# Tag Info

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Using the brakes on the front of the bike causes your weight to shift forward. Additional weight allows more force before the tire will slip (skid). If you brake hard enough the back tire of your bike will lift up and at that point all of the mass is distributed on the front tire. Remember the maximum force is $F_{max} = \mu F_{normal}$ and $F_{normal}$ ...

5

The braking force acts between the tyre and the road. The centre of mass is above this point so there is a rotational effect which increases the force going down through the front tyre and decreases the force going down through the rear tyre. Because the amount of braking force the tyre is able to produce is limited by the amount of force going down through ...

4

On most bicycles, your center of gravity is not halfway between the front and back wheel - it is closer to the back wheel (image source: http://www.esquire.com/cm/esquire/images/d1/bike-080210-lg.jpg). This means that the back wheel carries more of the weight. Now assuming that you inflate the tires with the same pressure, this means that the contact ...

4

There are some good reasons why you should not take a sharp turn at high speeds. 1) On a flat road, the force of static friction is what provides the centripetal force to accelerate you through a curve. Unfortunately, there is a maximum value for static friction that depends greatly on the mass of the vehicle. The heavier it is, the more static friction you ...

4

Problem: Given Newton's second law $$\tag{1} m\ddot{q}^j~=~-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \qquad j~\in~\{1,\ldots, n\},$$ for a non-relativistic point particle in $n$ dimensions, subjected to a friction force, and also subjected to various forces that have a total potential $V(q,t)$, which may depend explicitly on time. I) ...

3

As you seem (correctly) to understand, your free body diagram for the car should be as in the left of the drawing below. The force summation does not close to a polygon. You know neither $F_N$ nor $F_F$, the components of the force on the car from the road. But you know their directions ($\theta$ is the banking angle) and you know that all the forces must ...

3

Theoretical Answer Consider that you travelled in your car at $10~{km}/{h}$ for one hour, then at $100~{km}/{h}$ for the next hour. First hour of the Journey: You travelled a distance of 10km, so the work done is $$W=F\times s=10F$$ Second hour of the journey: You travelled a distance of 100km, so the work done is $$W=F\times s=100F$$ The work done is ...

3

You are wrong at assuming constant friction. Rolling Friction increases when you increase speed of the car (See the formulae at the bottom). Also, aerodynamic drag increases with the square of speed (See the formula at the bottom). So, at higher speed, the car engine needs to counter higher rolling friction and air drag to maintain that speed. While the ...

3

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as: rolling drag between the tires and the road surface, fluid drag from the air that the car moves through, and ...

3

Lots of excellent answers here, but for fun, lets think about this backwards. Imagine you have the worlds first and only FRONT wheel drive motorcycle, and your rev it up and pop the clutch. What kind of launch do you think you would get with very little weight on the front tire? The reverse is true during braking when the deceleration shifts the weight of ...

3

The theory of friction that is described in the source is the Prandtl-Tomlinson model. I'll explain it in two steps to answer your two main questions. Q1: What is meant by "partial irreversibility of the bonding force?" All bonding, including the bonding responsible for friction, is due to electrostatic attraction between atoms. Here is what that looks ...

3

Braking acts to stop the front tire. Friction acts at the contact patch under the front wheel to introduce a vector force directed towards the back of the bike. Since the force is not directed through the center of mass of the motorcycle/rider system, it introduces a moment or torque that acts to rotate the motorcycle and rider such that the back tire begins ...

3

Actually drag is NOT completely the friction between the object and fluid. Sometimes there can be almost no friction, but still highly significant drag. In a non viscous fluid, there is no friction between the object and the fluid, but there IS still drag. The phenomenon of ram pressure transfers momentum losslessly between the object and the fluid without ...

2

I suspect this question is rooted in the widespread misconception that some external force is needed to keep the Earth rotation. That is not the case. Angular momentum is a conserved quantity. A rotating object that is not subject to any external torques will rotate with a constant angular momentum. This is the rotational analog of Newton's first law. An ...

2

Dynamic friction is constant, it doesnt change with speed. That is why the trick works. If you pull the cloth fast enough, the friction force will act for such a short time that it will not be enough to pull the stuff above it. In the case of the weels, the frictions force is also constant, but you make it last longer per unit of lenght because the wheels ...

2

The rate of temperature change will be the power per unit mass times the specific heat. So if you have a certain mass of water $M$ flowing per second, at a velocity $v$, losing $\Delta P$ pressure per second, then work done is $v\Delta P A$ and $A = \frac{M}{\rho v}$ . Then with a heat capacity $c$ (about 4.2 kJ/kg/K for water), and the relationship between ...

2

If there is not enough friction to keep the vehicle in its circular path, it will skid. The force needed for the circular path is the centripetal force: friction (the force keeping the car on the road) must be greater that that. Now the no-slip condition (centripetal force < friction) implies = $\frac{mv^2}{r} < mg\mu$ . Your equation follows by simple ...

2

When you are moving along a banked curve, three forces are at play in your frame of reference: Gravity: pulling you down; and because of the bank, pulling you inward Friction: preventing you from moving either in or out. Magnitude and direction depend on the coefficient of friction, the normal force, and the direction in which you are trying to move ...

2

For unlubricated friction, the simplistic model is quite good when the surfaces are flat and macroscopic deformation can be neglected. There are three things in particular I would like to elaborate on. First - the question of the equation of motion that you wrote down. The force of friction, at a microscopic level, is actually a consequence of the breaking ...

2

Well, for starters, centrifugal force is just a pseudo force. Get it? Let me make it a bit simpler.. See, Centripetal force, which is mv^2/r is acting towards the center and centrifugal force away from the center. I thinking you're thinking why did we use the centripetal force in case of banking of road.. right? Well, we did that because, This is nothing ...

1

Tribology (not the study of tribes!) is the study of what happens when things 'rub'. This involves friction and wear when solids rub against other solids (such as in mechanical bearings) and the effect of liquids (such as 'lubricants') and other fluids. Friction at a solid-liquid interface is still called friction. It is a 'damping' or 'dissipative' force, ...

1

So, in the first case the frictional force have the same direction with the acceleration of the center of mass but it's not in the latter one. Can someone explain the difference between those 2. In your first case, you say "to make the wheel roll faster", but you don't say how this is done. Is is pushed? Do you apply a torque to the axle? If you ...

1

The friction between a solid and liquid is a function of viscosity. The best way to answer this is with a model setup called Couette flow where a fluid sandwiched between two plates is sped up by the movement of the top plate: Image source: University of Virginia, Physics 152 taught by Michael Fowler The friction force $F$ that the fluid exerts on ...

1

You can always imagine torque as being represented by two opposing forces at either side of the object. In this case imagine the torque being generated by two forces of magnitude F at the top and bottom of the disk, the top one pointing forwards and the bottom one, backwards. In order to generate torque $\tau$, these two forces must be of magnitude ...

1

You have ignored the car friction with air! If we assume the car as an aerodynamic body (air flows on the car surface smoothly and does not separate) then the air friction on the car is proportional to the square of car speed.

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The car is experiencing acceleration in two perpendicular directions. At the moment in question, the car is moving at the given velocity in a circle with the given radius. You have the formula for the centripetal force needed, and this is supplied by the radial force of friction. The car is also slowing down with the given acceleration. This requires a ...

1

Yes, it is correct. Kinetic friction acts when sliding only. Of course there is an instant of time where your feet slide. But you can despise that...

1

There are two separate questions here: why doesn't $F_f=\mu_k F_N$ describe the motion? (i.e. the velocity doesn't reverse at large $t$) is $F$ dependant on velocity? I can't improve on Floris' answer to (2), but it's worth making a few comments on (1). The problem is that (1) is not an equation of motion. For example both $\vec{F}_f$ and $\vec{F}_N$ are ...

1

If the applied force equals the max static friction then: $$\boldsymbol F_{applied}-\boldsymbol F_{\max \text{ friction}} = 0 = m\boldsymbol a.$$ From which it can be seen that $\boldsymbol a =0$, taking into account that the forces are opposite. Therefore the object does not move.

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