Tag Info

Hot answers tagged

23

Olin Lanthrop suggested a plausible approach but there was a lot of (inaccurate) guessing in his answer. I was going to write this as a comment to his answer but it got too long. Note - in the below I round to no more than 2 significant figures - the nature of the problem doesn't support more. Let's take the famous Sherman tank as our example. A brief ...


17

This is difficult to answer since you seem to have a major misconception about forces. The best answer is to go back your freshman, or even high school, physics book and read the section on forces. However, briefly: The pulling force isn't somehow split between the train and the ground. The rope will pull with the same force on whatever is holding each ...


16

I took a look at the clip and my take on it that the object of the exercise was not to slow down the tank, but to move it sideways and land in the lake a half a mile away. Hannibal says, "rotate the main gun to 82º" which I take to be sticking out sideways. Background Information The facts I could find were: M1A2 Abrams ...


14

Alright I'll throw my hat into the ring with an answer. The idea that it's an unsolved problem is totally bogus. When you start to fall to one side or another if you turn the wheel slightly in the direction you're falling the bicycle starts to follow a curved path. There is a force due to friction that deflects the rider's path into a curve: The ...


11

Do the math. I'm no military expert, so I'll guess at some parameters, but I think it will show the effect is so small that it doesn't matter even if the numbers were considerably more favorable. Let's say the tank weighs 50 tons, which I think is rather light for a tank. That's 100,000 pounds, which puts its mass at 45,000 kg. I don't know what the mass ...


10

I do not agree with the angular momentum theory: if you were to hop off your bike at speed and let it go by itself, it would not go very far before falling on its side, even less if you put a ~150 lbs sandbag on your saddle. There is indeed an effect caused by momentum, but this is negligible when compared to the actual contribution of the rider. I agree ...


8

I assume a steady-state universe and that the bodies have no velocity relative to each other. Yes, they will eventually collide. Gravity has an effect over any distance, including the ~46 billion light-year radius that constitutes the spherical observable universe (the actual size of the universe may be much larger). Of course, the force will not be very ...


7

The force on rope is equal for both of them at any time. For winning the game the force on ground is responsible.


7

The diagram is misleading. Look at this: At any moment in time, you have the following forces on the particle: Gravity Tension in the string When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string ...


7

No, NASA has not confirmed that. What NASA has confirmed, again, is that it has some rather nutty folks working for it.


6

Consider this for the upper ball: The angle of the inclined plane is simply derived from geometrical considerations. Looks pretty easy to solve, right? Once you know the normal force of the inclined slope, reflect and apply it to the bottom ball to complete the problem.


5

Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically. These two forces combine to give the ...


5

Yes, they will collide given the initial conditions. The speed at collision can be calculated. We can presume them to begin with virtually zero gravitational potential energy. We need an assumption of size when they collide. Let's assume a size of 50cm. That way when they collide, the centers will be 1m apart. $$U = -\frac{GMm}{d}$$ When they collide, ...


5

This is unphysical. There is no force or momentum that would make something unstoppable. First of all, any moving object can be made to diverge from a straight line path simply by applying a small force perpendicular to its motion. Secondly, no matter how large a momentum an object has, it can be completely halted by a similar object with the same momentum ...


5

Normally, suspension consists of a damping component and a spring component. For such a suspension, higher speed means higher acceleration and greater force. Driving faster will cause a bigger jolt. However, high end cars these days use active suspension - and that changes everything. With active suspension, you can either respond quickly to bumps in the ...


4

If $F_{ii}=0$ then you are right that the $i\neq j$ constraint is unnecessary although it does make the physical interpretation clearer: all atoms (other than $i$) act on $i$. However, in practice, the force is usually given as a function of separation, $F(r)$. And so when you evaluate $F_{ii}$, you effectively evaluate $F(0)$. The problem is that $F(0)$ is ...


4

The argument sounds perfectly reasonable. Consider arbitrary parcel of fluid in equilibrium. It exerts downward force equal to it's weight on the surrounding fluid, and it does not move. Therefore according to the second law of motion, the downward force must be balanced by upward force of equal magnitude, the buoyant force (otherwise it would start to ...


4

If the particle moves from the point $x$ to $x+dx$, and assume $dx\gt 0$ for simplicity, then its potential energy increases by $$ dU = \frac{dU}{dx}dx $$ Well, it increases if $dU$ is positive and decreases if $dU$ is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ...


3

The normal force does decrease with angle. This does not mean that the coefficient of friction changes: We can, depending on the angle $\theta$ of the slope, split the gravitational force $F_g = mg$ acting upon a thing with mass $m$ resting on the slope into the normal force $F_n = mg \cos(\theta)$ and the force pointing down the slope, $F_s = ...


3

You have two subproblems here. One is the motion of the train and the other is how to grab the rope. Let's focus on the train. It has a mass $m$ and there is some friction with the ground (assumed low because it is on rolls) $\mu N$ , being $ \mu << 1$. You apply a force $F$, assumed modulo constant. So, the horizontal acceleration is: $$\sum F = m ...


3

The answer is that, despite our best efforts, we still can't quite put a finger on it. The gyroscopic forces mentioned by bobie in his answer have been proven not to be sufficient to fully explain why a bike stays upright. It's indeed very surprising that physicists are not able to explain the mechanism behind such an (apparently) simple and ubiquitous ...


3

If the bodies are initially at rest, then the orbit will be a degenerate ellipsis of finite semi-major axis and eccentricity 1, i.e., a line segment. The semi-major axis $a$ is half the initial distance. Time to collision is half the period $T$. This can be directly derived from Kepler's Third Law. $$ \frac{T^2}{a^3} = \frac{4\pi^2}{GM} $$ $$ T = ...


3

Force on the ground can never be greater than force on the rope Each players individually exerts some force on the rope and on the ground, but that doesn't means that the same amount of work is done on the rope and on the ground, or that one players exerts the same total force as his opponent. The winner overcomes his adversary only if/when because he ...


3

So far, it is still an open question as to why bicycles are stable at all. There have been a few ideas put forward, but they have been disproved by construction of non-standard bicycles. The most common explanation is that the wheels on a bike act as a gyroscope, preventing the bike from falling over. A bike was constructed with counter-rotating wheels to ...


3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...


3

Let's consider this sketch of a man with legs asymmetrically astride: weight : ca.80kg (800N, g = ca. 10) height : 180 length of leg: 1m fore angle 45° rear angle ca. 15° Distribution of weight The rear leg is 73.2 cm and is 19 cm behind the center of mass. Let's assume that the legs are in compression only, (if the person were standing on ice or ...


3

That's rather an unusual question though interesting, but it amuses me to make an even more unusual answer. Disclaimer: this is in no way a rigorous answer, as many assumptions, simplifications have been made (even entire phenomenons such as gravity attraction if the velocity is not perpendicular, solar pressure from photons...) - only a reflection on drag ...


3

...truck will experience larger force. But Newton's Third Law of motion says that 'to every action there is an equal and opposite reaction'. So the force experienced by the truck (M) should be same to that experienced by the car (m), but negative, isn't it? Third law states that momentum must be equal and opposite if both vehicles must come to a ...


3

I think you made a minor boo-boo :) Remember that velocity is an integral over the acceleration with respect to time: $$ \vec{V}=\int_{t_0}^{t}\vec{a}(t')dt' $$ You had a force that is given by: $$ F(t)=F_0 \exp(-t/T) $$ So indeed the acceleration is given by $$ a=\frac{F}{m}=\frac{F_0}{m}\exp(-t/T) $$ and after integration, a factor of -T comes out thus: ...


2

"I found out that F applied to mC must equal to the tension exerted on mA" is wrong F=(m_a+m_b+m_c)*a T=m_a*a T_x=m_b * a T_y=m_b *g T^2=T_x^2+T_y^2 Replace and get F=[(m_a+m_b+m_c)*m_b*g]/sqrt(m_a^2-m_b^2)



Only top voted, non community-wiki answers of a minimum length are eligible