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The string contacts the point on two infinitesimally close points with different slopes. Imagine a small pulley end the two points are the entry and exit point of the string. If the string is between points A on the left and point B on the right (with B lower) then we call the angles of the string from horizontal $\theta_A$ and $\theta_B$. If the mass is ...

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Not sure i understand fully your question but in general friction can be seen as a force (vector) pointing in the opposite direction of motion with (if there is a motion). Moreover, the force is tangent to the surface of contact. Thus for an object with spherical symmetries (like a pulley, cylinder, sphere...), the force is perpendicular to the radius.

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To state it simply, friction is the resistance to motion of an object within a system, in this case a ruler on a desk. As you suggest in your question the normal force to the surface is important to friction, the equation is: Coefficient of friction = force required to maintain constant velocity / normal force however turning the ruler on its side does ...

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You didn't specify in what direction the force of hand is applied, so for simplicity I assume that you are applying the force perpendicular to the desk. Now there are four forces on the book: 1) Gravity ($mg$) is trying to take the book down; it has a component $mg\cos\theta$ that is perpendicular to the desk and a component $mg\sin\theta$ that is parallel ...

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If C is the center of mass of the ball, and A is the contact point then the velocity of the ball at the contact point is $$\vec{v}_A = \vec{v}_C + \vec\omega \times (\vec{r}_A-\vec{r}_C)$$ If you have infinite friction then you have no slipping which means $\vec{v}_A =0$. If there was slipping then only the speed in vertical direction should be zero ...

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Based on how you phrased the question, I think the answer comes from intuition and definition-chasing. Static friction: This is a contact force which acts against forces trying to slide two surfaces against each other. It is limited in maximum magnitude because we don't expect two surfaces to be inexoriably fused, at least in the scales of problems in ...

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Static friction is always less than or equal to some maximum total relative force, that if exceeded will result in the motion of the object. The static friction will change both direction and amplitude to keep the object still with respect to whatever is providing the friction. So long as the force applied to the object, from another means, is less than or ...

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Here are the steps you can take. Degrees of Freedom. There are 3 degrees of freedom, one for the base plate, one for the box and one for the mass. Hence there are 3 variables that you need to track, as well as their derivatives. I will name them $x_0=\gamma(t)$ for the plate, $x_1$ for the box and $x_2$ for the ball. Free Body Diagrams. For the moving ...

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The behavior of the system (not surprisingly)depends on the initial conditions. (For the sake of argument, we can assume the box starts stationary with respect to the table and $\gamma(0)=0$) I am assuming the problem is $1d$; this way we will end up with two coupled equations of motion. Let's show the box's coordinate with $\chi$ and ball's coordinate with ...

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I) In case of a point particle with mass $m$ (and no moment of inertia), the best one can do seems to be to model the friction/drag via a Rayleigh dissipation function ${\cal F}(v^2)$ with a friction/drag force $$\tag{1} {\bf F}_f~:=~-\frac{\partial {\cal F}(v^2)}{\partial {\bf v}} ~=~-2{\cal F}^{\prime}(v^2){\bf v},$$ i.e. the Lagrange equations read ...

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kinetic energy of the water molecules decreases with time because the water molecules transfer the acquired kinetic energy to the bowl . thus KE decreases . hence , the waves dissipate

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There is no easy way to model a spinning coin and calculate these observations. It slows down mostly because of air resistance and friction(here you must take velocity dependent friction-angular velocity in your case-) and it moves due to the combination of torque of gravity(a.k.a. precession) and friction. Velocity dependent frictions generally gives you ...

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Let's say you roll a ball (of mass $m$) down an inclined plane of angle of inclination $\theta$ and coefficient of static friction $\mu_{static}$. Then you know a force parallel to the inclined plane acts on the ball through its center of mass. Another force parallel to the surface acts in the opposite direction of motion as follows, The force $\vec F = ... 0 F=M x A= F 25 x 25=6.25 For the explanation is,the formula must be F=M x A=F For example 25 kg. is the mass and .25.00.00 is the acceleration of the speed limit.Now you must multiply the two 25 and the answer is 6.25 1 The contact forces with two blocks are$N_1 = m_1 g + m_2 g$for the bottom block (to the floor) and$N_2 = m_2 g$for the top block (to the 1st block). The available traction is$F^\star_1 = \mu_1 (m_1+m_2)\,g$and$F^\star_2 = \mu_2 m_2\, g$or $$\begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} ... 0 Yes the static friction of the body is zero if it is rolling. This is because, the velocity of the lowermost point (i.e. the point of contact is zero by definition of pure rolling). So, as there is no tentative motion of the lowermost point, the static friction is 0. But whenever a force is applied, the lowermost point has a tendency to slide. As a result ... 0 When one talks about static friction one generally thinks of the coefficient that enters in the static friction force (\mu_{static}). The force that you can apply to a body without moving it is proportional to this coefficient ||{\vec{F}}||\propto \mu_{static}. You should not confuse this friction with the friction that occurs for a body that moves ... 2 When a disk or other object is rotating on a horizontal surface with constant velocity, there is no static frictional force. Your logic is correct: if there were a horizontal force, the center of mass would be accelerating. If the rolling object suddenly encounters a frictionless surface, it would continue to satisfy the rotating without slipping condition. ... 0 Surely, friction is an internal force but that does not mean that it won't cause any motion. Friction may not produce any motion in some cases but in certain cases it does. The friction on the plank will act due to that by the block as shown below, As you can see, the plank is applying a frictional force on the block in the forward direction (which tends ... 1 In the cases where you have static friction, the forces will always be defined by the looking at the system and applying the constraints(in other words F_s\le \mu N will only give an upper bound). On the other hand when you are dealing with kinetic friction, it can be easily derived from the famous F_k=\mu N. As an example, let's solve this problem(As ... 2 All forces act in pairs, so let me start by matching them up: Force on M_1 = F = - M_1 on Some force providing device Surface on M_1 = F_1 = -M_1 on Surface M_2 on M_1 = F_2 = - M_1 on M_2 The values for the forces horizontal components are found using... F = Given (1 Newton)$$F_{sf} \le \mu_{sf} \cdot F_n F_1 \le \mu_1(M_1+M_2)g$$... 0 This is my attempt to illustrate what happens when the car wheel is turned: Focus on the bit of the car tyre marked with a red spot, and the bit of the road marked with a green spot. If we could look at the contact patch between the tyre and the road we'd see something like the rectangle I've drawn on the left. When the wheel is straight the red spot on ... 0 The tires (or ball) want to travel in a straight line and friction (traction actually) make it deviate from the line. Friction (acting on the direction of travel) is present on both cases. For the car there is the additional effect of traction which enforces the going around the bend part. The ball is going on a straight line and so no traction is needed. ... 0 It is all about the distribution of pressure under the contact. For a block of uniform weight the pressure can be assume almost constant under the area and so when traction is broken it will happen all at once all over with a force of$\mu N\$ as you stated. But for other geometries, or for elastic parts (like tires, or marbles or billiard balls on felt) the ...

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When you say underinflated tires experience more friction, do you mean static friction (i.e., resistance to slipping) or rolling resistance, which is something quite different? Afaik the origin of the friction law is very much phenomenological, and has it's limits of applicability (especially at the static - dynamic transition). My understanding as to why ...

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There are some vital considerations you are not including in your initial analysis. One is the performance and response (due to 'jiggling/vibration' of the tire at low pressure) of the tire depending on its shape.

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