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1

You are overcoming the static friction, which others have noted is generally greater than dynamic friction. But, the force needed at the point of contact is presumably the same when twisting versus when lifting, so why the difference in the behavior of the stand (lifting up, versus just sitting there)? The reason is because of the direction the force is ...

1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.

1

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2}$ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2}$ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...

0

For the same reason if you try to push a block or your coffee mug placed on the table it take some force before it move but when it starts moving you can lower the amount of force you applied but the block/mug keeps moving.

3

Actually, spinning the pens bring a relative motion between cap and pen. When the surfaces of two objects are at rest with respect to each other static friction force acts between them and a kinetic friction force acts between then when they are in relative motion. Static frictional force $>$ kinetic frictional force. You may want to read about the ...

0

This is not a complete answer, but by spinig the pen you give to your system the same energy as if you pull it out directly, in two diffrent forms.So the ''spinning'' energy overcomes the friction and the rest pulls out the pen.

1

The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing. If the center is below the ...

0

Previous answers are great and explain the dynamics very well. I'd like to point out that this can be explained just as easily in a static situation. Imagine the weight the shaft has to carry. You don't even have to imagine a curve to note that the weight will automagically center (and lower the center of gravity of) the train in the first image. In ...

1

Typically, the Friction force will be proportional to the velocity of the object it affects (or at least it is usually assumed to be proportional). The Force you describe is constant and pointing in negative x-direction. The situation you are describing resembles an object on a Hookean spring in a gravity field. At first, F1 pulls it upward (positive x) far ...

8

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...

-1

To see why the first configuration is used rather than the second, perform the following experiment: Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball. Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball. Which is more ...

21

Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...

-2

I can see a difference by thinking about the flanges... Given that 1) the flanges are on the inside of the wheels 2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail... compare and where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes ...

0

Mathematically, the friction is described by a couple of Heaviside functions $\Theta$, asuming your body is of length $d$ and the transition from non-friction to friction occurs at $x_0$ with $x_1 = x_0 - d/2$ and $x_2 = x_0 + d/2$. $~~~~~~~~~F_\mu (x,v) = -\mu \cdot v \cdot \left[ \Theta(x-x_1) \cdot \Theta(x_2-x) \cdot \cfrac{x-x_1}{x_2-x_1} + ... 0 This is a subtle problem. Assuming that the normal force that is involved in friction is strictly dependent on how much of the block is over that surface, it is seen that the friction force increases steadily as more and more of the block moves over that surface. This means that the amount of friction force and the acceleration are dependent on the ... 0 Assuming the body comes to rest before it completely enters the region, the body will trace out a partial sine curve in the space of$(t,x)$. Suppose the body is of length$L$has uniform mass$M$. We let$x$correspond to the positive horizontal distance from the interface. Then the force of kinetic friction (with coefficient$\mu_k$) is \begin{cases} ... 0 In general, the direction of a friction force is determined by which object you are analyzing, and you only have 1 object in your free-body diagram. The friction will be opposite the direction the object is or is attempting to slide across a surface. It is not necessarily opposite the direction the object is moving with respect to your coordinate system. ... 1 The coefficient of friction is the ratio of the frictional resistance force to the normal force. The coefficient of static friction is the ratio of the maximum amount of friction that must be overcome to start an object moving, to the normal force. The coefficient of kinetic friction is the ratio of the amount of friction that must be overcome to keep an ... 0 tok3rat0r probably has the right of it. You have not told us exactly how your timing data was acquired. If it was done by some sort of stopwatch (mechanical or electrical) you should assume an uncertainty of at least 0.1 seconds, and perhaps more for a situation where you have to push a button twice in 1/2 second. If you assume a worst-case error of 0.1 ... 0 There doesn't seem to be anything wrong with your calculations given the data and assumptions you've made. So lets consider how long you would expect the block to take in a frictionless situation. By conservation of energy,$mgh=\frac{1}{2}mv_f^2$. Putting in your values I get$v_f=2.63 ms^{-1}$. Then using$d=\frac{1}{2}(v_i+v_f)t$I get an expected ... 1 Other people have answered your questions about the signs; I want to address your comment, "why does this sound like the normal force?" By Newton's third law, pairs of forces always come in equal and opposite pairs. If the ground exerts an upward normal force on you, you exert a downward normal force on the ground. If sliding on the ground exerts a ... 1 What everyone said, but also: the way the question is phrased, the model it uses, is a 1-dimensional model. Imagine a point travelling along the real number line/the x-axis on a cartesian plane. The model in the problem assumes there is no vertical movement, or 'side to side', but only 'pure' 1-D movement along a straight horizontal line. There is only that, ... 0 As user37496 said, the proper way is to calculate the error, using the errors of your measurements. But there is also an alternative, that is you measure often and calculate the mean, and determine the standard deviation as an error. Three times is not really enough to give a reasonable result but it shall do. The mean time is:$t = \frac{1}{3}(0.41 + 0.44 ...

1

It's all a matter of how you've chosen your coordinate system. Think of how the velocities were calculated. They essentially looked at two points in time and the two corresponding points in space. If your positive direction is to the right, and the guy has advanced to the right, then its a positive number for velocity. Force carries the same logic here. The ...

0

The force/ acceleration of the skater comes out as negative because: the force or acceleration is in the opposite direction of motion. You can also think of it like this, frictional force always tries to oppose relative motion. In this case the skater is moving forward w.r.t the ice and the ice is moving backwards w.r.t the skater; so the ice exerts a ...

14

It's very common to get mixed up about signs. The only recommendation I can give is to establish a clear sign convention and stick carefully to it. To show what I mean let's consider your skater: I'm going to use the convention that positive is to the right and negative is to the left. remember that quantities like velocity and acceleration are vectors, ...

2

When the skater is on the ice, friction stops him/her in 3.52 seconds as you said. The molecules in the skates rub against the molecules in the ice, and the ice molecules absorb some of the skate molecules's energy, slowing the skater down. The reason the force is negative is because the friction is acting in the opposite direction of the skater's motion. ...

0

There are at least 3 answers to your question depending on what you mean by "world"? 1 - if you mean "universe," the answer is no, because there would not be "anything" that could walk, or a place to walk on. 2 - if you mean "earth," the answer is no, because there would be no earth. 3 - if you mean "surface," the answer is no, because a horizontal force ...

0

Newton precisely expressed it in the law of inertia. A body (in this case a normal person with shoos trying to walk on frictionless surface) maintains its state of motion unless unbalanced external force acts up on it. This means, the person on the frictionless surface maintains his state of rest, if he was at rest, or moves with the velocity that he was ...

1

Inertia is just another name for mass: it is the property that if you kick a rock so that it makes a certain arc, it will take twice the force to make that arc when you kick two rocks glued together (twice the mass, twice the inertia). You do not really "overcome it" in the sense of finally being victorious and never having to care about it again. In ...

-1

It seems to be your assignment, problem set or so: So instead I will not be spoon feeding you much, here is the method: Use this formula: Work = FxD = KEf - KEi where: KE - Kinetic energy F - Friction D - Distance Remember: KE = mxv/2 Note: Convert units to SI

2

The net drag force on the surface of the propeller is a function of the relative velocity of the fluid against the propeller's surface. The simple model for drag force that can be (loosely) applied to the question regarding the propeller is $$F_d=C_dA\frac{1}{2}\rho v^2$$ where $v$ is the relative velocity, $A$ is the projected area normal to the ...

2

If there is no friction, then you can not rely on it providing traction. However, other forces must still exist, or you'd fall through the floor! So, there is simple artificial solution to being able to walk in frictionless world. All that is needed is sufficiently contoured surface, something like this kind of knobs covering the surface: #### #### ...

1

Some people have suggested throwing something out--that's not walking. However, a world being frictionless does not mean that it is perfectly flat. You can walk by exploiting this fact--if a foot is in a depression you can develop a horizontal force no greater than the sine of the angle of the walls of the depression times your weight. Note, furthermore, ...

1

The simplest way to model friction to solve your problem is through something called the coefficient of restitution, which gives the ratio of the ball's speed after and before the bounce. $$C = \frac{v_f}{v_i}$$ So everytime the ball hits the ground, you can use the coefficient of restitution to calculate the new velocity based on the previous velocity. ...

1

Suppose the friction is given by $f$. For the sake of convenience It is taken as constant. Suppose the ball of mass $m$ is given an initial force of $F$ infinitesimally greater than $f$ so that it starts moving initially with infinitesimal acceleration but as there is friction, the extra kinetic energy gets converted to heat energy while the ball works ...

1

You can rearrange the equation by dividing m: $$F = ma ⇔ a = \frac{F}{m}$$ Now, knowing the force of friction, you can calculate the acceleration. The last remaining step is to calculate the velocity. To do so, there are two possibilities you could choose: The exact solution. To gain it, you will have to solve something called a differential equation: ...

2

Can it be shown mathematically using principles of mechanics that it is not possible to walk in a friction-less world, or is it only by experiment? It all depends on your definition of walk. If you mean, slide along a horizontal surface thanks to friction, then by definition no, it is not possible. Your critical angle is zero, so the horizontal ...

-1

When the pressure increases, the flow decreases thereby providing the same "power". You can have larger flow if you naturally decrease the pressure, or larger pressure if you restrict the flow.

4

YES walking is possible in a frictionless world with appropriate limb movements. The idea is to twirl the arms and legs to cause propulsion by the Magnus effect. The technique has been developed in a British government ministry https://en.wikipedia.org/wiki/Ministry_of_Silly_Walks Named for its investigator Gustav Magnus, the Magnus effect causes an ...

1

In addition to the answers given already which cover the momentum and traction sides of the problem, I'd also like to toss in magnetism. You could just use magnetic fields to either push or pull you in the direction you want to go. Bonus points for using simple electromagnets that could be switched on and off so you can accelerate/decelerate at will.

1

If you measure the box's acceleration $a$, then it must be the case that $$m a = F_\text{fric} + F_A - F_B \quad \Rightarrow \quad F_\text{fric} = m a - F_A + F_B.$$ In particular, in the special case where the box is no accelerating (i.e., it's at rest or moving with a constant velocity), then the frictional force will be zero if and only if $F_A = F_B$. ...

1

There are two types of friction $f_s \leq \mu_sN$ for static friction and $f_k = \mu_kN$ for kinetic friction. So the friction is balanced by the difference between A and B until $F = \left|A-B\right| = f_s$ at which point we get an acceleration of the block due to the fact that ususally $\mu_s > \mu_k$ which are dimensionless constants which depend on ...

4

The purpose of the question, as I read it, is related to an ideal condition (perfectly smooth surface) and a man with regular feet/shoes. In this case, normal daily walking would not be possible, as the horizontal force that moves the person ahead is coming from the friction between the feet and the ground. F = k*N; F : Force (horizontal) k: friction ...

2

Yes. You need ridges (about 2mm tall): _____|~~~~~|______|~~~~~|____ You can walk on that if your tread is the right shape friction or not. Boot tread will interlock with the ridges providing non-friction-based traction and so the ability to walk.

15

Although friction is not one of the four basic forces of nature, it exists because those basic forces exist. Friction is the resistance to motion of two objects held against each other. Friction that allows us to walk depends on gravity to convert our mass to weight which holds our feet against the surface where static friction enables the soles of our ...

5

You cannot walk at all if there is no horizontal component of the force of interaction between you and the ground; by this definition "no friction" is tantamount to "no ability to move horizontally". But to add to Hapa's ANswer: you can move by throwing stuff. How do you do this? Here's one way if you don't have a handy sack of hammers ready in your pocket ...

16

I assume by zero friction you mean no roughness or deformity in the ground. Perfectly smooth. Even so there is a way to walk. As you plunge your foot into the ground you compress it a little based on the atomic theory of matter. This impression allows your foot to be slightly lower than adjacent atoms and can therefore push away from them. On ice this ...

28

If there is no friction, you can still move by conservation of momentum. Take some stuff with you that you don't need. Throw it away in the opposite of the direction you want to go!

0

I agree with you. The existence of microwelds on the surface and the object is essentially what causes our feet/shoes/socks/whatever to push back and forward upon the surface to propel us forward. Without something to back up on (or more accurately, to hook on to), we can't expect to propel forward. Newton's Third Law says every force has an equal and ...

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