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8

If we find the force for a decelerating object (let's say a car on a road), we'll see that the force is negative as in it’s in the opposite direction of the motion of the object. Not necessarily. The sign of the force just depends on its direction; it has nothing to do with the direction of motion of the object. If the object is moving in the positive ...


8

Contact forces have no real distinction from electromagnetic forces, but that doesn't make the term irrelevant. It is shorter and easier to say "contact force" than to say "short range force based on un-modeled local electromagnetic interactions between nearby solids". So we use the term whenever it is convenient and whenever the meaning ...


4

The thing that one should keep in mind, is that forces gravitate. So if we eliminate dynamics by considering a static system of black holes that are “held fixed” then the thing that prevents those black holes from falling toward each other would necessarily contribute to the curvature of spacetime and would distort the geometry of black holes. Having said ...


4

then the force from the person (their weight) would act upon the egg. That's not true. You can push on the egg, but that doesn't mean that you're actually applying that much force. Push a wall, you can probably generate 100N or more against the wall. Now push a feather. You will not be able to apply that much force. However, if the egg can only handle a ...


4

A force (F) is a vector quantity. A vector has a magnitude (which cannot be negative) and a direction. The negative of (F) is a different vector with the same magnitude but opposite in direction. When working with vectors, it is usually best to work with their components. In a chosen coordinate system, a vector may have one or more negative components ...


3

You're kind of overthinking this. Newton's 3rd law does not say there have to be two forces for every interaction, it's stating a fact about the nature of what a force is. It is essentially saying there is no such thing as a "one sided" force -----> All forces are like this . <-----> In your example, what happens is the person ...


3

What you do not to happen is for the vehicle to aquaplane which will happen if there is a significant film of water between the tyre and the road surface. Because motorbike tyres are much rounder than car tyres the contact width across the tyre for a motorbike is much smaller than than for a car so any water between motorbike tyre and road surface is ...


3

There's a few pieces to this. The first is that it will be easier to understand forces if you don't think of them as "negative." Forces are vectors, which have a (positive) magnitude and a direction. Where you can get a "negative" thing regarding vectors is when you write them down in component form. When using components you pick ...


2

You're right, Frictional $F$ does no work on the runner in the ground frame. Running is a more complex motion. During the thrust phase of your step, the muscles in your leg are doing work on your moving torso. ($F$ holds your foot still so this work can be done). Then during the recovery phase, your leg muscles do work on your foot to lift and accelerate ...


2

You're right to be suspicious of this proof. It's wrong. The error is most succinctly stated in the fourth paragraph: [for] a body of mass zero, [...] Newton's second law tells us that $F_{\rm net} = 0a$, so $F_{\rm net} = 0$. If "bodies" are things that Newton's laws describe then there are no "bodies of mass zero". You can't push a ...


2

I think you have misunderstood the question. What you have calculated by $F=mg$ (with $g=9.81\text{ m/s}^2$) is just the vertical weight force of the vehicle. But the question is not about friction or vertical force. Instead the question describes the horizontal movement of a vehicle. You can set up the kinematic equations for position $x(t)$ and velocity $...


2

Impulse $\vec J$ imparted by force $\vec F$ will be equal to the change in momentum $\Delta \vec p$. $d\vec J=\vec Fdt$. Force being constant $\vec J=\vec Ft$. $$\Delta \vec p=m\vec v=\vec J$$ which gives us. $$F=mv/t$$


2

The equation you would use is $$v=v_0+at$$ Note that the constant acceleration can be expressed as $$a=\frac{F}{m}$$ if the object has a mass $m$. So $$v=v_0+\frac{F}{m}t$$ and so $$F=\frac{m(v-v_0)}{t}$$ and if the initial velocity is zero then $$F=\frac{mv}{t}$$ So if $t=2s$ then $$F=\frac{mv}{2}$$ If your target velocity is for example $10ms^{-1}$, then $$...


2

I think that the earlier answers haven't adequately explained why there isn't a force law for the strong force. It's not because it's relativistic or quantum (electromagnetism is both of those). It's because it's nonlinear. In the gravitational and electromagnetic formulas in the question, $q_1$ and $q_2$ (or $m_1$ and $m_2$) are the charges at exactly two ...


2

Torque equals the rate of change of angular momentum. Differentiate the angular momentum to get the answer. And your method is not correct. Don't treat vectors like scalars. That is simply not allowed. $$\vec{L} = I\vec{\omega} $$ Using your method, you would need to differentiate this way $$\tau = \frac{d }{dt} (\vec{r} \times m \vec{v})$$


2

Any task or problem that asks for initial or final values, or any other values for that matter, must clearly define those values. If you are asked to find final velocity, then you need the definition of "final" in that particular scenario. Sure, the final velocity could have been defined as the velocity just before impact. Then you'll need a few ...


2

When you apply brakes to a car the rolling motion of wheels is reduced. In day to day life you might have seen that a rolling body rolls farther than a body slipping this is because ground applies a resistive force to the slipping body but it applies no torque in the forward direction this causes the body to stop faster. Same thing happens when brakes are ...


2

Here, $\mathscr{d}m$ is inherently negative because rockets mass decreases with time. $$ \mathscr{d}m = -|dm| $$ And the rocket's mass is, $$ m + \mathscr{d}m = m - |\mathscr{d}m| $$ For reference, please look Young's University Physics with Modern Physics, Ch 8, Sec 8.6


2

As far as I see, it is an experimental fact. Applying the forces on an object through springs, it is possible to know the modulus of the forces by the spring displacements. Knowing also the direction of them, the forces can be modelled as vectors. It happens that adding them using the parallelogram rule works either for a body at rest (where the vectorial ...


1

If the spring balance has zero mass then the net force on it must always be zero (otherwise it would have infinite acceleration). So the two tensions on either side of the spring balance must always be equal and opposite even when the masses $m_1$ and $m_2$ are accelerating. Suppose $m_2 > m_1$ and so $m_2$ is accelerating downwards with acceleration $a$ ...


1

In general, you have fixed 3x3 mass moment of inertia tensor $\mathrm{I}_{\rm body}$ riding along the body coordinates and it needs to be rotated into the inertial directions to be used in the equations of motion. Give a 3x3 rotation matrix $R(t)$ this is done with $\mathrm{I}(t) = R\, \mathrm{I}_{\rm body} \,R^\top$ So the mass moment of inertia is changing ...


1

The most important thing here to understand is that action reaction pairs act on different bodies.In this case normal force and mg act on the same object so they do not qualify as an action reaction pair.


1

As others have mentioned, action-reaction pairs must act on different objects to qualify as one under Newton's third law. In this case, the reaction force of the gravitational force $m\vec g$ exerted on us by the earth, is the gravitational force exerted on the earth by our body; And the reaction force of the normal force exerted on us by the surface we're ...


1

The normal force simply does not qualify as a valid action-reaction Newton’s third law pair. A valid action-reaction pair requires that the action force and the reaction force act on different objects. For example, consider an object sitting on a table. The table pushing on the object (normal force) while the object (pulled down by gravity) pushing on the ...


1

How can work done be equal and opposite when the forces are unequal and object travels the same distance When you say “the forces are unequal” you need to distinguish between the action-reaction pairs of forces acting on the block per Newton’s 3rd law, and net force acting on the block, per Newton’s 2nd law. There are two action-reaction pairs acting on ...


1

Newton’s third law tells us that the force exerted by the block on the spring is always equal in magnitude and opposite in direction to the force exerted by the spring on the block. These forces are internal to the block/spring system. There may be other external forces acting on the block and/or on the spring, and the block and the spring may or may not be ...


1

$$F = T \sin \theta$$ $$mg = T \cos \theta$$ $$\Rightarrow \frac{F}{mg} = \tan \theta$$ $$F = mg \tan \theta$$ here $\theta = 30$ and $m = 25 \ \text{kg}$. So your answer should be about $14.43376 \ g$ which evaluates to about $141.6 \ \text{N}$, if $g$ is taken to be $9.81 \ \text{m/s}^2$. You have done this incorrectly because you have plugged in the ...


1

In the last olympic marathon, everytime they showed the front runners in slow motion, it was clear that each step was also a small jump. The center of gravity oscillates about 10 to 20 cm I guess. So they have to do work to increase its gravity potential at each step. Of course they "fall" just after, and by the same amount, but it doesn't help too ...


1

A fluid exerts a buoyant force on an object, and the object exerts a reaction force on the fluid (which causes some displacement of the fluid).


1

The cable will exert an impulse which changes the bar's angular momentum. The maximum force will depend on how much the cable can stretch. Treating it like a spring (with a high k) is not unreasonable (but measuring the k could be a problem), and there is likely to some bounce.


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