New answers tagged

1

They aren't. Frictional force down an incline is the coefficient of kinetic or static friction multiplied by mass and the acceleration of gravity and the sine of the angle of elevation. Meanwhile, the component of weight down the incline is all of the above without the coefficient of kinetic or static friction.


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the kick will have to be less than the kinetic friction (which is less than the static friction threshold...) Not quite. The kick will have to apply less force than the static friction. There is no need to consider kinetic friction at all. Only when the kick applies a larger force than the static friction limit, will the foot start to slide and kinetic ...


2

The things I learnt in years of pinewood derby racing: use the maximum weight keep it at the back make sure the car tracks straight focus on stability The weight is your "engine". Since you start at a slope, mass at the back has further to drop than mass at the front (really!). You can think about it like this: if the weight of the car is evenly ...


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Disclaimer: This answer was written before I found out that a pinewood derby is for miniature wooden cars that run in a track. Therefore, only part of it will apply to a pinewood derby. As I understand your situation, you will be traveling down an incline and then on to a horizontal surface. You are saying you want to maximize the effect of gravitational ...


1

If the blocks do not slip relative to one another you can treat them as one block. If they move at constant speed down the slope the component of their combined weight down the slope must be equal to the kinetic friction force up the slope. The kinetic friction force will depend on the normal reaction force between the two blocks and slope. If you do not ...


1

By the way, is the expression of $F(A)$ correct? Yes. If $F(A)$ is the parallel component of $A$'s weight, then only $A$'s mass should be included in the expression. Only if you wanted the weight-component of the entire system (both boxes as if they were one) should the total mass be used. Does each of these two forces implies an action reaction ...


3

I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static ...


0

Just recall that kinetic friction always opposes the motion; if this were not true you could use friction to generate free energy. Static friction opposes the applied forces.


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No, the minus sign is not used to show direction. A vector is just a direction and magnitude and cannot be negative. But it may be subtracted from another vector. This minus sign does not show direction but just a mathematical procedure. But it turns out that subtracting a vector $\vec A$ form another vector $\vec B$: $\vec A-\vec B$ is the same as adding ...


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After thinking over the answers and discussion, I believe that the problem is just that using sign to denote direction only really holds up in one dimensional problems (unless you're using complex numbers). Therefore, the solution is to use the formula $F_f=\mu F_f$ to deal with magnitude only, and if using sign notation, you'll just have to apply the sign ...


1

The small object exerts a force in the opposite direction to the normal force on the cart


1

It would probably be wiser to state the friction law as: $$|F_F|=\mu |F_N|$$ where $|F_N|$ denotes the modulus of the Normal force. Now consider the following diagram: Both blocks and slopes are identical. Left: some net force on the block causes an acceleration $a$ (left and up). The friction force $F_F$ points in the opposite direction: it opposes ...


0

In the relation $F_f = \mu F_{N}$, $F_f$ and $F_{N}$ are magnitudes of the frictional force and the normal force, and they are both positive. (I assume you are considering kinetic friction.) The direction of the frictional force is in such a way that it opposes the relative motion between the two surfaces. This cannot be inferred from $F_f = \mu F_{N}$.


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In laboratory experiments with bricks, the sliding angle comes to be the same irrespective of the orientation of the brick. However, when it comes to a child sliding there are some extra factors. The child usually is not sitting at normal to the plane especially at large angles of sliding due to the fact that he can topple more easily than slide. He wants ...


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My guess would be that if you're sitting, the surface of contact between you and the slide is smaller, thus the pressure is higher than if you were to lay on your back. Your jacket is a deformable material, its friction coefficient with steel might vary with applied pressure.


0

friction always tries to oppose the relative motion ... or you can also choose any arbitrary direction if your answer shows negative sign than assumed direction is the opposite direction of real friction force acting ...


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In general the energy of the system must be conserved. Heat and sound result from shuttle reentry because the kinetic energy of the shuttle is being reduced and that energy must go somewhere. The molecular collisions of the air molecules with the shuttle and with one another result in an average increase in the speed of the molecules (heat). Similarly, a ...


1

What would the force be if there was no spring? Each side of the spring feels the same force - so if you put a black box around the spring and only saw the string "going in" and a string "coming out" of the box, with the same tension on each, the force needed to move the box would be the same. This means your approach is correct.


0

F(x) = kx, according to Hooke's Law. This means that your friend is incorrect, and you got the question correct. Note: the potential energy of the spring is (kx^2)/2, so your friend confused potential energy with force. A bit of dimensional analysis would greatly decrease the chance of mixing the units in this way.


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The downward acceleration in addition to g is also due to air resistance.... For speeds such as those of the ski jumper the air resistance in Newton's is taken to be proportional to velocity. ie -KVy hence the total downward acceleration is... -(KVy/m + g) So we see higher the mass m the lesser the downward acceleration and fitting this in the trajectory ...


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Most interestingly it is the opposite of what I find when riding a bike down a hill and what was hinted at above. Long distance ski jumpers benefit from maximizing their surface area while simultaneously decreasing their weight. The less they weigh and the more drag they can produce, the farther they go. Their bodies are the primary source of weight and, as ...


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Increasing mass means that wind resistance has less effect. Consider the extreme example of a ski jumper who only weighed 1 gram, in still air. How fast would the jumper be moving by the time they left the slope?


2

As John Rennie points out, there are in practice always small effects that lead to dissipation. Some of these effects may be involved in an essential way such that they cannot be removed. Gravitational waves are such an effect discussed in the comments of John Rennie's answer. There exists another effect that is much larger. First we note that gravity cannot ...


0

Untested Hypothesis: Instability When all of the cards are exactly equally spaced each card's base would shift the exact same amount every time they are flipped back and forth. However, if one card's base is shifted slightly, it would feel more force trying to displace it in the direction it was already displaced causing the base to be even further shifted ...


0

your two equations are correct when the velocity is going to the left, and right respectively. If the velocity is zero then the acceleration could be anywhere in between. This means if the accelerations have opposite signs the blocks will decelerate to a stop and then stay stopped. If they have the same sign then the friction won't hold the system in place. ...


4

So for starters you need to make sure the center of gravity is in the middle of the tray (so in you scenario of lighter/heavier cups, balance it out). This is assuming that you make as many right turns as left turns. If this is not the case, say 70% of your turns are right turns, then you would want to slightly offset the orientation so that it's not quite ...


1

If you have no means to change the inclination of the tray while you are driving, then the answer is: flat. You'll make as many left turns as right turns, so you can't favor left tilt vs right tilt. And you will speed up as well as slow down, so you can't favor forward tilt vs backward tilt. The distribution of cups ... which cup goes where ... ...


2

There's another component to the solution: nonuniform friction. The blade, even if untaped, has different coatings than does the shaft. The butt end may or may not have a tape roll applied. All this means that, even if you throw the stick with zero spin applied, when it hits the ice it's almost guaranteed that there'll be a nonzero net torque due to the ...


1

It is called conservation laws, conservation of momentum and conservation of angular momentum. Because friction on the ice is very small, the geometry of the stick is a line with non uniform mass, there will be angular momentum and linear momentum that will be transferred to the ice at the points of contact. To only rotated there should be no linear ...


1

If your objects were both 100% perfectly rigid and perfectly hard, then any nonzero amount of kinetic energy which is enough to have the ball hit the cube (without being stopped by its own friction before it hit) would be enough. Though of course such a small kinetic energy would not make the cube move much, so kinetic friction would stop the cube before it ...


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Mostly because our flesh burns at a much much higher temperature than leaves, we cannot get that high a temperature using friction because we don't have enough energy (we usually stop rubbing them after a few seconds). Another problem is that even though the flesh will burn at 6-800 degrees Celsius, we still have a lot of water inside our body which will ...


1

The effect the ball has on the cube does not only depend on its momentum or kinetic energy. What might or might not move the cube is the force exerted by the ball. When the ball hits the cube, it takes some time to be stopped, depending on the deformation caused. The shorter the time the higher the force and the more likely it is large enough to overcome ...


1

If we idealize the scenario enough, this is a simple exercise in differential equations, so let's get to work. First, we know that it's initial speed is $150 \text{ m/s}$, but that is by no means its final speed - obviously, the bb slows down as it travels through air! Let's suppose that the moment the bb exits the barrel, it is no longer being pushed (as ...


0

The actual behavior of forces that may be categorized as friction are very messy indeed whether they are static, dynamic, or a combination of the two. There are no models that suitably allow one to obtain a 'deeper' understanding of friction, only models that partially meet practical needs, in an approximate matter, and mostly they have no basis in ...


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What I try to say, is that it makes equally sense if a ball rolls down of tray or up (without friction) in a newtonian system. The "proces" is then reversible. In thermodynamics a proces will only go one way in a given situation therefore irreversible. – Hamid Mohammad 18 hours ago When you think of a ball rolling up or down a hill it is easy to ...


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You have to be a bit careful about what you mean when you ask about Newtonian mechanics being reversible or not. As stated in one of the comments newtons mechanics is only a set of rules that tell you how objects accelerate if they are subject to some sets of forces. It does not necessarily tell you the nature of these forces of where they come from. To ...


1

Force is the rate of change of momentum, $$ \vec F = \frac{d\vec p}{dt} = m\frac{d\vec v}{dt} = m\vec a. $$ If you have an external force (like friction), you are exchanging momentum with the outside world. In this case, conservation of momentum demands that the momentum in your system change. Only with zero net external force can you observe conservation ...


0

You have a different situation when the bb is inside the barrel of the bb gun. Assuming that the bb is a tight fit in the barrel (and it should be), there is pressurized air pushing it. The air is doing expansion work on the bb as it does so. Due to this, you need to use the thermodynamic relationship for the process that is involved. If you are using a ...


0

The way i ended thinking about it is as follows: I convinced myself friction is forcing opposing motion. Now imagining a ball rolling down an incline, I considered the point of contact of the ball with the incline, call it x. This point intends to move in a direction that is opposed to the general translational motion of the ball as a whole (to be more ...


1

There seems to arrive much confusion on this topic. I've updated the answer here to give a clearer picture of what goes on. Consider a star instead of a ball rolling down the incline: For it to roll without slipping, whenever a leg is touching the ground it must stand still (it must not slip or slide). That means that during the time of contact for one ...


0

The math of the underdamped oscillator $$\tag{1}\ddot{x}+2b\dot{x}+\omega_0^2 x~=~f\qquad\Leftrightarrow\qquad -\underbrace{(\omega^2+2ib\omega -\omega_0^2)}_{P(\omega)=(\omega-\omega_+)(\omega-\omega_-)}\tilde{x}~=~\tilde{f}$$ is easy to work out. The characteristic frequencies $$\tag{2}\omega_{\pm}~=~\underbrace{-ib}_{\text{exp. ...


1

You already have a good mathematical answer, so I will focus on an answer with almost no equations. I take it you understand the basic mathematics of the simple harmonic oscillator. When you add damping, the amount of energy you lose per cycle depends on the velocity: the faster you go, the more energy you lose (at the same amplitude) because the force ...



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