? If so, how to incorporate friction into the equations of motion? If not, what method allows friction to be accounted for?

Ideally, it should look circa like this: But in reality, it looks often like this: Is the formula valid also in this case?

I reason the frictional force on the plate from the ice is doing work first 3 meters (while the motorbike is moving on top) and then an "x" distance after the motorbike has left it. Does anybody have an idea of how one might solve this problem?

I’m writing a presentation for younger students explaining how science can help them satiate their curiosity in a fun way. This presentation’s goal is to apply as much science as possible to a ridiculous question in hopes of promoting STEM fields. My example is, how many capybaras --the world’s...

I have a flat planar part made of crystalline sapphire (about ~2k weight, and polished to a mirror finish) that rests on three ball bearings, and I want to minimize the static friction at these 3 interfaces. The ball bearings are fixed so they cannot roll, and the sapphire part can only slip...

I'm unsure on where to begin with this question, i've tried many different formulas that aren't giving me the right answer. I believe to start I need to convert the kilo newtons to newtons. I tried w = fs + mgh w = 10500 x 8.9/sin(13.9)+(1845.69 x 9.8 x8.9) = 549986.46 J and then convert to...

First, I assumed that the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N## It's the first problem...

Hello Everyone I want to model forces affecting on syringe plunger , but I do not know how to calculate terms like friction and damping coefficient. What I imagine is that : F_driving = ma + cv + f ----------------(1) where: f: friction c: coefficient of viscous damping m: mass of plunger (is...

Hi everyone, i have been trying to find an answer to this problem I have but couldnt find any good answers... (I dont know much about this stuff, but need a formula for a Project I am currently working on). So The problem goes as follows: Assuming we have a ball with a mass of m which is...

Ok, logically, it must be that the static friction force of block A equal to the force of gravity on block B, so mass of block A is: m_A * 9.8 * 0.30 = m_B * 9.8 m_A * 2.94 = 2 * 9.8 m_A * 2.94 = 19.6 m_A \approx 6.7 kg. However, when I look at block A individually, there is one thing...

Young & Freedman 13th ed, Exercise 7.81 Starting with the crate, here is its free-body diagram: In accordance with Newton's First Law: $$ \Sigma F_y = 0 = n+(-w_c \cos{\alpha}) $$ Thus ## n = w_c \cos{\alpha} ##. And according to Newton's Second Law: $$ \Sigma F_x = m_c a_x = w_c...

I don't understand part (b) In part (a), I need to calculate the coefficient of the static friction: mg * \mu_static = 35 58.8 * \mu_static = 35 \mu_static = 35 / 58.8 \approx 0.6 So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot...