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9

Why doesn't something break, or why doesn't the aircraft just skid down the runway, wheels locked, tyres smoking? Basically, the aircraft you're accustomed to simply don't have all that much thrust, particularly compared to the force required for hard braking. An example: a commercial airliner will typically have a rollout time of about 30 to 35 ...


6

Mathematical demonstration It's straightforward to see why this happens if you use a bit of linear response theory. Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says ...


4

If you apply the front brakes, and then increase thrust along a line that is some distance above the ground, you increase the downward force on the nose wheel. This helps to prevent skidding - although it would not be sufficient if the thrust was sufficiently large. You can compute this with a simple diagram: I drew the limiting case where the thrust of ...


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 ...


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 ...


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 ...


2

I'll try to answer your question on a conceptual basis, because the math involved would tend to obscure the concept. For a mass that is bouncing up and down on a spring in air, the system is almost totally undamped, and the spring will oscillate at a frequency that depends on the spring constant and the total mass that is oscillating. If I take this same ...


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 ...


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

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 ...


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 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 ...


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

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 ...


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 ...


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 ...


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 ... ...


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.


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 ...


1

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


1

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 ...


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.


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 ...


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 ...



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