# Tag Info

2

The following diagram might be helpful: It is a diagram of the possible trajectories of the cannonball, fired at different angles. In green is the "critical" angle - the one that would just reach a height of 250 m. Do the following calculation for that height: A cannon ball that reaches a height of $h$ must have had a vertical velocity $\sqrt{2\cdot g\... 0 ;angular displacement = angle rotated by the body with respect to its axis angular velocity = rate of change in angular displacement angular acceleration =rate of change in angular velocity The acceleration of the rotating body have two components : tangential acceleration (acting tangentially to the path which changes the speed of the object) ... 0 Angular velocity and angular acceleration are both in a direction parallel to the axis of rotation. If the magnitude of the angular velocity is constant, and the rotating body is not precessing, then the angular acceleration is$0$. You define your angular velocity$\omega$to be perpendicular to your radius at all points, because it's defined as the cross ... 2 The mass loses kinetic energy and potential energy due to friction (with the tube). The amount of kinetic energy and potential energy lost is equal to the work done by friction on the block. 0 This is usually a pretty fun thing to watch on a slo-mo camera. What appears to happen is that the bottom of the slinky doesn't move at all until the top catches up and the slinky is full compressed. However, I am here to prove to you that this is not the case. The bottom of the slinky does not remain stationary the entire time; it actually experiences an ... 2 Let's start at$t=0$at the top of the tube. Let's also posit that for each full revolution$\frac{W}{n}$friction work is done on the block. When it's at its lowest point, after half a revolution, it's total energy is: $$T=K_0+2mgR-\frac{W}{2n}$$ Then it climbs back up and total energy is: $$T'=K_0+2mgR-\frac{W}{2n}-2mgR-\frac{W}{2n}=K_0-\frac{W}{n}$$ ... 0 As you know, a moving bicycle stays upright because the rider turns the steering wheel "into the fall". The interesting thing about a well-designed bicycle (and I hint at this in my answer about the curved fork of the bicycle is that when the bike tilts out of the vertical plane, a torque is generated that will turn the front wheel into the fall. This is why ... -4 Elliptical planetary orbits are apparent paths of planets about their central body, which is considered static in space. By simple mechanics, it is physically impossible for a free macro body to orbit around a moving central body, in any type of geometrically closed path. Sun (central body of solar system) is a moving body, planets (and other macro bodies) ... 1 In your solution consider what happens when$t=0$. The projectile is at position$(0,0)$and the jet is at position$(0,250)$ie at a height of 250 m directly above the projectile. If this was the initially condition then the projectile would never hit the jet. I have changed the annotation of the graph which I hope will help you understand how the ... 1 Projectile's path is a parabola, whose apex (point of maximum height) depends on$\theta$. For particular values of$\theta$, apex does not even reach up to the altitude at which enemy-jet is flying. So for all those cases where apex is equal to or greater than the altitude of enemy-jet, there is a$\textit{possibility}$of hitting the enemy-jet (the gunner ... 2 Your friend is wrong, and seems to suffer from a serious case of anti-intellectualism, not to mention arrogance and authority issues. It's true that the models of physics we know are approximations, but the reason we're using them - after centuries of research - is that they're extremely good approximations. To suggest that our physical laws can't cope ... 6 Newton's First Law is invalid because friction exists in real life. Let's review what Newton's first law says: When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a net force. Your friend is right that friction exists in real life. But your friend is wrong that ... 0 A free body diagram is your friend ($c= \frac{\ell}{2}) The general case above yields the following equations of motion \begin{align} A_x & = m \ddot{x}_C \\ A_y - m g & = m \ddot{y}_C \\ A_x c \sin \theta - A_y c \cos \theta &= I_C \ddot{\theta} \end{align} Since the pivot does not move, the acceleration of the center of mass(\ddot{...

1

The formula for gravity is $F_g=\large{\frac{Gm_1m_2}{r^2}}$. So store variables for your $x$ and $y$ velocities (or an array or whatever method you wish) and change your position by your velocity each frame. Also each frame, use a loop (I don't know what language you're using so I can't give exact code) to look at each planet and plug everything into the ...

1

Your friend actively misunderstands the Newton laws and the ideas behind it. He is about to argue that formula $E=mc^2$ is invalid because it states that Youngs modulus ($E$) is equal to mass ($m$) times hypotenuse squared ($c$). And he is supporting his arguments by several argument fouls. First law states that when there's no forces applied to the body it ...

1

Well, the "object" has two kinds of energy at all times: kinetic (movement), and gravitational potential. This calculation you want to do is easiest if there is only one significant body, like the sun. In this case, at the beginning your KE (kinetic energy) is 0.5mv2, and your gravitational energy is Ug = -Gm1m2/r. Now, an object can never completely leave ...

1

CASE -I:Consider the acceleration of a body with mass M when a force of 100 N is applied on a body. CASE-II : Consider the acceleration of the same body when a force of 1000 N and 900 N respectively are applied on it simultaneously in opposite directions. The acceleration in both cases will be the same as the net force is 100 N. Now consider the ...

2

You don't have all equations, and one is not correct. The usual assumption in these problems are: There is no friction. Ropes are glued to pulleys. From 1. it follows that $T_1=T_2$ You forgot, that $m_2$ is acted on by $T_2$ twice: ${\ddot{x}_2} = {\frac{2T_2}{m_2} -g}$. $T_3=T_2+N$, where N is force which rotates the big wheel. ${\ddot{\beta}} = {\... 0 Cut out a map of your country drawn on a uniform thickness aluminium plate of homogeneous density. Take it to your physics lab and calculate practically the moment of inertia at multiple points in various axes. Do you think with the results alone you can redraw the shape of country i.e the shape of the aluminium object in your hand. 5 I will only discuss rigid bodies here; I do not understand fluids well and I doubt you were thinking about fluid bodies anyway. The first thing you need to understand is that the concept of "moment of inertia" can best be understood in terms of the inertia tensor. The word tensor might deter you, but you must face it to understand many concepts in physics. ... 4 I have noticed this effect often in cars and sometimes in trains. This is the reason I think it happens though I can't claim to have done any research. The car stops because the breaks are applied, the wheels stop turning and there is a force of static friction between the road and the tires. In the car frame of reference I experience this backwards ... 6 As the saying goes: "All models are wrong, but some of them are useful." For me, laws of physics are actually just models, i.e., simplifications of reality that give satisfactory answers to certain questions. Newton's laws are useful to build bridges, build skyscrapers, land rockets on the Moon, etc. You could also argue that: In reality, a force is ... 4 Short Answer: Your friend is wrong because our models of friction (within certain parameters) are derived from Newton's laws. Tidal forces, for example, are are a form of friction. The argument that Newton's "laws" are "invalid" in an pure philosophical sense, better rest on Newton simply pulling the concepts of gravity, force and inertia out of thin air in ... 3 The two forks and the toothpick wedged in them are rigid. For a rigid body to balance at a single point, the centre of mass of the body must lie directly below that point. Then, any change in orientation of the body will raise the centre of mass, adding to its gravitational potential energy. The system prefers to be in the lowest energy state, and so the ... 15 For the bicycle to handle well, it needs a certain trail - an offset between the point where the wheel touches the ground, and the point where the line through the steerer tube (axis of steering) touches the ground. If this distance is too large, the bike is hard to turn; if it's too small (or negative), it becomes unstable. In principle it is possible to ... 1 A curvature means a constant redirection of the force exerted on one end of the front wheel fork. Straight front wheel forks would need some kind of sharp "turns" at some points. This would result in a sharp change in the cross section which means sharp change in the resulting forces at that cross section. Sharp changes are usually a weak point in mechanics ... 17 Wolphram johnny gave good explanation but for trampoline case there can be more explanation Whenever you jump from earth, according to "third law", the force you give to the earth is equal to the force (reaction force) given by earth to you which makes you go off from the ground, the force you gave to earth actually moves earth (negligibly due to its high ... 2 There are two quantities that are relevant to this discussion: kinetic energy and momentum. The change in an object's kinetic energy is equal to the work done, which is forces times distance. $$K_2 - K_1 = W = F\cdot (x_2 - x_1)$$ It does not matter whether the object is a bowling ball or a tennis ball. If you apply 1 newton of force over 1 meter, the ... 3 Mass is accounted for already in the force.$F = m \cdot a$Then work formula could also be$W = m \cdot a \cdot d$where the mass$m~ [kg]$is accelerated at$a ~[ ^{m}/_{s^2}]$over the distance$d ~[m]$which is equal to$W ~[N \cdot m]$or$[\frac{kg \cdot m}{s^2} \cdot m]$Always look at units they are probably the most important thing to use to ... -1 Force is mass times acceleration. From this you can get work as mass times acceleration times distance. 7 Newtons laws are a good approximation for how the world works when the velocity is less than 1% the speed of light, the gravity isn't too strong, when the number of elementary particles an object is composed of isn't too small as an object needs to be large enough for quantum uncertainty to be insignificant, and the amount of space is small compared to the ... 49 Regardless of relativistic effects: Newton's First Law is invalid because friction exists in real life. False, the first law talks about the case when no forces are present, if forces are present go to the second law. Newton's second law is invalid due to the same reasons. False, you add friction to the total force. Newton's ... 1 Newton's laws are valid for all situations where velocities are small (compared to the speed of light, ie relativity is not important) and where quantum effects are negligible (mostly where objects are much bigger than elementary particles). The problem with your argument is that you and your friend are using idealized expressions for Newton's laws, not ... 0 In SI units:$1~\mathrm{N} ~\mathrm{m}=\mathrm{1~kg~ m^2 ~s^{-2}} $and$1~\mathrm{J}=\mathrm{1~kg~ m^2 ~s^{-2}} $. So the units Newtonmeter and Joule are the same in SI-Units and there dimensions are actually equal too. This is a fact and just a matter of definition. Work along a curve$C$is$W=\int_C \vec{F} \cdot d\vec{s}$. So in the simplest case where ... -1 It is true that$Work = Force * Distance$. Picture a mass,$M$, in empty space away from all gravitational fields. In other words, it's not moving at all, it's not accelerating at all. Then, an acceleration,$g$, suddenly appears and now we have the situation where this acceleration results in a force being applied to the mass. By one of Newton's basic laws,... 0 I would think the second situation causes more damage. The total energy of the first situation is $$2\cdot\frac 12 m v^2$$ The second situation total energy is $$\frac 12 m (2v)^2=4\cdot\frac 12 m v^2$$ After the collision, car plastic deformation energy can be approximated by the total kinetic energy. Well you can argue there are other way to consume the ... 7 The gravitational potential field can be found by a full volumetric integration from the overall volume, or planet, or whatever: $$\Phi(\mathbf r) = -G\int_V\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}dV$$ This just comes from having a distribution for$M$in the potential formula: $$\Phi = \frac{GM}{r}$$ Also, its easy to see that, if you are ... 0 @Numrok 's analysis makes certain assumptions which are then mentioned in the last paragraph of the answer. For simplicity assume that the speed of the cars is measured in m/s rather than km/hr. In the head on collision with both cars of the same mass and speed the final kinetic energy will be zero if the cars interlock as a result of the collision. ... 0 No the car cannot move... even if it is in the friction less surface. and the force you give to the car with a hit in inside is an internal force if you consider car and road as the system. as the internal force have newton third law counter forces within the system they cancel out and the net force is 0, lets say you are hitting the dashboard with force F ... 0 in this answer im not substituting values with proper units as it is all just about comparrison there are two ways of refering impact: force and impulse and for each body impulse is different impulse on an body = change in its momentum ultimately if you measure impact as impulse, first case both car have same change in velocity (30 to 0) if masses are ... 4 The car can be moved provided it isn't on a frictionless surface, or more precisely the centre of mass of the car/occupant system can be moved provided the car isn't on a frictionless surface. As you say in the question, assuming no external forces are acting then because the momentum of the car + occupant is conserved the occupant cannot move the centre of ... 1 seeing the person and the car as one system you can say that the center of mass always stays at the same point if no external force is applied. However, if the person jumps in one direction inside the car, the car will move in the other direction. The center of mass stays at the same position. Of course the floor applies an external force to the system and ... 2 This is a two-body problem with the center of mass at rest, much like a stationary nucleus emitting an alpha particle. Therefore if the person were to move in one direction the car would go in the opposite direction. 3 The force the fluid does on the bottom piece does not depend on the height of the water column of the reservoir. It does depend on the height$h$of the water column in the plate. This can be easily seen by the fact the water is static so the pressure at any horizontal plane is the same. The pressures in$a$and$b\$ are the same. The force of the water on ...

0

Yes you could rise up the rope. You need to exert a force more than your weight, f = your mass x g. But you have to rappel exceedingly faster. Basically you could look at this as a rocket which is constantly refueling. If you exert less than your wait you break your fall to the degree of your force. You'd need lots of rope though depending on its mass.

0

I think you should look at Newton's 3rd law in a frame where forces are balanced and the initial impact transition has settled. e.g. when you hit the wall with 50 lbs force, even if you have good muscle control and apply close to 50 lbs to your hand, it does not move in a linear acceleration because it has to fight its way through a complex multi degree of ...

51

You've caught a non-intuitive part of Newton's 3rd law. It's actually applying in the case you mention, but because the objects involved are of dissimilar hardness it's easy to perceive the impact as a violation of the law. Impacts are actually really complicated. Consider this slow motion video of a punch to the gut. We won't be able to cover all of the ...

15

What makes you think that the maximum force you applied to the dry wall was anything like the maximum force you applied to the brick? It certainly wasn't. The dry wall gave way much before you were able to attain the same force as applied to the brick. Try punching the air and see how much force you are able to apply. The experimental evidence that the ...

7

TL;DR: The physics of hitting things are not as easy as exerting a constant force on something. What I am trying to say with that is that Newton's law of course applies, but it would be more obvious to see it if you were just pushing/leaning against the wall with your weight. Then I'd say the two walls probably feel roughly the same. So what is different ...

11

There is no doubt the Newton's third law holds in this case. The source of confusion is the fact that you are neglecting the time interval of the collision as well as the momentum change the colliding body. As we shall see it is incorrect to assume you applied the same force in both cases just because you started with the same initial conditions, i.e. the ...

Top 50 recent answers are included