# Tag Info

1

First, you can have currents caused by something else that are pushing you one way while the wind is blowing the other. Second, as the other answer says, the ripples from the wind can be deceiving. Far from the shore, it's difficult to get a sense of speed relative to trees or docks. And the ripples on the surface don't always extend into the water itself ...

0

I have timed a similar phenomenon swimming between two landmarks in the Corinth gulf in Greece and since then I have observed it in a small sea lake off it. It is due to currents below the surface, not caused by wind, but by water entering or leaving the gulf/sea lake. This in my case is a continuous flow one way or the other , due to the tides sometimes ...

1

You can't measure speed over the ground when you are moving in a boat unless you carefully observe nearby shore objects and measure your progress with a timepiece, or unless you have an instrument that measures speed against the bottom of the lake. It's quite likely that you saw wind waves washing past you, and attributed their apparent speed to your boat's ...

1

The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from ...

1

If it is static friction, then the two blocks are stuck together and both have the same acceleration. In that case, the top block has a net force of 40 N (100 N pull - 60 N friction) and the bottom block has 60 N (just from the friction). Since 60N < 80N max friction, then the ansatz that this is static friction is consistent.

1

Normal force between blocks will be $N_1 = mg$ and the normal force between the road and the larger block will be $N_2 = ( m + M ) g$. So the friction forces are, $$F_1 = {\mu \over 2} N_1 = {\mu \over 2} m g$$ $$F_2 = \mu N_2 = \mu ( m + M ) g$$ The total force on the bigger block will be $F_1 + F_2$ to the left. So the larger block will ...

8

To add to Rick's answer: Rick is right that it is only the front wheel that is skidding. Most of the weight is over the main wheels (it is essential that the centre of gravity be close to the main wheels, to ensure the plane does not pitch when the wheels strike the ground on landing.) Only a small amount of weight is on the nosewheel. It would seem that ...

0

The person only needs to be able to counter the force acting on the plane to slow down the movement. The force on the plane is from the wind - it'll be a reasonably big force because the plane presents a reasonably large surface area - but if the person is wearing ice spikes, it's not inconceivable that they could make a difference.

1

I propose redefining this problem as follows (because I'm not sure it has a solution the way the OP has defined it). Let $y=f(x)$ be some symmetrical (around $y$) function like $x^2$. Let the point mass experience a friction force acc. to the usual simple model $F_f=\mu F_N$, with $F_N$ the Normal force acting on the point mass in the point $(x,y)$ ($N$ ...

39

Simple (Wrong) Analysis Shoes Assuming the coefficient of friction on the ice is approximately the same for the tires and shoes. It would do just as much good to get into the plane as to try to push it. Both would increase the frictional force by at most $\mu\,m\,g$ Having established an upper bound for the effectiveness of pushing we can compare this to ...

2

When considering whether two surfaces will have a high friction or a low friction when rubbed together, more important than whether they are individually smooth or rough is what the barrier to their passing actually is. Consider first two perfectly flat plates. Even if the two plates are made out of wood, which is rough, they slide relatively easily past ...

0

Kinetic friction is in the direction opposite the moving object's velocity relative to whatever surface it is sliding on. Static friction opposes a stationary object's tendency to slide relative to the surface it is resting on. In other words, it is opposite the direction in which the object would move—relative to the surface—if there were no friction. ...

0

Friction is always tangential to the object's surface. If it's propulsive (causing the motion of the object) then it acts in the direction of motion. If it's resisting, then it acts opposite to the direction of motion.

1

As @rmhleo pointed out in his answer, the frictional force doesn't depend on the surface area, because no matter which part of the object is in contact with the other surface, the total normal force (and thus the total frictional force) is unchanged. However, that assumes a couple of simplifying conditions: namely, that the two surfaces are consistent in ...

1

Yes, frictional force does not depend on the area. This is clear from a simple mental experiment, think of a block resting on a surface. Assume it is a prism whose faces have different areas. Whichever face it is resting on, the friction force will be the same: when on the smaller surface, the contact is reduced compare with the case where the resting face ...

2

The idea here is that static friction is larger than dynamic friction. This is something that depends on the nature of the materials in contact and is not in general true because friction is not a fundamental force: it is a result of some very complicated phenomena at a microscopic level. An explanation that does make sense to me is to think of the two ...

0

So why if applied force is increased the normal force can't withstand the increased force? The other answers answer this well. Also is there any possibility that in a situation the frictional force could always cancel the applied force? In that case you would need a material (surface) which is infinitely strong. If you put a heavy stone on a ...

0

To add a little to Asher and Gert's answers: The atoms on surfaces are constantly in motion. If you're used to thinking of solids as being really rigid and difficult to deform, try this viewpoint instead: think of them like a bunch of grains of sand held together by tiny Slinkies. The atoms can move out of the way and slide back into place, or even be ...

0

Why can't friction withstand any force? Because the amount of friction that inter-surface interactions can provide is limited, not unlimited. In the idealised force diagram below, only the object's weight $mg$ provides a vertically downward (aka 'Normal') force, causing a friction force between the object and the plane. This friction force is ...

2

Your image shows what is going on at the microscopic level between two surfaces. To understand why friction works, you have to look smaller, at the atomic level: and when you get to that point you're no longer taking about "friction" as we know it, but about physiochemical interactions between atoms and molecules. Those interactions are mediated by ...

-2

The magnitude of static friction adjust its value to the applied force hence static friction is called self adjustable....

0

If friction is constant, then the path followed doesn't matter. This is true because the only force involved is gravity, which is a conservative field. And that means (or is the definition of :-) ) that the energy expended to go from point A to point B is path-independent. Now, if you're interested in like the required initial velocity, you'll see that ...

0

In this system: Around half circle #1, gravity is a component of centripetal force. The normal force between coaster and path is due to centrifugal force. Around half circle #2, gravity is also a component of centripetal force, via the normal force between coaster and path. The normal force is a combination of gravity and centrifugal force. Around half ...

0

In theory there is no torque from friction once the cylinder is rolling. Imagine that there is no supporting surface but no gravity - then the cylinder will just continue to "roll" forever. So all the supporting surface needs to provide to the cylinder is a force through its centre of mass to counteract the gravitational force. This is a force perpendicular ...

0

Assume a simple model of friction where the friction force is given by $F=\mu mg$, with $\mu$ the friction coefficient. To determine whether the force $P$ can cause any movement we need to look at the balance of forces along the x-axis. For the top body with mass $m_1$, if $P>\mu_2 m_2 g$ the top mass will slide over the bottom one. For the bottom ...

0

Friction is present in the macroscopic and microscopic world as a resultant force. It is an important (nonconservative) force in classical mechanics because it makes certain types of motion, like walking, and the absence of motion, like resting on a chair, possible. Friction is possibly absent in the quantum mechanical world, although there doesn't seem ...

0

Without friction there would be so many different things! We walk because of friction, we can write on paper because of friction. To put it simple, it is attached to us in our everyday life. While calculating the amount of force required to love a particular body, we have to take the account of limiting friction. If it is greater than the limiting friction, ...

1

A year has passed since this question was posed, and as luck would have it, this week a paper appeared on the science preprint repository 'ArXiv' discussing exactly this phenomena. The preprint (PDF) is available here. Key to the phenomenon of spectacular friction forces between two interleaved phone books is a simple geometrical conversion of the traction ...

2

They produce heat because the surfaces on small scales are rough like canyons rather than flat like the ocean. As these rough surfaces come into contact with each other they repel. When two atoms are brought very close together they store potential energy. When they move apart that energy becomes kinetic. However, this kinetic energy generally isn't enough ...

0

Here's how I see it: Assume the friction force is proportional to surface area $S$, via (all other things being constant): $F_\text{friction} = \mu S$ with $\mu$ the coefficient of friction. Then for an infinitesimal ring with thickness $dr$ at $r$: $dF_\text{friction} = 2\mu \pi rdr$. And $dT=rdF_\text{friction}=2\mu \pi r^2dr$ Integrating between ...

8

Assuming terminal velocity of 200 km/hour, the scenario seems equivalent to stepping out of a car that's travelling at 200 km/hour. In that case it's not the fall (hitting the road) that kills you, it's the friction (i.e. sliding or tumbling along the road). There might be a minimum of friction initially (when you're falling parallel to a vertical wall) but ...

14

I have slid down a much smaller version of this at Burning Man. Paha'oha'o was a 30 foot tall volcano art piece which you climbed and then "sacrificed" yourself by dropping into a pit featuring a slide just like you mention. The drop features a 10 foot free-fall, just enough to take your breath away, after which the careful curve of the slide gently catches ...

7

Probably the closest to what you are asking about is the story of Ivan Chisov's survival (see Ivan Chisov); but there have been several other similar cases (see for example 10 Amazing Free Fall Survivors).

35

The answer is Yes and your thinking is correct. You try to differ between impact and sliding on a curve. In fact the impact is just a sudden large force, while a curved (e.i. circular) motion similarly applies a force, just much smaller but also over a longer period of time. The key in surviving any fall is to reduce the force on your body at "impact". A ...

26

Let's make life easy for ourselves by assuming that the slide is an arc of a circle: We also assume the slide is made out of something with a very low friction, so the skydiver maintains a constant speed $v$ all the way round. The reason that using an arc of a circle makes life easy is that the acceleration felt by the skydiver is simply:  a = ...

2

Yes. In fact it would be better to imagine that you skydive towards a "track" that you can strap a "chair" onto, and then the chair is stuck on the track. the acceleration to keep you in a circular orbit of radius $R$ is only $v^2 / R;$ with terminal velocity being about $v \approx 56 \text{ m/s}$ a $1~g$ acceleration will be accomplished by a radius of ...

1

we often confuse or loosely say that friction opposes motion but it should be kept in mind that it opposes the "relative motion" so if two bodies are moving opposite to each other then we need to change the frame of reference from earth to the body and check what is the direction of relative velocity. once determined we can clearly say that the direction of ...

0

Let the angle of the slope be $\theta$, $m$ the mass, $a$ the acceleration, $\mu$ the friction coefficient and $g=9.8 ms^{-2}$. On the mass acts a vertical gravitational force $mg$. Decompose this into a component along the line of the hill, so that is $mg\sin\theta$, and one perpendicular to that, so $mg\cos\theta$. The latter provides a friction force ...

0

The retarding force from friction is $50g \cdot 0.05$ Newtons, acting up the hill. If the particle is sliding, most brakes will not be effective as you have already accounted for the sliding friction. If the brake is effective, it will add a force of $260$ Newtons up the hill, as you are given the force. You can then resolve these into vertical and ...

1

There are four forces to consider: 1) Gravity, acting downward 2) The push from the person. You said an acute angle, but that could mean either upward or downward because there are two directions from the horizontal. In the end, the analysis procedure is the same (but the results aren't). 3) The normal force of the wall on the block, acting ...

1

When one analyzes an object using a free-body diagram, one generally considers the possibility of all reasonable forces and eliminates the ones which are not likely to contribute to the acceleration of the object or are summarized by other forces. For the object resting on a horizontal table in a gravitational field we immediately recognize there is a ...

1

$N$ is always perpendicular to the surface exerting it (that's what the "normal" in "normal force" means). A friction is always parallel to the surface. So, yes, $f_s$ is always perpendicular to $N$. If the surface is horizontal then $N$ and $mg$ are both vertical. Any $f_s$ would have to be horizontal. If the forces add up to zero (if the object is not ...

2

Friction always opposes motion, so for example if the block is moving up the wall, the force of friction retards the motion and acts downward. If the block moves down, then the force of friction acts up. Basically, which ever way the block moves, the friction acts in the opposite direction. You pretty much said that yourself. You said in absence of a force ...

1

Friction exerted by wall on the block acts in the opposite direction of the impending motion (or motion) of the block relative to wall. Yep, that's your answer. In absence of F and fs, the object will move vertically downwards. (F is required to keep the block in contact with the wall) So fs acts vertically upwards? Why are you discarding F? F is holding ...

0

In the case of sphere when it roles you have to analyse in which direction is the force acting. Consider a force acting along a line passing through its center of mass, now this force is not going to roll the sphere it's going to slide the sphere. For rolling a sphere an external torque must be given to it and as we know friction always opposes relative ...

0

One remark: I understand the case on the right, but not completely the on the left, so I need clarification with it ... One hugely important thing about the case on the left is that the speed of box 1 is stated to be faster than the speed of box 2, both of which are presumably being measured relative to the ground. That means that relative to box 2, box ...

Top 50 recent answers are included