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This tag is for the classical concept of forces, i.e. the quantities causing an acceleration of a body. It expands to the strong/electroweak force only insofar as they act comparable to ‘classical’ forces. Use [tag:particle-physics] for decay channels due to forces and [tag:newtonian-mechanics] or one of the other subtopics of [tag:classical-mechanics] for the dynamics of classical systems.

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"Tension" is a simple special case of the state of stress in any solid object. But you usually start learning about mechanics using simple situations where common-sense ideas like "tension" are all y …
answered Jul 14 by alephzero
43
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The position you are taking seems to depend on hindsight. Put yourself in the position of Newton being the first person to state these laws. The first law was a flat-out statement that Aristotle was …
answered Aug 4 '16 by alephzero
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"The internal strength should push back upwards" is wrong. If you cut the beam at any point, to hold it in position you need equal and opposite forces on the two sides of the cut, one upwards, the … other downwards. Suppose your beam is length $l$. Measure all the forces as positive downwards. The total downward load is $ql$ so the reaction force acting on the beam is $-ql/2$ at each end. Note …
answered Apr 25 by alephzero
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A car is pretty stable on only three wheels, so long as you are driving fairly fast in a straight line. However if you lose a front wheel on a front wheel drive car, you will lose the power to the ot …
answered Nov 1 '17 by alephzero
3
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I would argue that the book is wrong. There is a boat-shaped "hole" in the water. If the boat moves, the "hole" also moves. Ignoring all the local details of how the water moves, the net result is …
answered Apr 25 '17 by alephzero
0
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You can always do one integration of an equation of the form $$\frac{d^2}{dt^2}x(t) = F(x(t)).$$ Write $dx/dt = v$, so $d^2x/dt^2 = dv/dt$, and multiply both sides of the equation by $2v$. $$\begin{ …
answered Aug 13 '16 by alephzero
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force" then the idea of "some kind of force in the rope" doesn't really make sense. You can only talk about forces acting on the ends of the rope. If you imagine that you cut the rope somewhere along … its length, you then have two more "ends" which have equal and opposite forces acting on them. The "correct" way to deal with all this to replace the idea of "tension in the rope" with the notion of …
answered Nov 24 '16 by alephzero
0
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Your picture is fine so far as it goes, but... If the helicopter rotor has 2 or 4 blades, now draw the picture for the blade that is opposite to the first one. The slope of that blade is in a differe …
answered Apr 27 '17 by alephzero
1
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You are missing at least two things here. First, the total braking force is not directly applied by the driver's foot. All modern road cars, and F1 cars, have a servo assisted braking system so the d …
answered Jun 27 by alephzero
0
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particle of the body. Those forces are balanced by an internal stress distribution in the body, which has nothing to do with whether the body is rigid or flexible. For example if the object was a …
answered Sep 10 '18 by alephzero
1
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I only watched the video to the point (about 15 min after the start) where Susskind he explained that according to Aristotle's "law of motion", the mass in a spring-and-mass system would move towards …
answered Jul 25 '16 by alephzero
0
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Unless Newton's laws of motion are wrong, the answer is clearly "no". I'm assuming the OP is thinking about Newtonian mechanics, not General Relativity - i.e. I'm considering gravity to be a "force" …
answered Oct 4 '16 by alephzero
1
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eventually stops because the damping forces take away energy from the system as it oscillates. If you haven't studied damped oscillations yet, you can see your mistake by supposing that you lower …
answered Apr 23 by alephzero
1
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, you need some sort of motor and brake which would apply the correct time-dependent moment at P. However since the question only asks for the forces at P, I suppose a pedantic examiner could argue …
answered Jan 12 '17 by alephzero
1
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It can't be applied to that problem, because the beam from A to the load is one continuous part of the structure. If it was a pin jointed truss, there would be two separate members, one from A to C …
answered Oct 7 '18 by alephzero

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