My question is : Suppose a block of mass $m$ is kept on an ideal floor. To lift it, why I need to apply an upward force which is of magnitude $mg$? Force $mg$ is acting on the block in downward direction which is the force of gravity. If I apply an upward force $mg$ it just makes net force acting on the block zero. Then why the it would move upward? And why with zero acceleration
57331
Consider there is a box of mass $m$ at rest on the floor. Most books give an example that we need to do a work of $mgh$ to lift the box $h$ upward.  If we analyze this work done, the external force acting on the box by us should be equal to the weight of the box. Therefore the net force is zero which in turn there is no acceleration. If there is no acceleration and the initial velocity of the box is also zero, how can the box move upward?
54254
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).
102620
suppose i jump from ground then work done by normal is zero as point of application  displacement is zero(work is doneby internal forces) but during the pushing part  COM has upwards displacement then shouldn't work be done if apply force on COM  ??
66525
I did a weird activity and now I am trying to figure out the physics behind it:  While sitting on a chair I kept my feet above the ground and tried to move the chair forward. I was able to. Initially, I thought that perhaps the center of mass won't be changing as I was pushed backwards. But I moved the chair for a while and saw that it had moved a considerable distance and I was still sitting on the chair. Hence the center of mass did change for sure. According to Newton's second law, there must be an external force acting on the system(me and the chair). But I am not able to figure out what force is it. I have a feeling that it is somewhere related to the fact that the system is not completely isolated but I am not sure how.     Note: I am not in contact with the ground or any other object close to me. I have lifted my feet and have my hands on my lap.
3124
According to the Third Newton's law of motion:        For every action there is an equal and opposite reaction.   So, I understand that if I hit a brick wall with $50\, \mathrm{lbs}$ of force, the brick wall also hits me with $50\, \mathrm{lbs}$ of force (usually painfully). In this instance, Newton's third law makes sense. What I'm confused about is, if I hit a patch of drywall with $50\, \mathrm{lbs}$ of force, it's probably going to break, and due to the lack of pain in my hand, I can tell it did not hit me back with $50\, \mathrm{lbs}$ of force.   How does Newton's third law apply to situations when one object or the other is destroyed? It certainly seems like at that point it is incapable of delivering the full force of my blow back to me. What happens with the energy?
59540
There are several conditions that must be met in order for a force to be . One of them is that the curl of that force must be equal to zero? What is the physical intuition behind this? If you can, please explain it to me via the magnetic force fields because i have read that time-varying magnetic fields are not conservative because they do not meet that condition.I do not understand this.
118480
The  and  and  forces require particles like  and . But in case of gravity there is no such particle found.  Every mass bearing object creates a gravitational field around it, and whenever another mass bearing object enters its field the gravitational force comes into operation.  If all other forces of nature have some particles associated with them why should gravity be an exception?  And if there is no such particle, what exactly is the  and how does it spread over an infinite distance and cause the gravitational force to operate?    Note: I am a high school student and have not studied .
4134
Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:   Force $F = mg$ acting towards the center of Earth. Normal reaction $N$ offered by the surface of Earth.   If no other forces are acting, we say $F=N$. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?
122995
A block of mass 90 kg sits on a slope with an angle of inclination 12° above the horizontal. The coefficients of friction between the block and the slope are µk = 0.03 and µs = 0.2.  a.  What is the maximum frictional force that will allow the block to stay in place?  b.  At what angle of inclination will the block start to slide down the slope?  My Question: I am not looking for these two questions to be answered, however, the question has caused me to doubt my understanding of static and kinetic friction re: incline planes.  When i calculated the  the angle at which the block would begin to slide, using the arc tangent of 0.2 i got an angle less than 12 degrees. However, the question is suggesting that the block is at rest with when the incline plane is at 12 degrees. What is my conceptual error in thinking?  My next question:  Wouldnt the minimum angle of the incline plane (tan inverse of 0.2) be needed in answering the first question? that is, having it placed in the formula for the static friction formula = 0.2 x 90kg x 9.81 x cos of (tan inverse of 0.2)? assuming that tan inverse of 0.2 was found already? Is there something wrong with the question or how am i thinking is conceptual flawed?  My Final Question: Is it that the maximum frictional force that allow the block to stay in place is based on whatever angle the incline plane is set at? So, for example, if the incline plane is lowered to an angle of 5 degrees, the max friction force will be adjusted, even though the block would slide at 11.31 degrees? Is it that if lowered to 5 degrees, that the max friction force to overcome would be different?  In other words: If a block is at rest on the incline plane  at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for  fs(max) = UsMgcos (minimum angle needed for block to slide)?
108749
When drawing force diagrams I am really struggling to understand why we put the weight of an object as a force acting on it, rather than as a force acting on whatever else there is around it.  For example, if we have a book on a table, and are drawing two separate force diagrams for the forces acting on each object, then surely the weight of the book is a force acting on the table, not the book?  Or if you model a diagram of a person standing, it feels wrong to label the weight as acting on the person, when surely it is acting on the floor? And the force acting on the person is the reaction of the floor to the weight?  It all just feels a bit off! Please can someone offer some help to correct my intuition?  (I should add that I kind of understand that the weight of an object does act on it, insofar as gravity does pull things down...)
10573
I have been looking for this for quite some time now. A simple pendulum behaves in SHM. Let's put that pendulum in an upward accelerating elevator. The component of the force that acts in SHM $(\text{mg}\sin\theta)$ still stays the same in my head.  However, websites and books tell me to use $m(g+a)\sin\theta$ where $a$ is the acceleration of the elevator.  I tried to look up Free Body Diagrams, but I can't find any for the case of accelerating frames. Can someone explicitly prove this without using the flimsy argument of "Think it's a noninertial frame with a new effective g"?
36282
I took this question from Quora because it seems interesting and this community would have some fun with it. I would assume that one would use the modulus formulas for force to counteract the gravity, but given that the question details allow for "a solid cube, although internal structure is fine (honeycomb, etc)," this might get more complicated.
73298
This is the question:      A rope brings a 50 kg object downward at an acceleration of 0.75. What is the tension force of the rope?    I thought that, since the tension and the weight are both "going" downward, the net force would be Weight + Tension = Mass x -Acceleration. The acceleration is negative because it is going downward. But if I do it that way I get the wrong answer. The answer is only correct if net force is Weight - Tension, and if acceleration is positive, neither of which makes sense to me. Can someone explain this?   It could be a silly arithmetic error that I haven't noticed.
18076
So I came across this picture on facebook   My logic says it should show 200N, because you are pulling it with 100N on each side. On the other hand (as one of the facebook comments also stated, also saying is he is wrong he wasted 20k$ on physics degree), if you replace the right side with a wall you get this situation:    In this case I guess everyone agrees that the dynamometer would show 100N force. But if you isolate the dynamometer and "cut" the ropes on either picture, you would get the same thing. Because the system needs to be balanced, there would be 100N on each side.    And I study engineering too but I am confused by this. Why do we get the same distribution of forces, but dynamometer shows a different force?
62225
On , it is written that      So, we see things falling with an acceleration which we call the acceleration of gravity,and thinking that we live in a straight line , uniformly moving or stationary inertial frame, we attribute that acceleration to a force, the force of gravity. Whereas in reality,objects falling towards the Earth are moving along geodesic paths with no acceleration and according to modified version of law of inertia, have no force acting on them. They fall simply because the curved space-time near the Earth ...   Now, why do the objects falling towards the Earth move along the geodesic paths with no acceleration? That means the objects don't have any force acting on them, but why? A body in a free-fall moves with acceleration $g$, so, why is it written like that? Why does the author use Law of inertia on freely falling body?? Law of Inertia can only be applied when no external force acts on the body. So, is the freely-falling body accelerates under force of gravity or moves uniformly while moving through geodesic paths as quoted by the author?
8237
My calculus text (Swokowski, Olnikc, Pence, 6th edition) gives the  formula for work as $W = Fd$ and then goes on to explain that if the force varies over the distance the formula becomes an integral.    As part of an example, it then shows that the work to lift a 500 lb beam 30 feet would be 500 * 30 = 15,000 ft-lb.  But don't we have to exert an upward force greater than the weight of the beam somewhere in our model in order to get the beam to move up?  Then the work would be greater than 15,000 ft-lbs according to the definition of work.  It seems that in setting up integrals or just using the formula  W = Fd examples always use the weight of the increment or object to be  lifted without taking into account the fact that more than that force  must be exerted at some point in order for the thing to move upwards. By the formula W = Fd, if we greatly accelerate an object upwards, the  work done lifting the object will be greater than if a lesser  acceleration is applied across the same distance, but examples in my  book don't seem to take this into account.  (I am assuming that  F = ma, and of course the mass stays constant.)
81967
Consider a particle on the ground. This particle is raised by a force of magnitude $mg$ to a height $h$ above the ground. At this point, the work done on the particle by the force is $mgh$, which is equal to the potential energy of the particle. But, during this period, the force of gravity also acts on the particle and is displaced by $h$, and so does a work of $-mgh$ on the particle. Shouldn't  the two cancel and no net work should be done on the particle?   If they don't cancel, then where did the energy that came from the work done by the force of gravity go?