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If there is kinetic friction it will act tangential to the circular path, not perpendicular to it. So if you are wanting to analyze the centripetal component of the net force kinetic friction will not contribute to it. If you want uniform circular motion with kinetic friction then you will need some other tangential force to counteract the kinetic friction ...


3

The Work-KE theorem says the net work on a system is equal to its change in KE. But the net work includes work done by conservative and nonconservative forces, and the work done by conservative forces is, by definition, the negative of a change in PE: $W_{net} = \Delta KE = W_C + W_{NC}$ $=-\Delta PE + W_{NC}$, so $W_{NC} = \Delta (KE + PE) = \Delta E_{mech}...


3

When you are dealing with the surface of macroscopic objects, it is very important to learn about asperities. These describe the high spots on the surfaces, which are very important when understanding friction between these surfaces. When you touch two macroscopic surfaces together the asperities (high spots) touch and the surfaces cold weld at these ...


2

Why is the component of contact force is not like $F_{contact} \cos \theta$ and $F_{contact} \sin \theta$ but here they have no trigonometric values like we normally make components for a force. Like $F \cos \theta$ and $F \sin \theta$. They are still the components of the contact force and that's why you get a relation between the normal force and the ...


2

A jack is a block sliding on a slope problem. You use a free body diagram to balance the forces and make sure to include friction in the direction of opposing motion. If you unwrap a thread groove and the corresponding tooth on the nut you can find the balance of forces, which axially balances the applied force/weight $F$, and tangentially balances the ...


1

If you have the acceleration $a$, you can calculate the distance traveled $s$ using the equation $v^2 = u^2 + 2as$ You have your initial velocity which is $u = 30 ms^{-1}$ and you need your final velocity to be $v = 0$. And if this distance is shorter than the distance to the traffic light you will have your answer.


1

Your logic is solid. Indeed there is room for some ambiguity in the answer. The thing is tension and static friction both could be called self-adjusting forces. They adjust themselves according to other applied forces. In this case, both these forces are applied on a single object. There are 2 unknowns and 1 equation. $$F_0 = T +f_{static}$$ This system is ...


1

When you say “higher moment of inertia” or “higher angular momentum” I assume you are referring to a rotating rigid object correct? This might be the fly wheel or torque converter (correct me as to what since my knowledge about cars is minimal). It seems to me that you should be using the idea of torque then asking how varying this will vary other things. ...


1

You need to solve this as a statics problem twice: once with the maximum friction acting with F and again with the maximum friction against F. In each case you are looking for F.


1

Friction is still a mysterious effect and it surely involves electrical charges. When two dielectric material gets into contact for some short time they develop charge accumulation in the contact areas. Charge accumulation has an important aspect of creating attractive and repulsive forces between the surfaces. This charge accumulation between surfaces has ...


1

Indeed, hard to find references. Nerveless, this gives static friction coefficient wood/glass around $0.13-0.14$, so for kinetic friction coefficient reduce down that value for about 25%. But this gives it as $0.2$. So summing up results seems that kinetic friction coefficient is in range $0.1-0.2$.


1

Friction opposes relative motion between the surfaces, which is not always the impending motion of the bulk object. Indeed, the ball's bottom surface would try to slide down the ramp, so friction opposes this by acting up the ramp on the ball.


1

In an inertial frame of reference an object at rest will remain at rest unless acted on by a force and an object in motion will remain in motion at a constant velocity unless acted on by a force. This is according to Newton's first law of motion. So an object considered to be moving with no friction or other forces acting on it will continue moving with a ...


1

When you send a bowling ball down the alley, it starts by sliding. The kinetic friction force (predicted by the coefficient of kinetic friction) on the bottom of the ball produces a torque which causes an angular acceleration. When the backward tangential velocity of the bottom of the ball (measured relative to the center of mass), matches the forward ...


1

When a wheel is rolling without slipping or sliding, it has static friction with the ground not kinetic friction. When a wheel is slipping or sliding, such as on ice or mud, it has kinetic friction not static friction. The words slipping and sliding are fairly synonymous, but we might be more inclined to say that when a wheel has lost static friction and has ...


1

I suppose that when the system was assembled, some small tension $T$ was applied on the string to attach it to the wall, so that it can be straight. That is before $F$ comes to play. When some $F$ is applied, while $F - T < \mu mg$ nothing moves. And the tension in the string doesn't change because the string is not affected by $F$. As soon as $F - T = \...


1

Now here I want to ask that was this way of thinking correct or was it just a coincidence that I reached this solution? Your thinking is seems OK and solution correct, but it could have reached more quickly with applicable free body diagrams and realizing that a constraint is that both blocks must move at the same time. I will respond to your response to @...


1

I am asking that both the pairs of friction force which are action-reaction pairs on body A and B are opposing the force F and is that actually what is happening? Friction opposes relative motion$^*$ between surfaces, not other forces (at least not explicitly). Pulling on block A to the left will tend to cause block A to slide to the left across block B. ...


1

This is similar to damped harmonic motion (where the damping is caused by air resistance and friction). The cosine term in your equation represents the oscillatory motion and the exponential part of the equation ("modulates") determines the decay of the amplitude over time. Assuming you have access to one, a sonic motion detector can be used to ...


1

Yes, but. What happens with the air of the old boundary layer? It is now trapped between two surfaces, experiencing friction from both sides. All that air will quickly come to a rest and this impulse transfer will slow the airplane down. In other words: Where there was one boundary layer before, you produced three, two of them producing double the friction ...


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