# Tag Info

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To offer insight on the comment about potential energy being defined for a system, rather than just the body itself. This stems from the definition of potential energy. For a body in a gravitational field, the source of the field had to do work in order to bring this body to its current position. Generally, potential energy is defined to be zero at an ...

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Check out the solution in the image. It will be dependent on the inclination of the plane. Here, a is the horizontal acceleration of the inclined plane.

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Nothing is probabilistic in physics as far as I know. I think it starts sliding when the inertial force of m2 exceeds the force of friction k.

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The block would not necessarily start sliding at the maximum acceleration of $m_1$, but rather as soon as the acceleration exceeds $\mu\,g$, because at that point, the force required to accelerate it would exceed the maximum frictional force. If you want to simply check if the block will start sliding, then you can compare the maximum acceleration to ...

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The stiffness $k$ of a coil spring can be expresses as: $$k=\frac{E\,d^4}{16\,(1+\nu)\,(D-d)^3\,n}$$ Where $E$ is the modulus of elasticity of the material, $d$ is the diameter of the wire used in the coil, $\nu$ is the poisons ratio of the material, $D$ is the outer diameter of the coil, and $n$ is the number of wraps in the coil. Now $\frac{D-d}2$ is ...

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Your equations for tension are all correct. You have three equations and three unknowns $T$, $a$, and $a_M$ so you can solve for that system. If you'd like to know the normal force between $M$ and $m_2$ you can calculate that as: $$N=m_2\,a_M$$

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You probably experienced that a thick sheet or bar of material is more rigid than a thin one : cross section count. You will find the theory and complete formula getting the spring "constant" here (it is pretty involved).

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When an object, moving at a constant velocity, hits something, it will either stop, decelerate (or accelerate in some cases), or bounce back. During this collision time, there is a change in velocity (acceleration/deceleration) of the moving object, hence force is exerted on the wall (or in this case, on you), due to change in velocity. Hence we can use ...

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The relation is empirically observed across a wide range of materials and conditions. But it is not exact. Attempting to derive a similar equation just from lower-level physics such as solid surface molecular interactions would be horribly complex.

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Friction is the force parallel to the surface of contact while the normal force is perpendicular. On the micro scale, both are related to the electromagnetic force. This is why $F_{fric} = \mu_k F_N$ So when the normal force is increased by adding the crate on the box, so does the friction. Note that the normal force is always perpendicular to the ...

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In the first case when the box is stationary your statement is correct and you asked no question about that case. In the second case, the box is moving and only the kinetic or dynamic friction is relevant. Assuming the crate you add on top of the box weighs the same as the box, the normal force doubles, and therefore the dynamic friction force doubles. This ...

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Short answer: The theorem of conservation of total momentum (for the entire system asteoroid + the two bottles) can be applied successfully with no contradiction in both reference frames. In both cases, the result is that the velocity of the asteoroid does not change and there is no net force acting on the asteroid. To compare the same situation in the ...

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The technical answer is that the force "comes" from your buddy and you, who spend (chemical) energy to throw the bottles. If we consider an idealised scenario, though, where a system of three glued points breaks suddenly down, then I will argue that there is actually no change in the momentum => no need of force: 1) Before the break-down the system's ...

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Here are some simple cases applied from How to get distance when acceleration is not constant? Constant Power $P$, Constant Friction $F$ \begin{align} a(v) & = \frac{P}{m v} - \frac{F}{m} = \frac{P}{m} \left( \frac{1}{v} - \frac{1}{v_{final}}\right) \\ v_{final} &= \frac{P}{F} \\ t = \int \limits_{v_1}^v \frac{1}{a(v)}\,{\rm d}v &= ... 6 The mass of the asteroid changed, but the mass of the asteroid + bottles did not. Your outside observer would need to include the bottles in calculating total momentum; otherwise the system is not closed. This is the same principle behind operating rockets in vacuum. We can change the momentum of a rocket by firing out mass (exhaust) in the opposite ... 9 Your analysis in the frame of the asteroid is correct, and the asteroid does not change its velocity. Unfortunately your analysis in another frame is too simplistic and is incorrect. Force are applied to objects. When the momentum of an object changes, a force is involved. Here the question is: what's the object? If the the object is the combination of ... 3 The caveat here is that the second law is stated that net force is equal to the change in momentum. Assuming you and your buddy are not too wasted and are able to synchronize throwing the bottles off with the exact same force, exactly in opposite directions and through the center of mass, the net force is zero, and therefore there is no change in momentum ... 3 Momentum is really always conserved. Truly. Throwing something upward (accelerating it with your arm) causes your feet to push harder on the ground. The increased down-force causes the ground under you, and ultimately the entire Earth, to shift direction downward. Fortunately, the rock and the planet attract each other gravitationally, causing the rock to ... 1 While not quite an inherently "physical interpretation," the technique of inertial imaging allows one to use higher order mass moments for characterization and identification of biological molecules and molecular complexes. (See: Inertial imaging with nanomechanical systems) The basic idea here is that sticking ("adsorbing") a molecule ("analyte") onto a ... 2 If we apply a force F to a mass m and a friction force (drag) F_d also acts on it the force diagram becomes: With a the acceleration the object experiences, the equation of motion becomes: F=ma+F_d. As the mass moves towards the right, say for an infinitesimal distance dx, an infinitesimal amount of work dW is performed on m by F: ... 1 This may not be the ultimate answer as I don't know the quantitative relations between variables. But I can say the following: The ultimate velocity is determined by the power (P) of the car (or other objects, let's use car for example) and the friction (\mathbf{f}). Now thatP=\mathbf{v}\cdot \mathbf{f}$$It means that all the power of the car is ... 0 You're very close. Instead of changing reference frame so that an end is stationary. Change reference frame so that the center is stationary. The rod will then spin about the center. Your equation for \dot \theta is correct (though you're using different coordinate systems for your two velocities so the signs are strange)$$\dot \theta=\frac{\Delta ...

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Following the general approach given in this link, we can write down the equations of motion and solve for the normal modes. I am using $x_1$ and $x_2$ as the displacement from equilibrium since it just removes a few $m\cdot g$, $-L_1$ and $-L_2$ terms but otherwise doesn't change the result in any fundamental way. You can then adapt this approach to solve ...

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The normal reaction between the spider and wall will decrease as the speed increases.so the friction acting will also decrease.at a particular speed friction will be equal to component of weight along the wall.at that moment spider will start going down.the speed will be max for this situation

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my answer is also zero. because resultant become zero. let a force f1 and f2 pull the rope having equal magnitude. then, magnitude of f1= magnitude of f2 let tension on rope be T1 and T2. then, =(f1+T2)- (T1+f2) (since f1=f2) =f1+T2-T1-f1 =T2-T1 T1=T2 THEN AT THE END OF THE ROPE ARE EQUAL THEN TENSION AT THE MID OF THE ROPE WILL BE ...

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Straight Newton's Laws plus kinematics solution An Atwood's machine develops an acceleration of \begin{align*} a &= \frac{m_2 - m_1}{m_2 + m_1} g \,, \end{align*} in the direction of the heavy mass (here $m_2$). Both objects have this acceleration though in opposite directions, and (assuming the rising one doesn't hit the pulley) it persists over ...

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In addition to james large's answer, you can also see the center of gravity as the intersection of all the vertical lines when you hang your solid by any point. (link from http://www.splung.com)

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The center of gravity of a rigid body is the centroid of its mass. https://en.wikipedia.org/wiki/Centroid Basically, it's the weighted average position of all of the mass in the body. That is to say, if you scale the position vector of each particle by the mass of the particle, and then compute the average all of the resulting vectors, you'll get the ...

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From your description, I think the quantity you are talking about is similar to the concept of inertia where the mass is abstracted into a matrix through its distributions in space. Mathematically, you can understand the matrix properties as following: For a positive semi-definite matrix, the Eigen values are always real and non-negative which makes sure ...

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By Newton's 3rd law, the force on each component is equal and opposite which means the acceleration of the center of mass has not changed (because there is no net external force other than gravity) and also, by conservation of momentum the CM's velocity has not changed due to the splitting. You can conserve momentum for the tiny interval of time starting ...

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First of all, the system has not been subjected to any external force other than gravity. As a result, the ball breaking into 2 parts due to internal stresses will have its center of mass unmoved. The entire system, in this case, the ball has not experienced any net external force other than gravity. So there is no net external force acting on the C.O.M. of ...

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Newton's Third Law tells us that the momentum imparted to fragment #1 is equal and opposite to the momentum imparted to #2. So if I take as my system all the atoms of the original object, we see that the momentum of the system hasn't changed at all.

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So, there are two distinct things that we're talking about here. It looks like you don't mind some advanced notation so I'll try to use that to illustrate the mathematical side of the physics I'm talking about. Rigidity and the axis of rotation One of the things that we're talking about is that the object is rigid, meaning that it's composed of a bunch of ...

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Torque? Why do you think you need to think about torque? Is the center of mass over the base of support? It is if $0 < M D_1/2 + m D_2 < D_1$. I.e., if $\frac{-M D_1}{2 D_2}< m < \frac{D_1}{D_2} (1-M/2)$.

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You are right in saying that $I$ allows one to relate angular velocity and angular momentum in a linear way. It is just not as simple as the momentum and velocity case. An intuition for why things get complicated is that $L = r \times p$ involves a cross product which makes it very sensitive to the choice of a specific set of orthonormal bases(with fixed ...

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Here is my derivation of this result. I hope you find it helpful: Say we have n different forces $F_1, F_2, F_3... F_n$, applied at n different points. Now we pick two centers $P$ and $Q$, and express the radial vectors (1) from point $P$ to each of the n points (where forces are applied) as $r_1, r_2, ... r_n$ (2) from point $Q$ to each of the n points ...

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I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever ...

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If there is no torque, then $$\sum \tau = \mathbf{r_1}\times\mathbf{F_M}+\mathbf{r_2}\times\mathbf{F_m}=0$$ Therefore, $$\mathbf{r_1}\times M\mathbf{g}+\mathbf{r_2}\times m\mathbf{g}=0\tag{1}$$ where $\mathbf{r_i}$ denotes the position of the center of mass of the combined system relative to the force applied. If we give the box dimensions $h$ and $l$, the ...

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Energy is conserved. So loss of PE = gain of KE. PE lost $=mgh$, KE gained $=mv^2/2$ where $v$ is the speed at the foot of the second ramp, as the body started from rest. So solve $$mv^2/2=mgh$$ for $v$ where $h=15m$ and $g=9.81m/s^2$.

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But, what if it is in a friction planet and in order to overcome the force of friction is 4N, then our equations contradict. W=(4N)(10m)=40J. Why do I get answers that contradict each other? If it's on a surface with friction, then the final and start velocity will no longer be the same. You calculated $40\,\mathrm{J}$, so the kinetic energy must ...

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My equation is $$\frac{x^2}{a^2} + \frac{x^2}{b^2} = 0$$ That's not the equation you want for a satellite. That equation describes an ellipse with its center at the origin. You want an ellipse with the origin at one of the foci: $$r = \frac{a(1-e^2)}{1+e\cos\theta}$$ where $r$ is the distance from the origin to a point on the ellipse. $a$ is the ...

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You get different answers because these are different situations. let's start with the first case. In the frictionless planet, speed is constant if the sum of all forces acting over the body are 0. No difference in kinetic energy, no work done. In the second case, the work should be negative (the friction force opposes the displacement, work is the dot ...

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Time of flight is determined only by the vertical component of velocity - it is the time interval between when the projectile was released ($y=y_0$) and when it reaches the ground ($y_t=0$). As the collision is with a vertical wall it acts in the horizontal direction (assuming no friction during the short duration of the collision) and so has no effect on ...

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If the collision is perfectly elastic then there is no change in the kinetic energy during the collision. If there were a change in the kinetic energy during the collision then the flight times would be different. This is analogous to a ball being thrown and encountering no collision.

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The 2nd solution you wrote down appears to be the correct solution. Offhand, I see two problems in the first solution. First, I think that a problem with the first attempted solution is that you made a subtle mistake in assuming that F=ma means that $F=mr_1α$. That seems like a plausible step at first but if you examine this step more closely you'll realize ...

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$m_1g - T = ma$ $3T -m_2g = ma$ Solving these, and taking $m_1=m_2$, We get, $a=g/2$ Which is same as Daniel's answer.

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The mass m2 is hanging from what is known as a Luff tackle, which has a 3:1 ratio. This means that, from the string's viewpoint, the system acts exactly like a mass of size $m2\over 3$; the inertia is one third of m2's, and the force generated by gravity is equal to $gm2\over 3$. Since m1 and m2 are equal, this means that the system is equivalent to a ...

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Surely the gravitational potential energy lost by the mass should equal the elastic potential energy gained by the spring? On the contrary, if something fell down doesn't that imply it gained some downwards velocity and hence the change in kinetic energy should be related to the net work done on the object? Here is a warning. Let's say you measured k ...

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The batter in both cases is connected to the Earth. Momentum is conserved but you have to add the amount given to the Earth--which is impossible to detect but it is there. If you drop a ball and it lands on the Earth and stops, momentum has not disappeared, it is transferred to the Earth. The answer just above notes that the bat will give some spin to the ...

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When you hit the falling ball, friction between bat and ball will momentarily stop the side of the ball that is hit from moving down. However, since this force of friction is not applied at the center of mass of the ball, this will not result in a complete arrest of the vertical motion: instead, the ball will acquire some spin. However, it will not lose all ...

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