My physics tacher told me that to calculate the weight that you feel from negative g forces you have to devide your mass by the negative g force, but then my friend told me that it did't make sense and now I'm confused... Please help!
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My physics teacher told me and some other students that when you experience negative g-forces, your weight equals: $$\frac{mass}{g_{force}}$$  So when the g-forces equal -4 g your mass will be: $$\frac{mass}{-4}$$  Now I don't think this is true, because this doesn't work when the g-forces are positive. And also to follow the pattern (example pattern):)  $45 \, \mathrm{kg} \times 3g = 1350\, \mathrm N$  $45 \, \mathrm{kg} \times 2g = 900\, \mathrm N$  $45 \, \mathrm{kg} \times 1g = 450 \, \mathrm N$  $45 \, \mathrm{kg} \times 0g = 0\, \mathrm N$  $45 \, \mathrm{kg} \times (-1g) = -450\, \mathrm N$   $45 \, \mathrm{kg}\ /\;(-1g) = -450 \, \mathrm N$  $45 \, \mathrm{kg} \times (-2g) = -900\, \mathrm N$   $45 \, \mathrm{kg}\ / \;(-2g) = -225\, \mathrm N$  Which one is true?
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I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!  We assume that in the $n$-body problem, the force between particle $i$ and $j$ can be derived from a potential i.e  $$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$  where $\vec{\nabla_i}$ is the gradient with respect to position vector $\vec{r_i}$   $V$ would usually only depend on $\Vert\vec{r_i}-\vec{r_j}\Vert$  My lecturer then defines the total energy of the $n$-body system as  $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$  However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read   $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$  For the second equation we also get:  $$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}}  + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0 $$  So this seems reasonable to me. Is this actually correct?  We also have an example of the two-body problem in the notes where we only use $V_{12}$ as the potential, not $V_{12}+V_{21}$
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Is there a difference between Weight and Force of Gravity or they just mean the same thing? My textbook does not clarifies this point.
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This is a problem that has bothered me for a couple of weeks now, and I can't seem to wrap my head around it and understand it.  Let's say we have a planet with a mass of m. We also have an object of relatively small mass (so small that its gravitational field would not affect the planet), and we know that at time 0 it is at position h.  If we know the acceleration due to gravity (g), we can calculate the position y  at time t:  $y = h - { g \; t^2 \over 2 } $  And we know that the acceleration due to gravity could be calculated:  $g = G { m \over y^2 }$  But since the acceleration depends on the position of the object relative to the planet, and the position of the object is dependent on the acceleration, this would obviously mean that the first formula won't work when we are working with displacement so large that it would change the acceleration significantly.  So, in essence, my question is, what formula can we use to calculate the position of an object, taking into consideration the change in acceleration due to gravity?
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This is a question on a worksheet that I'm giving to a student:  A man of weight $W$ steps into the lift. The lift then moves downwards with a constant speed. If $R$ represents the force acting on the man by the lift floor, which of the following statements is/are correct? (3) $R$ and $W$ are an action-reaction pair according to Newton's third law  Why isn't this statement (3) correct?
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Gravitational potential at a point in a gravitational field is defined as work done per unit mass in bringing a small test mass from infinity to the point.  I know that:   By integration, we find that gravitation potential is negative. The maximum value of gravitational potential which is at infinity distance is 0, therefore minimum value should be negative.   But, what I do not understand is force is in the same direction as displacement so why work done is not positive?
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I was solving a true or false question regarding total mechanical energy and the following was the problem.     It is possible for a moving object to have negative total mechanical energy.   This is my claim.  Total mechanical energy is defined as   $$M=K+U$$  where $K=\frac{1}{2}mv^2$ and $U=mgh$.  No matter which direction the object is moving $K$ must be positive due to the $v^2$. (I am assuming that mass is also never negative.)  Potential energy is also positive because $g=9.8$ and height is a distance, which is also non-negative.  And yet, the problems says that the answer is true, as in the total mechanical energy is allowed to be negative.  What am I not seeing here ?
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I was confused with this equation like why force is equal to $ma$ and not $mv$? Then I read the same question here on stack exchange, it's answers almost made sense but I still can't visualise the situation where a thing is moving at a constant velocity, would it then have no force? But if I think this scenario in my head, it would definitely have force when it hit something. So I wonder if I am mis-associating the terms in the equation?
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Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?
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In the diagram above why is the tension of the string attached to the pulley at "A"(the string attached to roof) equal to 2T?  Why is it not Mg+(M+m)g?(considering that the pulley is mass less) I have trouble understanding
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Using classical mechanics, the formula for gravitational attraction is   $$F = G\frac{m_1m_2}{r^2}.$$   This formula does not work for photons, and we need to use Einstein's theory of gravity to account for that, as the photon is massless. What if, instead of using the mass of the photon, we rather used the inertia of it?  The momentum of a photon is     $$p=\frac{h}{\lambda}.$$  Thus, using that, we can calculate its supposed 'mass' (inertia) as momentum is also equal to mass times velocity. So 'mass' of a photon is     $$m=\frac{p}{v}=\frac{h}{\lambda  c}.$$   If we use this mass as the mass of a photon, can we use Newton's equation for gravitational attraction?
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For a constant force, $P=Fv$. I understand the mathematical derivation of this, but this seems to me, intuitively, to be nonsense. I feel that my discomfort with this comes from a fundamental misunderstanding of force and Newton's Second Law, so I'm not really looking for any mathematical explanation. So, to begin:  How is it that a constant force does not add energy to a system at a fixed rate? Consider a rocket burning a fuel at a constant rate. The chemical potential energy should be converted to kinetic energy at a constant rate, that is, $(1/2)mv^2$ should be increase linearly. The magnitude of the velocity of the rocket would then increase at a less than linear rate, implying a nonconstant acceleration and therefore, a nonconstant force/thrust (F=ma).  If force is indeed a "push or a pull," shouldn't that constant rate of burning of fuel yield a constant "push or pull" as well? Clearly not, so I would have to think that, somehow, a given force applied to a certain object at rest would in some way be different than that a force of the same magnitude being applied to that same object in motion. In this sense, is force merely a mathematical construct? What does it tangibly mean, in physical terms? Would a given force acting upon me "feel" differently to me (in terms of tug) as I am moving at differing velocities?  Force being defined as a "push or pull," which is how it has been taught in my high school class, seems rather "handwavy," and maybe that's the issue. It's been troubling me for a couple of weeks and my teacher hasn't really been able to help, so thanks!
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I'm a high school student and I'm studying Newton's second Law. While my teacher is calculating a net force of an object he always treat all vectors as positive numbers. I think what he meant is the magnitude of the vector which is always positive, but the problem is that he didn't use the magnitude symbol, for example like $||Fg||$. He said the acceleration due to gravity , g, is positive, but shouldn't it be negative since it is pointing down. This also cause a problem: every time when I calculate the net force I can't add all forces together, instead I need to choose to use minus sign when the vector is pointing down. So in general which way is correct?
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When drag force ($bV$) equals to object's weight (mg) then upward and downward force becomes equal. As a result the object comes to rest. If this is true, how is a body moving with constant velocity?
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I have demonstrated that  only is measured based on the gravitational pull of where you live.  For example, the gravity on the surface of Mars is three times smaller than the gravity on the surface of Earth.  So my question is simple: If we take into account that weight is only measured based on gravitational pull on that , if there was no gravity how would we define mass?  That is, would mass weigh anything if no gravitational pull acted on it? Then what would be mass?
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As I was trying to work out the expression for the apparent "g"-in case of the rotation of Earth-I was considering the centripetal force (m$w^2$r) and the weight (mg) all the time, thinking these were the forces that were acting on the body.     But later I found out that it should have had used centrifugal force instead of centripetal force. But why? We should be considering the forces acting on the body right? And it's the centripetal force that is acting on the body (with direction towards the center of the small circle).
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Non-metal objects are attracted to the Earth due to gravity. So the weight of non-metal objects can be only dependent on their mass.   On the other hand metals can be attracted to the Earth's magnetic field, with the Earth acting as a giant magnet; causing metals to weight more than they normally should.   In other words the weight of metals and magnetic objects would be the sum of the magnetic force plus the gravitational force, is that correct?  If that effect is true, would it be more clear as we go closer to the magnetic poles, and metals would weigh different in different locations on Earth?