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0

according to what i have got about the law (maybe wrong or right): the breaking of the glass is the reaction of the glass to the excess force you have exerted on it. and you fill as much as the glass can react when it is not broken and is in touch with your hand.


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The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...


1

First find the force of gravity on the block, in coordinates relative to the inclined plane. "Down" relative to the plane, $F=ma$ yields $$F=10kg*10\frac{m}{s^2}*cos(\pi/6)=70.7N$$ "Left" relative to the plane, we have $$F=10kg*10\frac{m}{s^2}*sin(\pi/6)=50N$$ Since the tension on the string means the force of gravity on the hanging block is the same as the ...


2

Considering that the $10kg$ block does not accelerate from its initial position (ie the mass does not move), you need not have kinetic friction coefficient $\mu_k$ to solve the problem. Just use the static friction coefficient alone. Free body diagram: where $R=10 g \cos (30^\circ)$ and $\mu=\mu_s$. For equilibrium (zero acceleration), $$Mg-T=0$$ $$ ...


0

To first order, air resistance falls into two regimes at subsonic speeds. At very low speeds it can be modeled with a linear response to velocity, while at any higher speed you generally observe a quadratic response. The ratio of Reynold's number approximates the relative contribution for both components and has derivable values depending on the geometric ...


0

Yes, there is. Usually air resistance or other kinds of resistant forces can be considered as $bv^2$ or $bv$ where $b$ is a constant that depends on many things. For example, pressure, density, and so on. These functions are just an approximation and derived experimentally. You know that friction is an actual complicated force! They are usually neglected for ...


0

First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...


2

How accurate do you need to be? The problem with these calculations is that massless, frictionless pulleys are usually out of stock at Acme Mail Order. High school physics will give you the tension in the rope, a different set of high school equations will tell you how much extra tension is needed for a certain acceleration. A rough metric is add 10% for ...


1

It depends on the coupler design, and wether the coupler is in tension or compression. If you have a buffer-and-chain coupler there is simply no way you will uncouple the train when it is running without using explosives. Maybe a link and pin coupler can be released if you jiggle the throttle while extracting the pin. If you have a Janey / SA3 / AAR / ...


1

The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...


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Put the arrows after one another. Then draw a new arrow to the point they reach. The net force vector is the new arrow. The direction its' angle. The maginute is its' lenght (just as the magnetude of the two original forces were the lenghts of each).


0

I'll leave the problem for you to solve, but here's a hint. Remember that forces are vectors, and "net" means "sum of components". You can certainly use angles when summing vectors - just be careful with the signs so that you cancel what chould get canceled and add what should get added.


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UPDATE (Solution): The center of mass of $B_1$ has a displacement of $$\vec x_1 = \vec x_{1,1} + \vec x_{1,2}$$ One due to RWS and the other due to its contact with the other body. Differentiation (twice) leads to $$\vec a_1 = \vec a_{1,1} + \vec a_{1,2} \xrightarrow{algebraic} a_1 = a_{1,2} - a_{1,1}$$. But $$T = m a_1$$ and $$\sum \tau = I ...


0

The answer to the above question is the definition of work i.e. how we define it. When we apply force on a body, it is displaced with respect to its position. Work is said to be on the body. The definition of work says, "Work is the product of the component of force in the direction of displacement or vice versa". So the case I stated above has it's force ...


1

You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...


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While considering 3rd law, forces act on different bodies , and not on same bodies. So the body which is hit is under the influence of applied external force only. The force which the hit body applies back to the hitting object is acting on the hitting object, so no point of cancelling of forces as they are acting on different objects. I too used to think ...


0

The net force on raindrop plus wagon is zero. Consider a single rain drop. Let the momentum in the direction of travel of the combined wagon/raindrop system be p. Now p = p_wagon_before + p_raindrop_before where p_wagon_before & p_raindrop_before are the momentum of the wagon and the raindrop before the drop hits the wagon. We have then: ...


0

This is the opposite of the rocket problem. In a rocket, acceleration occurs becaue mass is thrown out the back end. F = d/dt(mv) = m.dv/dt +v.dm/dt. If F= 0, then m.(-dv/dt) = v.dm/dt <-- note the negative term with the acceration. In a "typical" problem mass does not tend to change significantly, but in the rocket this mass term is highly ...


0

But if the velocity of the wagon changes, the net force can't be zero, right? Only true if the mass is constant (it's not if a wagon is filling up with water). If mass and velocity both change, you can't say anything about the force. Record the experiment and play the video backwards. You will see a wagon moving backwards. The wagon is spraying water ...


0

Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum ...


0

Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected.


1

The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on ...


0

If you look up centripetal force i think you will find your answer. I wont post any links because i dont think i'm allowed to at the minute but a diagram will help you understand it better than i can explain it


0

The Anthropic answer to this question is that if gravity were a lot stronger, then the evolution of the universe would have proceeded in a different way, it would have collapsed just after the Big Bang. One can speculate that all possiblities really exists, but we can obviously only find ourselves in those universes with laws of physics that are compatible ...


2

The static electrical force between two particles of electric charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb’s law of electricity, which says that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force is repulsive for like ...


5

This question comes up again and again, in different guises. If you pull on a rope with a force of 50N, then the tension in the rope is 50N, not 100 N. It's no different whether the other end of the rope has a weight of 50N on it, or whether it's attached to an anchor point in the wall, or to another boy. Newton 3 talks about the fact that A exerts a force ...


0

Electric fields and electric currents are very different things a good comparison is like what moisture in the air is to a river if you think of it this way things will make more sense.


1

Force a alone will produce both downward motion (as if the force were applied at the center of mass) and rotational motion, because it causes torque. But the linear acceleration is the same, regardeless of where you apply the force. For traslation, both c and d balance a, so the disk will not translate in either case, but it will rotate with d because there ...


45

To deal with this type of problem, you must be careful to define exactly what system you are dealing with, and then not change that system part way through the problem. This definition allows you to be very clear about whether the "system" has any external forces acting, and thus whether the momentum of the system is constant or not. In this case, you seem ...


32

When the raindrops hit the wagon's surface, they aren't moving relative to the tracks. Friction is required to accelerate the raindrops to the wagon's speed. By Newton's third law, there must therefore be a reaction force on the wagon surface by the raindrops.


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The mass of the cart is changing! This is the variable-mass system, which says, $$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ where $v_{rel}$ is the relative velocity of the escaping/entering mass. In your case, there are no external forces so, $$ v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ So the change of velocity comes from the change in the mass.


0

what you need to do is replace the springs with equal and opposite forces on both ends and do a Free Body Diagram. The answer comes straight out of this.


0

Hard to answer this without just suggesting an answer. Consider the forces on the mass. There is gravity that is downwards, but I guess the mass is not moving so think about the forces on the mass - you could also think about the forces the point where the two springs joint together. Hope this is helpful.


1

If pulleys are not weightless and maybe rope is not weightless. Weight $Q_1=m_1 g$ accelerates whole setup. Tension $T_1$ is responsible for acclereating two pulleys, rope and mass $m_2$, $T_2$ accelerates one pulley less to the same acceleration, so that it is lower. By the same reason $T_3$ is lower than $T_2$.


1

An external force $F_{\rm ext}(t)$ appears as a source term $qF_{\rm ext}(t)$ in the Lagrangian. For example, if the equation of motion is, $$\tag{1} m\ddot{q}~=~-\frac{\partial V(q)}{\partial q} + F_{\rm ext}(t), $$ then the Lagrangian reads $$\tag{2} L(q,\dot{q},t)~=~\frac{m}{2}\dot{q}^2-V(q)+ qF_{\rm ext}(t).$$


0

In the case that the force is conservative I would model the force by adding an extra potential term $\psi$ to the Lagrangian such that: $$\vec{F} = -\nabla\psi$$ If the unforced Lagrangian was $$L_{\text{unforced}} = T - V$$ the forced version is now $$L_{\text{forced}} = T - (V + \psi)$$ As far as I am aware modelling non-conservative forces in the ...


2

Force was defined to be proportional to acceleration because that definition makes description of classical physics simple. For example on Earth we have a downward pointing constant force - gravity. If we define another quantity, one which is proportional to velocity, lets call it push, it would be quite useless in describing gravitational interactions. An ...


5

You mix the relations between the things. A force produces changes in the linear momentum. The acceleration $a = F/m$ produces changes in velocity $v = p/m$. So, your question should be either why don't we take the force proportional to the linear momentum, or why don't we take the acceleration proportional to the velocity. Now, the second thing is ...


3

Well, technically we do: $$ F_{net}=m\frac{dv}{dt} $$ The net force is proportional to the rate of change of velocity, which we call acceleration. Note also that it's not $F\propto \Delta v$ (force proportional to the changed velocity because the changed velocity occurs over a period of time, $\Delta t$, that is also important--consider the difference in ...


0

If you mean force proportional to velocity, that restricts the second law to only the specific case when the force is proportional to the velocity (the object will feel a drag or will accelerate exponentialy with time, depending on the sign of the proportionality constant). Such a law will not decribe the dynamics of any other objects.


6

Suppose a glass could sustain 100N force and that my muscles can exert up to 200N force: if I went all out, I couldn't punch the glass with a 200N force because the glass would break, which means it's not able to apply a 200N force on me. I apply F = 200N and the reaction is only f = 100N. The glass table-top in the picture can support a weight ...


1

Since the direction of friction is specified by the direction of motion, it must be the opposite direction, right? Wrong. You can go back to whoever told you that and yell at them. :-P Seriously though: the direction of friction actually has nothing to do with the direction of motion. One really obvious way to see this is that you can make the ...


1

The same question in the OP has been asked and replied in this question (already linked in one comment here). Probably, if you had made the example of the glass, your post would not have been misunderstood. I have just observed a simple fact there: if a glass can take, support only 100N if we exert a greater force, it will break and offer no support, the ...


2

The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show $T$ anywhere.


2

Your mistake is that the two tensions are different, because of the presence of a pulley with non-zero $I$. What you have missed is: 1) connect the two tensions to the torque: $(T_1-T_2)R=I\alpha$ and 2) link the accelerations $R\alpha=a_1=a_2$ NOTE (from comments): If the pulley had a zero $I$ (moment of inertia), then the two tensions would be be ...


0

Your reasoning is alright. "This, I understand. My problem is that I don't understand that if the angle θ is 90 degrees how can the work done by F⃗ on the object is zero. For example; say you have a particle and the direction of the displacement is directly to the right, and you also have a force vector acting on the particle that is straight up(like the ...


0

Escape for a moment from the definition "work is equal to the force times the distance ('times' meaning scalar product of course)” The meaning of "work" may be interpreted as a transfer of energy, or specifically, the transfer in or out of a system's kinetic energy [1]. Your Question: "If I lift some object from a ground, the force to be put in above ...


0

I agree with Wolphram jonny's answer that even if your fist has the same momentum in both cases, the force your fist exerts on the feather will be less than the force on the glass, and in each case the force your fist exerts is matched by an equal and opposite force from the object you're punching. But I thought I'd also add a simple argument for why this ...


-1

Let me admit that the lift goes down with an acceleration greater than g, and let me also assume that the lift has no ceiling. The bucket will remain in the air. From the point of view of the observer on the ground the bucket will fall gravitationally. So the water in the bucket, so the block. Now, with respect to the water, the block has no more weight. ...


2

It doesn't have anything to do with the torque on the lug nuts. The torque spec is there to make sure the nuts are tight enough not to loosen on their own, but not so tight as to damage the nut, stud or wheel. To a first approximation (given four wheels, five studs per wheel, and a weight of 3000lbs), each stud will have to support 150 lbs of vertical ...



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