# Tag Info

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The pulleys make it trickey. You can see how this works by looking at the lengths of the individual rope pieces. When the cart moves 1 meter to the right the bottom part gets 1 meter shorter. The top part also gets 1 meter shorter. Since the rope can't change length the left part (that is attached to $m$) gets longer by 2 meters. So for every meter that the ...

9

When spun around the balls are set in motion. That motion would continue in a straight line if it wasn't for the wall. The wall causes the motion (the velocity) of each ball to turn all the time. (This is where the centrifugal effect comes from - the balls "feel" swung outwards because they want to continue in a straight line but the wall is in the ...

7

I made a classical experiment to demonstrate centrifugal force. Just to clarify, It's only centrifugal in a rotating frame and centripetal in an inertial one. Confusing?Here is a good distinction between the two. As per your current understanding, Il explain it to you in the frame of the rotating stand. Picture it as the stand being still and the entire ...

5

At the bottom of the floating mass its weight would be pushing downwards on the water, and the water pressure on the bottom would be pushing upwards on the object. This would be the up and down action reaction pair from its buoyancy. There is also a sideways action reaction of water pressure on opposite sides of the object but they cancel each other out. As ...

5

The floating object exerts exactly the same forces on the fluid as the fluid it displaced would have exerted. This is the essence of Archimedes principle: the rest of the fluid does not notice that the displaced fluid is gone.

4

Right so I understand your confusion because I've been there before. First of all have in mind that, in this situation, if you push the X box, all boxes will move together thus having same acceleration $a$. If what you say was true (that all boxes should be under the same force), since each box has a different mass, their acceleration would be different (...

4

You need to do a free body diagram (FBD) on boxes X and Y applying Newton’s 2nd and 3rd laws with the constraint that all boxes have the same acceleration. For example, take box X. An FBD on X will have $F$ acting to the right and $F_{yx}$ acting to the left where $F_{yx}$ is the force that box Y exerts on box X. Then per Newton’s second law net force on X ...

4

My legs were pushing the earth back but I was not being displaced. How do you know your legs were pushing back against the earth? Did you have a force meter? What you would actually find, were you to measure the situation properly, is that you did not actually push back against the earth. You only pushed up and down, effectively jogging in place. The ...

3

Yes, there is a net upwards force from the fluid. It is called the buoyant force.

3

This problem is in Introduction to Classical Mechanics by David Morin as problem 9.23. Let the rate of precession of the coin be $\Omega$. Let the moments of inertia be $I = \frac 14 mr^2$ and $I_3 = \frac 12 mr^2$ respectively. In this situation, it is most convenient to find $\mathbf{L}$ about the center of the coin. The important thing here is to (...

2

Weightlessness is when your proper acceleration is zero. The proper acceleration is an important concept in general relativity because it is a scalar invariant but, despite the fearsome reputation for complexity that general relativity enjoys, the proper acceleration has a simple physical interpretation. To determine your proper acceleration simply drop an ...

2

Of course there is a mathematical way to know the answer as given in Nihar Karve's answer but I would like to give you some physical insights of it. Actually , Normal force arises due to electromagnetic nature of materials (also linked with quantum mechanics but just ignore it for simplicity). Since electromagnetic forces depend on the distance between ...

2

Sum of forces in the $y$-direction $=$ $$F\sin \theta + N - mg$$ (using the standard Cartesian axes). Although you haven't stated it explicitly, I will assume that the block has no motion along the $y$-axis (if it does, $N = 0$). Hence $$F\sin \theta + N - mg = 0$$ So $N = mg - F\sin\theta$, where $mg$ is implicitly assumed to be positive.

2

Can't the acceleration of second body be larger than the first body momentarily such that it loses contact with the first body for a while? Not the acceleration, but the velocity. For example, if the initial push is not gentle, but sudden. The vibration can lead to a momentarily loss of contact. But just after losing contact, the second body by inertia ...

2

It is an exercise called "running in place" Running in place doesn’t provide the same benefits as running since you use different muscles and movements, but many of the benefits are similar From this article on the physics of running, to run in place you must use your feet in such a way that there is no forward momentum due to friction of your ...

2

Your left forefinger is exerting a force on one end of the band. Your right forefinger is exerting a force on the other end. Each end of the band is exerting a force on the finger with which it is in contact. You can see this, because the ends of your forefingers have turned white and have temporary grooves in them! [Try not to suppose that the band is '...

2

My answer is going to be a bit different from others but give it a try. Newton's third law says that if a body $A$ exerts a force on a body $B$ then the body $B$ will alao exert the same type of force of same magnitude and in opposite direction. The origins of the both the forces are same. What do you think why are all water molecules in a container not ...

2

To really understand this qualitatively you should first convince yourself that the pressure in a liquid only depends on the height and density of the liquid above it. It does not matter which shape the container has. Therefore the total mass above it and volume do not matter, only their ratio does. Pascal's law is also important to understand. Now consider ...

2

The proof is valid for a cylinder. You are right, though, that it is incorrect to talk about 'upward pressure' and 'downward pressure'. It is fine, of course, to talk about the upward force and downward force due to fluid pressure. There are two well known ways to derive A's Principle for a general shape of solid. The first is a generalisation of the method ...

2

The question asks for an explanation "in layman terms", so I will try to provide one. The term "centrifugal force" is problematic. But the other answers have covered this already, so I won't. In fact, I won't mention forces at all. Imagine you suddenly remove the curved stand while the balls are moving. The balls will move away from the ...

1

It is easier to understand if we compare with the situation of the balls in a plane rotating disk as described in this question. In that case, the balls would increase continuously its velocity, while migrating to points with higher disk tangential velocities. In the present case, instead of a continuous increase of the kinetic energy, part of it becomes ...

1

you have two external forces that act on the ball , the centrifugal force ,$~m\,\omega^2\,y~$ and the weight force,$~m\,g~$ the ball move towards the instantaneous tangent on the curve due to the resulting force (centrifugal force plus weight force) towards the tangent , but only if the resulting force greater then zero. Remark: the normal force is ...

1

I am looking for a force similar to the normal force, but as far as I can tell, the normal force can only exist between two SOLID objects, not a solid and a liquid. Not quite; liquids can also apply/receive a normal force. The catch is that additional normal forces must act in the other two directions as well. (For example, a normal force in the x-direction ...

1

The forces exerted by the springs on the mechanism (which is defined as the combination of the four bars connected by joints) are internal forces. They can not cause the mechanism as a whole to accelerate. Only external forces can do that (such as gravity). I'm not sure what your teacher means by "cancelling forces". There is no acceleration of the ...

1

why don't we include the Normal force acting on the fluid due to the bottom surface of the container keeping the fluid in the derivation of pressure variation with increasing depth in a fluid ? We do. Pascal’s law for a hydrostatic fluid can be written $\Delta P = \rho g \Delta h$. This applies to all points in the fluid, including those in contact with the ...

1

Is this proof valid ? Yes, this proof for the cylindrical body is valid. But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out). How can I write a proof with any general solid ? (not just cylinder) Archimedes' ...

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If the speed is increased, the horizontal component of the normal force must increase to provide the extra centripetal acceleration. If the normal force increases, then its vertical component will exceed the weight of the car.

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Yes, it can happen but the block on the left will catch up very soon, so for any practical purposes you can assume both move together, even if there is a small periodic motion between them. To see this, imagine that the contact force between the two blocks is like a spring (but only when they push into each other, because the block on the left can never ...

1

There's a missing force you didn't consider. The 2 arms of the ladders exert horizontal forces on each other at their point of contact.

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Your object consists of atoms composed of charges (electrons and protons). Likewise, the ground is as well. As the object hits the ground, the atoms in the ground are displaced and pushed closer together. The electrons in the atoms begin to repel each other and therefore resist this compression and as a consequence your object is then pushed by the atoms in ...

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