Thanks for clearing my doubt about freely falling bodies. Am SHIVAM from INDIA, preparing for JEE 2023 and entered the world of physics. Really, I'm fascinated the most by physics after seeing your lectures. Thanks again for your wonderful explanation and demonstration Sir.
Weight is defined as the measurement of gravitational force acting on a body, and in the case of freefall, the body is said to have zero weight? Doesn't that mean no gravitational force is acting on the body? And if this is the case, then why is the object pulled down? I can understand weightlessness in case of satellites orbiting in the orbits, because in the frame of reference of the satellite, the centrifugal force balance the centripetal force provided by the gravitational force, hence the net force becomes zero. How can it be justified in case of freefall?
all objects in free fall - that means that gravity is the only force acting on the object - have no weight. Astronauts in orbit around the Earth have no weight. *Place an astronaut on a scale; it will read ZERO!* When you jump off a table, during the few seconds that you are not yet on the ground, you have no weight. I cover that in great detail + demos in my 8.01 lectures - *watch them*
@@lecturesbywalterlewin.they9259 we feel no force? If any force acts on us, we feel it. But in free fall we don't feel Gravitational force. Whyyyy???? I think there is no such force. But then why the obj is Falling? BTW, your student from India. Namaste Sir🙏
@@umeshshende7540 you wrote: . *NOOOOOOO not if the only force that acts on you is gravity*. Gravity is the only force that acts on the astronaut, thus (s)he) is weightless. An astronaut is in freefall and everyting in the ISS is in free fall (thus weightless). The g in orbit of the ISS is about 9 m/s^2. That is needed to keep it in orbit. A glass with water in the ISS has no weight, the glass "floats around inclluding the water" The glass with water could be rotating in the ISS, but water never falls out of the glass as the water is also weightless.
Why do we feel weightless/No acceleration while free falling, BUT we're accelerating at g at squaring distance per sec?PLZ explain through the perspective of Einstein General Theory of Relativity?
anything that is in freefall is weightless.Watch my 8.01 lecture in which I explain this and do a demo which shows very drametically that an object dropped from about 3m is weightless during the time that it is falling. Astronauts are weightless, the Earth is also weightless as the Earth itself is in free fall. Use google for the definition of freefall.
@@lecturesbywalterlewin.they9259 thanks for the reply Sir...So, basically your definition of Weightlessness/Weight is directly dependent on Normal Force...But isn't the Weight Depends on g...So, astronomers also has weight since they're falling but still under g but lower than that on Earth's surface...So, my argument is this, if something is falling on Earth, it falls with an increasing acceleration of g...So how does that falling changing accelerating body has Zero Acceleration/Constant Acceleration since it is falling under acceleration "g"??
Hi Walter, having arrived in the wonderful world of physics, I'm watching this after reading the section in your book (for the love of physics) on falling in an elevator (pg 47). I really like your demonstration to illustrate the concept of weightlessness but my understanding is that weightlessness and zero weight are not the same thing. When you accelerate downwards in free fall you experience weightlessness because there is no normal force acting on you but your weight is still mg. Is this correct? Thank you very much 😊
weight is what your bathroom scale indicates. When you jumpof yoiur table your weight is zero during the time that you free fall. Your mass will no change. I cover all this in my 8.01 lecrures
And why all when talking about free fall They talk about movement and time They do not talk about what happens to weight in terms of distance and time And what is the strength that reaches the weight when colliding with the ground ?????
the impact force varies with time. It all depends on the elasticity of both the ground and of the object. Whether you drop the object in mud or on a marble floor makes a HUGE difference - Whether the object is made of puddy or of marble makes a HUGE difference - use google.
the impact force= w or not ........what happens to weight in terms of distance (w1) fall in distance (x) .and (w1) fall in distance (y) is not to same impact force....... time is not The only factor could you explain more
w=m*g ---f=m*a------if a=g------->f=w now ---the impact force= F or not ........what happens to (F)in terms of distance (w1) fall in distance (x) .and (w1) fall in distance (y) is not to same impact force....... time is not The only factor could you explain
first do your homework on the definition of weight I define weight as the normal force exerted on an object if it is attached to an imaginary bathroom scale. Thus you are weightless in free fall and thus also in the ISS. Thus according to my definition at impact weight will change - *it's the force which the ground exerts on the object and that changing force depends on the elasticity of both the object and the ground* This is my last msg on this topic. >>>w=m*g ---f=m*a------if a=g------->f=w>>> *this is nonsense*
+SUNIL PATIL when a body is in free fall the gravitational force acts on it. How else would it be in free fall? However, according to my definition of "weight" a free falling body is weightless. Astronauts are weightless (a bathroom scale will show ZERO) yet the gravitational force acts on them all the time.
Think of "weightlessness" more as constraint-less-ness. A body in free fall or orbit has a force of gravity acting upon it. But, due to the fact that it accelerates precisely according to that force of gravity and by definition of free fall, no other forces act upon it, there are no constraint forces necessary to keep its internal masses in position during the free fall.
@@zinodidine5088 It depends on the sign convention you assign to a. In the event that you assign a as positive when upward, then w = m*(g + a) is correct. In the event that you assign a as positive when downward, then w = m*(g-a).
When an object is in free fall, by definition there are no forces acting on it other than gravity. When the object is impacting Earth at the end of its free fall, it is no longer in free fall. It is being acted upon by the impact force. For a simple example, an impact force would directly follow Hooke's law, and be proportional to the deformation distance. So consider a 1 kg object falling at 5 m/s toward a 1 meter spring at its unstressed length, with a 100 N/m spring constant. After it compresses the spring 9.8 cm, it will receive an impact force from the spring that is equal to its weight at rest. Neglecting the change in gravitational potential energy during the compression of the spring, it will travel 50 cm along the spring before stopping momentarily and rebounding. If you account for gravitational potential energy during the spring impact, it will compress the spring 60.75 cm, before stopping momentarily and rebounding. At this turning point, the object will seem to weigh 60.75 Newtons, or about 6.2 times its weight at rest. An object will always weigh greater than its rest weight, upon impacting the ground. The shorter the stopping distance, the greater the impact force will peak. Which is why you need to extend the stopping distance to safely bring something to rest. It isn't always the case that the impact force is proportional to the compression distance, even if all materials obey Hooke's law at the microscopic level. For instance, a spherical ball impacting the ground will have a non-linear relationship between force and deformation distance. It is a graduate level concept to derive the theory behind contact mechanics that show this relationship. For the example of a sphere contacting a flat surface, the deformation distance is proportional to force raised to the 2/3 power.
