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You do not need a unit for force when measuring inertial mass in Newtonian Mechanics. The only things you really need are the Newton's second law and the concepts of inertial frame and acceleration. The way you shall proceed is the following. Take a collection $\{m_i\}$ of (unknown) masses and a spring. Use the spring horizontally to accelerate the masses ...


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If you know velocity, you also know acceleration. If you know acceleration, you can set up Newton's second law and all is known except the push on the track. This push is missing in your statement about the centripetal force, as @lucas comments.


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I don't understand the image either. But I'm going to guess that the pivot of the pendulum is taken to be at rest. In that case, the normal force takes on whatever value is needed to achieve zero net force on the pivot (so that it does not accelerate).


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Why do we not include the surface area of the container in the formula? Because it is not needed. Pressure $p$ is force $F$ per unit of surface area $A$: $$p=\frac{F}{A}$$ The pressure a gas exerts on the walls of a container is the collective force collisions of the gas molecules exert on the container walls, per unit of surface area. If we look at ...


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Pressure is a measurement of force/unit area. It doesn't measure the total force exerted over the entirety of the surface, but the force exerted on one "unit area" of the surface. One unit of area depends on the measurement system you're using, but by doing that surface area can be disregarded.


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Let's look at: $$F_g=Gm_{g_1}m_{g_2}/r^2$$ For an object with mass $m_{g_1}$, on the surface of the earth, then: $$F_g=Gm_{g_1}M_E/R_E^2$$ Where $M_E$ is the mass of the Earth and $R_E$ the radius of the Earth. You can now verify that: $$GM_E/R_E^2=g=9.81\:\mathrm{m/s^2}$$ So we could have written the second expression as: $$F_g=m_{g_1}g$$ An object ...


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There are a lot of very physics-y answers here, so I'll give a more real-world example to help you wrap your head around the idea. Imagine (or better yet, try!) you have a weight on the end of a string and you are twirling it around your fingers. Once the weight is rotating, you let it wrap around your finger. You can see that as the object comes closer to ...


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The centripetal force is directed radially inwards . Work done due to the centripetal force is 0 (as, S(displacement in one rotation)=0). Work done = F.S.cos(\theta)=m . a . cos(\theta).The accln. is thus directed tangentially outwards. Thus, tangential velocity increases.


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Orbital insertion is hard enough with ships where the initial stages can be aimed in directions favorable to the final orbit. This design calls for a fixed launch direction, which would be a terrible waste of on-board fuel for all but those satellites whose orbits coincide with the gun's trajectory. Even that can be overcome with fuel. The real issue is ...


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The rod exerts an upward shear force on mass on the left. This goes along with @lemon's comment regarding bending, although the shear force is not a tension. There is also a bending moment applied by the rod to the mass, with axial tension within the top half of the rod, and axial compression within the bottom half.


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When rotating, the force needed to keep the two masses in circular motion is the centripetal force, (in this case, the tension). Now, you get it.


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One more nasty factor: What is the expansion speed of your propellant. Take the Jules Verne approach and your spacecraft falls far short no matter how much powder you put in the gun because the expansion velocity is too low. Your craft will never exceed the expansion velocity of the propellant. Note, however, that you don't have to use explosives (or ...


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The terminology you use seems a little loose, air resistance, a type of friction, is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid, so it is not measured in mph but in force units. But to answer your question, when the airflow leaves the amplifier it produces a force or a change in momentum onto ...


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I want to focus on one thing, here. The nature of the normal force. You write Doesn't normal force oppose any other force? which is a easy impression to get when you are introduced to the normal force in the context of things sitting on other things in a gravitational field, but that's not the best way to think about it. The normal force keeps ...


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Wanted to give a different angle to thinking about this. Since you mentioned you wanted to consider this from the inertial frame (non rotating), then there is indeed no (fictitious) force. But in that frame, we can consider the water is indeed falling; however, the rate at which the bucket is also "falling" is such that the two stay together - in other ...


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In the bucket experiment when the bucket reaches the top of the circle why will it have a normal force acting on the water downwards? The normal contact force is exerted by the bottom of the bucket as explicitly mentioned by the author. So, it is acting downward when the bucket is inverted. Doesn't normal force oppose any other force? There is no ...


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Imagine a scenario where the bucket is rotated at just the right speed so that the centripetal acceleration required to keep the water on a circular path is exactly 9.81 $ms^{-1}$. Then at the top of the rotation, all the centripetal acceleration is supplied by gravity. However the bucket might be rotating faster in any given scenario but it still rotates ...


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Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...


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I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...


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Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.


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Let's say you have got such gun. Next logical step will be to install on the satellite a smaller gun that would shot-back several small shells and so accelerate the satellite, indeed? If this small on-board gun would use really many small shells (size of molecula) then your are getting just a traditional rocket. Apparently it is not much difference for ...


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Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...


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But the acceleration is not a partial derivative! Its a total derivative, $\frac{\mathrm dp}{\mathrm dt}$, with a $\mathrm d$ instead of a $\partial$. Anyway, I guess you might want to read about the Hamilton-Jacobi equation.


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The definition of force is $F = m a = m \frac{dv}{dt}$. The average force is $<F> = \frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt$. Write $F= m \frac{d v}{d t}$ to find $<F> = \frac{1}{\Delta t} \times$ momentum change.


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I would look at this in a slightly different way. Rearranging it: $$ m \ddot{x} = -(a|\dot{x}|+k) x = -k_{eff} x$$ If you look at it that way, it is really a variable, non-linear stiffness $k_{eff}$ that depends on the velocity, rather than a damping that depends on the position. In this respect (assuming $a > 0$), the stiffness coefficient has a lower ...


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If by "solve" you mean to get an answer that is independent of θ, the answer is... Nope. You need θ, it doesn't cancel out. However, here's a slightly similar problem that you CAN solve: You've got a similar hillside, and a sled at the top of it. The height of the hillside is h, the friction coefficient is μ, and the angle is θ. You let the sled slide ...


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Put a coordinate system on the center of mass and place each leg i at $$\vec{r}_i = \pmatrix{x_i, & y_i, & z_i}$$ where $z_i = z_{c}+\theta_x y_i - \theta_y x_i$ describes the vertical deflection of the point, given the center of mass vertical position $z_c$ and the two tilt angles $\theta_x$ and $\theta_y$. Add vertical loads the each point ...


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Let's look at what happens right when the falling jumper passes that equilibrium point (EP). At that point, as you correctly pointed out, the force on them (up, from the bungee) is equal to the force down (gravity). So, there is a total net force of zero. However, they already have momentum from falling. Newton's first law tells us that if there's no net ...


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The vector sum of the forces will be zero: $$\Sigma \vec{F} = 0$$ If you can assume that the bottom of the block is planer and the weight acts perpendicular to that plane, (i.e., the lengths of the legs are identical) things are much simpler. The weight acts downward and the forces from the legs act upward. The sum of the torques (moments) about any point ...


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The work done onto the spring is $dE/dt=F(t) \dot x(t)$. You should not look at the direction of $F$ alone, but at the the direction of the motion as well.


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If the pipe is cylindrical, there is no equilibrium height. All the air that goes in at the bottom must come out at the top, so the force balance will not depend on the height of the ball. In a ball flow meter (rotameter), the pipe is slightly conical, so that the gap between the ball and the pipe increases as the ball rises. If you want a very coarse ...


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The energy transfer is the work: force times distance. With a strike, the force is bigger and the distance (and time) is shorter. You could tune the two scenarios such that the work is the same. The big difference: when you cease to lean on the table, the table performs the same work back onto you (assuming reversible, elastic deformation), with zero net ...


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This is a tricky question. If the average force is really the same, what is going to change is impulse, the integral of force over time $$J = \int_0^{\tau} F(t) dt \simeq F_{av} \tau$$ When you push the surface you interact with it for a longer time ($\tau$), while when you smack it you interact with it for a shorter time. So if the average force $F_{av}$ ...


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Blood pressure is measured in mm-Hg. If we take a typical blood-pressure 120-80, that can be adjusted to height above or below the heart provided we make a very bad assumption that the human body is a bag of water. In reality, the human body is nowhere near that simple. 120-80 mm-Hg where Hg is a density of 13.59 g/cc and blood is about 1.06, then every ...


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Let's suppose first that the astronauts are tethered by a rope to their suits, and the rope is tight, in that initially there is no slack. Put another way, the length of the rope is the distance between them. They both grab the rope and pull with their respective forces and keep holding those points on the rope. Now astronaut A pulls with 20lbs of force ...


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You can calculate the air drag along your arms using the formula for air drag, $F = \frac{1}{2} \rho u^2 c_D A$, takin $u$ the speed at which they'll effectively move relative to air, and $A$ their projected area on the plane perpendicular to motion. $\rho$ is the density of air, that's why it's so much less efficient than in water. $c_D$ will be close to ...


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In your example you have the total load applied as $$ F = \int \limits_0^\ell w\, {\rm d} x = w \ell $$ where $w$ is the linear force density (in Newtons per meter). To get the torque you just include the position of the force $$ \tau = \int \limits_0^\ell x\, w\,{\rm d} x = \frac{1}{2} w \ell^2 $$ The rule for distributed loading is to find the total ...


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In the case of the gate, F=200N is the distributed force, which is applied from x=a=0m to x=b=2m from the axis. The force applied on an element of length dx is (F/L)dx Newtons where L=b-a. The torque on the gate due to the force on this element is dT = (F/L)xdx. We integrate this from x=a=0m to x=b=2m to get the total torque : $T = [\frac{F}{L}\frac12x^2] ...


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A force applied uniformly to a uniformly dense object (or with a uniform-force-per-unit-mass even if the object is not uniformly dense) is equivalent to the same total force applied at the center of mass of the object. The equivalence means the same total force and the same total torque. In particular, if the center of mass is chosen as the origin of the ...


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The physics in the 1st approximation are worked out here: https://en.wikipedia.org/wiki/String_vibration with the result that frequency depends on the length, tension, and mass density as: $ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $ The 1st factor is why bass strings are long and treble are not (on a piano). The lower 3 strings on your guitar have a ...


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Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ...


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Force The tendency to make something accelerate. Newton's 2nd law: $$\sum \vec F=m\vec a$$ Physical strength Can be different things. E.g. hardness $H$: a materials resistance against "bumbs" and indentations in the surface, yield stress $\sigma_y$: the stress a material can withstand before starting to deform ultimate tensile stress ...


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Well the difficulty is when the ball in contact with wall. Few things you can consider. elastic impact. In this case, you can treat the ball as an elastic spring. When it impacts the wall with a speed, the kinetic energy will be converted to the spring's potential energy. And then, after it reaches the maximum deformation, the ball will spring back into ...


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Force, energy, and work are all just different ways for us to describe motion. Energy (more specifically kinetic energy) is a measure of how fast an object's motion is. If an object has non-zero velocity, it has energy by definition. Force is a measure of how fast an object's rate of motion is changing. If an object's velocity changes, it's experiencing a ...


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Visual Solutions (Now Altair) makes VisSim Software. Here is a demo block diagram that they have used to simulate a bouncing ball: The $1/s$ blocks are integrator blocks from the VisSim library. The plot of the bouncing ball with these parameters is shown below. If you are interested in running this, the demo comes with the install and you can download ...


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We have to talk about Physical Strength last, for two reasons: (1) we have to clearly define the other physics terms first, and (2) Physical Strength gets us into biophysics so we'll have to talk about how muscles generate force, etc. "Force" is a push or a pull or a twist (if a twist, then it is also called "torque"). Force is what is required to ...


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From the equations of static equilibrium, N - mg = 0 in the vertical direction. The tipping is caused by the imbalance in the moments applied around point P. As long as mg * (L / 2) > F * H, then the book case won't tip over. H and L are the dimensions of the book case, so you can't do much about changing them. You can either add more books, which ...


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You almost solved it. Actually when you increase your force steadily the normal reaction shifts it point of action from directly under the centre of mass of the object to the point P. In other words in the limiting condition the normal force acts at point P. And hence it's torque is 0 about P. Well done.


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To determine the coefficient of friction, (in an ideal situation, of course) push a known mass of the object with a given force F. Also the frictional force has a numerical value of (approximately) coeff. times normal force. So you can calculate the coeff. this way. There are momentary 'bonds' formed between the surfaces. These bonds arise due to Van der ...


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Half of your questions are concerning Newton's law: Why is it non-uniform? Because the object would have different densities in different parts, so weight would be greater in some parts? Yes. Think of a car. It is in contact with the ground in four places and pushes down causing four normal forces. If the car is heavily loaded with bagage in the ...



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