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6

Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ...


2

The direction of the motion at any time $t$ is the direction of the velocity vector $\textbf{v}(t)$ as derived by solving the equations of motion; likewise $\omega(t)$ gives you back the direction of rotation according to the right hand rule. friction is the force that causes rotation is not entirely correct. Any force with non-zero torque generates ...


3

What matters here is how the value of $c$ compares to the value of $k$. Let us choose a $\zeta = \frac{c}{2\sqrt{mk}}$ One can show that when $\zeta =1 $ the system is critically damped, and will not exhibited any oscillations and will return to the origin in the shortest possible time interval. When $\zeta > 1 $ the system is over damped and will take ...


1

Friction is essentially the name people assign to the electromagnetic interactions on the everyday scales. Your question then comes down to asking what the universe will be if no electromagnetic interaction were present, which is the same as asking what the universe will be if we replaced the interactions we currently have with something else. The answer is ...


0

It's impossible to imagine Earth of that sort. We would all be dead. We have evolved with Friction, and there's no way, absolutely no way, our bodies could do the bodily functions (e.g. blood pressure would be so high, it would burst out arteries!). The question is ill-posed.


0

In order for an object to travel on a curved path of any type (circular or otherwise), it must experience a component of acceleration perpendicular to the path. The magnitude of that sideways acceleration must be, instantaneously, $$a_{\perp}=\frac{v^2}{r},$$ where $r$ is the instantaneous radius of curvature of a circular path, and $v$ is the instantaneous ...


0

I think you are looking at the math, and not thinking about the physics. Not all values of $\mu_s$ and $\theta$ make sense. If the curve is banked at $90^\circ$ and $\mu_s=\cot 90^\circ=0$, it would require infinite speed to make a turn without skidding. You could not make such a turn. You probably could have figured that out without the equation. It's ...


4

The coefficient of static friction and the normal force together allow you to calculate the maximum force that friction can apply, not the actual force. A block sitting on a shelf with no external horizontal forces will have a friction force of zero. If you apply a force from the side that is less than the maximum friction force, then the actual friction ...


0

Direction of frictional force is perpendicular to the motion. Coefficient of friction cannot exceed certain value. If the centripetal force normal to motion exceeds the permissible value, car will skid. The equation you are using will measure speed for maximum frictional force. You are assuming that it does not change, which is not the case. Coefficient of ...


0

The thing is that you cannot measure friction between the tires of the car and the road and then calculate its speed.There are two kind of friction. The first is the static friction and the other is kinetic friction. the maximum static friction is measured by $f_s \le f_{smax}\left( = \mu_s mg\right) $ In order for the car not to slip and make the turn, ...


-1

Clearly by logic if we will go lower than the velocity obtained we can make an easy turn.....So we need to find the max velocity .. If u will increase Ur velocity by this max velocity then the two tires will lift from the ground and then Ur car will get damage.


1

Friction occurs whenever energy is dissipated as movement happens. Suppose we move a distance $x$, and some mechanism dissipates an energy $E$ as we move (we won't worry about the mechanism of the dissipation for now). That means to maintain a constant speed we have to put in an amount of work $W = E$. But since we know that work = force $\times$ distance, ...


1

You are correct. Kartik's answer is also correct, but he just didn't directly answer your question (well, to be fair, it presupposes a complete rewriting of the laws of physics). A positive coefficient of friction opposes the direction of applied force, reducing or preventing movement. With a stationary object, the force is called static friction, and for a ...


1

The sign of friction depend on our choice of axis. If I choose right side to be positive and I push an object to the right then friction will act to the left and so the friction will be negative. But the magnitude of friction is always positive. It is positive by definition. The magnitude of any force (or any vector) is positive. In your question you ...


0

There is no fundamental physics reason why carrying a bag should use more energy than wheeling it. Whether you be carrying a bag or wheeling it, you are essentially sliding it along a line of almost constant gravitational potential (aside from a little jiggling up and down with your stride), so the bag's total energy isn't changing and in theory does not ...


3

Your first quote is correct for an idealised model. There is no rolling friction then. Both wheel and surface are considered completely rigid. Ideal model - no rolling friction Non-ideal/more realistic model - rolling friction comes into the picture These pictures are from this link that gives a very good graphic view on this. Going away from an ideal ...


0

Firstly, friction is the resistance to lateral motion between two surfaces and so is required for there to be no motion at the point of contact. There may be confusion between rolling friction and rolling resistance. The former being the friction between the rolling object and the rolling surface required for rolling motion to occur. The latter being the ...


1

It is an implicit function which you have to solve numerically. Typically you would use "fixed point iteration". You can do this in Excel with some user defined functions where for a given geometry you start with $f=1$ and then use the above $\frac{1}{\sqrt{f}} = L(f)$ a few times until in converges to a value $$f \rightarrow \frac{1}{L^2(f)} $$ In my ...


1

dmckee's comment is correct, this problem is hard (impossible?) to solve with pen and paper. But it's not too hard to set up, which hopefully will answer your question. There will advanced calculus by high school standards (including differential equations), no way around that, but I'll try to explain the equations at a high school level, I hope... I'll ...


1

The volumetric flow rate has to stay the same, because the same amount of water flows out of the tube as flows into it. The only way for the volumetric flow rate to change along the tube would be if there was a leak. Friction (i.e. viscous drag) can't change the volumetric flow rate, all it can do is change the pressure gradient along the tube. So you are ...


1

I would treat the car and the Earth together as a system. No internal forces can change the momentum of that system. However, the brakes can change the relative motion of two parts of that system - stopping the motion of the car with respect to the Earth.


0

Not sure anyone will look back at this, but I'd like to give an answer anyway! How do you not disturb the dishes when pulling a tablecloth out from under them? You're exactly right: this is about the inertia of the dishes and the forces on the dishes from the table cloth while the cloth is being pulled. Remember from physics that if we plot the velocity ...


1

When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is: $$ W=\int F \,dx$$ where $F$ is a force applied. It must be equal to the total kinetic energy of you and the block (if ...


0

In the pure hypothetical situation you pose, you, the block, and no forces from gravity, friction or wind resistance - the moment you 'push' the block you will impart a 'packet' of energy for only as long as you can extend your arms and maintain contact with the block. For at that moment the block will move away from you, and you will move away in the ...


1

Do a free body diagram and you will find for a horizontal plane that $$ F - \mu m g = m a $$ $$ (100) - \mu (50) (9.80665) = (50) (0.1) $$ $$\boxed{ \mu = \frac{(100)-(0.1)(50)}{(50)(9.80665)} = 0.1937\ldots }$$


0

Newton's 2nd law in the x-direction. There are only two forces that together equal the $ma$, and since you know one of them, it is straight forward from here.


0

I would approach the problem a bit differently. Firstly, there is no relative motion between the two blocks, and hence the force at play is the static force only. i.e. $F_{sf}$. This force accelerates the bottom mass, $m_2$. So, $F_{sf}=m_2.a$..... eq.1 Now, consider the top mass m1 to be in an accelerated frame of reference, which has an acceleration $a$. ...


1

You are right. In an optimal system, there isn't any friction. But in real life there is, because there aren't any balls or surfaces that are so perfect that they touch just in one point. Also the surface of the ball is never perfectly flat, so there will always be friction between the surface of the ball and the surrounding air. So for calculating the ...


0

Substitute $x$ by $A(e^{wt})$. Thus your equation becomes: $$ A(w^2)(e^{wt})+A(\frac{b}{m})(w)(e^{wt})+(\frac{k}{m})(A)(e^{wt})=0 $$ Simplifying: $$ (w^2) + (\frac{b}{m})(w) + (\frac{k}{m})=0 $$ Find out the roots . Here you understand that $(b^2)<=4km$ for real values of $w$. Let the roots be $w_1$ and $w_2$ Finally your solution to the ...


0

A good rule of thumb to quickly see friction directions is: Friction always tries to prevent the relative motion. [...] a ball sliding on ground(with friction :) Here, kinetic friction on the ball points backwards against the direction of motion. When things are standing still even though a force tries to move them (e.g. if you push on your dinner ...


0

There are forces like gravity and spring forces. These are in proportion to the mass or distance the spring is stretched. Then there are forces like the reaction of a surface. When an object sits on a surface, it doesn't penetrate. If the object pushes on the surface, the surface pushes back just hard enough to stop the object. The reaction force adjusts ...


0

What Haliday describes is in an "Ideal" situation, where the wheel and surface are perfectly solid and there are no other forces like air resistance. Then the only force between the wheel and surface is an upward force equal to the weight of wheel (assuming we have gravity) Actually if there is any friction, then energy will be lost and the wheel will not ...


0

Newton's first law states that any object undergoing a constant velocity or is at rest will continuously be at that state of motion ( either at rest or constant velocity ) Using this law, and coming to your situation if an object is moving at a constant velocity and if it encounters a frictional force it will indubitably decrease in speed and gradually fall ...


0

Your pressures at each are identical, while the valve is closed. The real difference is the fact that 2" volume of water will easily supply two 3/4" outlets. While the other is diminished from loss due to physics. Its called friction loss, due to pipes, lengths, elbows, tees, etc.


5

This is an interesting paradox! I suggest resolving it in the following way: First, note that the coefficient of friction between the two blocks allows friction "up to 30N" to occur - but if the two blocks are not moving relative to each other, the friction will be "whatever it is". Second, look at the force balance on the top block. We have the force of ...


2

In the second calculation, the $10N$ force is not acting on the m1 block. The only force on the block is the friction force from m2. The frictional force will be exactly the amount required to keep m1 and m2 moving together. The frictional force will be given by: $$F = m_1 a = (5 kg)(1.25 m/s^2) = 6.25 N$$ Checking with just the top block: $$a_2 = (F_{pull} ...


9

You are making a major flaw here. The friction between the 2 blocks is not going to be $10N$. It is going to be something, but we will have to calculate it. Assume the friction force to be $f$ such that the acceleration for both the blocks is same. Now, equations for $m_2$ and $m_1$ separately are: $$F-f = m_2*a = 3a$$ $$f = m_1*a = 5a$$ Solving both ...



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