New answers tagged

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The first term derives the torque for the curved surfaces and the second term derives the torque for the top and bottom of the cylindrical surface. I don't really know why but apparently the cross sectional area at the top keep changing as r->R where R=Dc/2. so dA=pi(r+dr)^2-pi.r^2 dA=2pi.r.dr+pi.dr^2 I also don't understand why but apparently dr^2 is so ...


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From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...


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As skier slides down, the normal force between skies and dome reduces. It is maximum at top but reduces further as he slides down. It is the normal force which is keeping skier adhered to dome. As the speed of skier increases, there is an increase in tangential speed which gives an increased centrifugal force. Centripetal force tends to skier adhered ...


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In accordance with the homework policy, here are some things to think about in order to solve this. There is a force holding the skier attached to the snow globe: that is gravity. As the angle gets steeper, the component of gravity pointing towards the center of the sphere gets smaller. This force needs to be strong enough to keep the skier moving in a ...


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Short Answer - No Refer to the figure below: If you draw free body diagram of the particle at a point beyond C you will notice that there is no force acting towards the surface of sphere. Hence it won't be able to complete the circle beyond point C whatsoever.


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Under normal conditions and size of the sphere the answer is no because the gravitational pull of the earth would attract the particle towards its centre. If the size of the sphere is bigger than the size of earth then the gravitational pull provided by the sphere will cancel out the g effect of the earth and the net gravitational pull will provide the ...


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If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.


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For First Case: the rotation produced by the torque at the centre of wheel will rotate the wheel in the clockwise direction, but here friction is present, as friction opposes the motion of particle that's why it acts in anticlockwise direction and helps the body to move. For Second Case: here $mg\sin\theta$ will act along the line of centre of mass, which ...


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UPDATE IN RESPONSE TO YOUR COMMENT I apologise : as you suggest, there might be a lift force on the sphere if there is a shear flow in the fluid (see Discussion in 1st Link). However, this force is likely to be much smaller than the drag on the particle. ...


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It is by simply erosion. The higher the water velocity impacts the surface, the faster the erosion or, if by drips, the longer time span it will require. High pressure water jets will erode rock very quickly while drips of water can take eons. But the process is the same. By erosion.


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Assuming that the rings share a central axis A: You'll have a double integral, with the outer integral being a sum of the component of force parallel to the axis A on each tiny chunk dQ of the first ring. The inner integral finds that force component on dQ by adding up the component parallel to A of the force between dQ and each tiny chunk dq of the second ...


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I agree that friction in the drive mechanism reduces thrust, rather than opposing the motion of the car. However, this is not the case for wheels which are not in the drivetrain - ie where there is front/rear wheel drive instead of 4-wheel drive. Friction in non-drivetrain wheel mechanisms are then sources of resistance to motion. If the car has rear-wheel ...


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The frictional forces try and reduce the relative motion between the water and the spheres. If the water is travelling faster than the spheres then the water exerts a frictional force on the spheres to try to make the spheres move faster and the spheres exert a frictional force on the water to try and make the water move slower.


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If the parts are perfectly solid and they are at equilibrium, then the pressure is constant (ideally). $$q(x,y) = \frac{\int p(x,y)\,{\rm d}A}{\int \,{\rm d}A} =\frac{\text{applied force}}{\text{area}}$$ But if either the floor or the body are ever so slightly elastic then the pressure distribution is given by a non-hertzian contact which concentrates all ...


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Drag force opposes the motion of a body relative to the surrounding fluid. In this case the surrounding fluid moves to the right and relative to that the solids move to the left. The drag force is opposing the motion to the left, hence it is towards the right. The solids are being swept away by the fluid.


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Motion is a very diffuse concept :) you have to add a frame of reference to make it meaningfull. In the frame of reference of the surrounding water the force definitely tries to stop the particle. So if you have a stone rolled along the ground by a swift stream, the force goes in the direction of motion (in the usual, external, frame of reference), since ...


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A free body diagram will show a car in motion has air drag force, gravity force and friction force on it. The net force keeps the car accelerated, decelerated, or moving at constant speed. Friction force is due to relative motion between wheel and ground. Engine output spins the wheels (torque from power train system balances the torque produced from the ...


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Because hydrostatic pressure depends on depth, the problem requires integration. The pressure as a function of $y$ according to Pascal's Law is: $$p(y)=p_0+\rho g y\sin \theta$$ Where $p_0$ is the atmospheric pressure and $\theta=45\:\mathrm{degrees}$. $y\sin \theta$ is the depth. On an infinitesimal piece of door of length $dy$, at position $y$ and ...


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what's the backward force to balance $F_{fs}$ so as to keep the car moving uniformly? Is that the Force produced by engines? The forwards force comes from the torque produced by the engine of the car and is transferred into the ground via static friction. The retarding force that keeps the car moving at the same speed is mostly air resistance (as well ...


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There's a physical reason why the block slides down along the slope of the inclined plane. If you perform this experiment, and if frictional force is small enough, you will observe that the block slides down along the slope of the plane. It will never move in a direction perpendicular to the slope. For it to never move in the perpendicular direction, we ...


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The normal force is not playing a role in this case because the force is perpendicular to the moving direction of the box. By "not playing a role" I mean, during this motion, $\ F_N$ is only be used to balance out $\ mgcos\theta $ so that the object won't be able to move "into the inclined plane". To find the force causing the block to accelerate down, we ...


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Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration $g\sin\theta$ along the plane. This is why the block accelerate in this direction. Notice that along ...


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Even though capillary forces may be the dominant force in this situation, hydrostatic (gravitational body) forces are still there. And that results in the equilibrium seen in the left most figure as you start this experiment. So as you bend the tube over, the head (force due to height of the fluid) is reduced and that would allow the surface tension of the ...


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If there was no air resistance, the object would still slow down, stop momentarily, then return to the ground - because it is pulled downwards (towards the centre of the Earth) by the force of gravity (ie the object's weight). More than one force can act on an object at the same time. Here, both gravity and air resistance act on the object. Gravity always ...


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Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$ \frac12 mv^2=mgh $$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...


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The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have ...


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I disagree with both other answers. It's not enough that the sum of forces is zero This is not a vocabulary problem The work-energy theorem applies also to systems of interacting bodies, where the total work of all forces (internal and external) equals the variation of the sum of the kinetic energies of all bodies. Now to answer the question. ...


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If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...


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In a rigid body, the particles remain at their positions irrespective of the body's motion. So Newton's third law is applicable here. Here the two particles exert equal and opposite forces on each other (which you call the internal forces). So the resultant force acting along the line joining the two particles is zero. Since the net force is zero, there will ...


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Your confusion might be coming from not clearly understanding that you need to define a system, and then all of your quantities are referenced to the system you have defined. If your system is A and B, then the force that A applies to B is by definition an internal force, and no work is done by that force on the system. But you can choose your system ...


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As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the ...


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1)For net wt 100kg to be lifted in air ,lift has to be greater than equal to 100kg. 2)To produce minimum of 100kg lift ,it depends upon shape,size,weight,angle of attack of wing and lastly speed /velocity of wing/plane.Further velocity depends upon thrust ,total mass and drag(shape of flying machine). In simple words only Forward thrust,total weight and ...


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Okay, so I figured it out myself. Here's what I think: Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular ...


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At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...


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I can do it by considering $m_1$ and $m_2$ to be a system, which would give me $a=F/(m_1+m_2)$. How can I use a free-body diagram instead to calculate the acceleration? But this is by the use of a free-body diagram. Otherwise, how would you know that it is the force $F$ you should include? Because, your acceleration expression comes from Newton's 2nd ...


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The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ...


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Concerning your wording "force is transmitted (and maybe decreases because of loss of energy)" - no, no, the decrease of force is not easily connected to the loss of energy. Force can be decreased because there is friction, but this does not imply a loss of energy (not if nothing moves). And also energy can be lost (plastic deformation of the rope) without a ...


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You can use the same explanation : "The first molecule pushes its nearest neighbours (which also push back), the nearest neighbours push their nearest neighbours, and so on until the force is transmitted throughout the fluid." The total force transmitted to or by a surface is in proportion to the number of molecules pushing, which is proportional to the ...


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Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container. To explain this, the hydraulic jack is a better example. Car jack works on mechanical forces. A mechanical jack employs a screw thread for lifting heavy equipment. But hydraulic jacks use force ...


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The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...


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You will need to use two free body diagrams, one for each mass, one FBD will include the force of A on B and the other one the force of B on A. Both mass have the same acceleration (do they), so you end up with a little system of three equations and three unknowns.


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Although the average force applied during a collision might be small enough that an object can take it, the peak force applied can be much higher. In physics this is called impulse. Calculating the impulse for real world collisions (like a car crash) is very complicated. This is because cars have many structural members and the materials are not uniform. ...


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To understand this, use the definition of force $\frac{d{\bf p}}{dt} = {\bf F}$, namely the force is equal to the rate of change of momentum. Something like a collision can be very complicated to model, but the average force is approximately given by ${\bf F}_{average} = \frac{\text{change in momentum}}{\text{time taken}}$. Typically, in a collision, the ...


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On page 68 the paragraph headed Calculating displacement given a time and acceleration includes the text: Assume that you’re on your traditional weekend physics data-gathering expedition. Walking around with your clipboard and white lab coat, you happen upon a football game. Very interesting, you think. In a certain situation, you observe that the ...


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There isn't a "centripetal force" vector. As the car goes around the banked curve, the normal force on the car increases relative to what it would be on an un-banked straight road. The vertical component of the normal force supports the weight of the car, and the horizontal component of the normal force provides the centripetal force necessary to cause the ...


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I'll give you a couple of ways to think about this. First, geometrically, the circle you are thinking about drawing should contain the entire circular path of the car. If we're assuming that the car is remaining at a constant "elevation" on the banked surface, then the center of that circle has to be at the same elevation: otherwise you'd be drawing a cone ...


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The whole point in components is that when you add them, they must must give the original vector. The two components you've drawn don't. Their sum is not the original gravity vector. Remember that components are supposed to follow coordinate axes, so they are perpendicular to each other (in that way they take care of distinct directions so we can treat ...


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Gravity doesn't have a horizontal component. The component of gravity normal to the plane in your diagram can be said to have a horizontal component, sure (and a vertical component of magitude $mg\cos^{2}\theta$). But there is also a component of gravity parallel to the plane of magnitude $mg\sin{\theta}$. That component can be resolved into a vertical and ...


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Generally bones are under compressive stress and are optimized for this. The HUman body is an example of Tensegrity Tensegrity, tensional integrity or floating compression, is a structural principle based on the use of isolated components in compression inside a net of continuous tension, in such a way that the compressed members (usually bars or ...


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Newton's 2nd Law of Motion gives the impressed force as $F=dp/dt$, so a physical theory for a massless particle exerting a force requires that the particle have momentum, $p$. First we will discuss mass, momentum, the force law, and Special Relativity. In Newtonian physics mass is identified in two ways: by it's inertia, or as the quantity of matter. The ...



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