# Tag Info

## Hot answers tagged friction

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) $$F_{sf} \le \mu_{sf} \cdot F_n$$ $$F_1 \le \mu_1(M_1+M_2)g$$ ...

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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. ...

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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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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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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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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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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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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 $\vec F = ... 1 The contact forces with two blocks are$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} ... 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 ...

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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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