New answers tagged free-body-diagram
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According to the second Newton law body in the end will achieve speed :
$$ v = v_o + \frac 1m \int_{0~\text{sec}}^{10~\text{sec}} F(t)dt $$
Not necessary force should be constant,- it can vary with time in the same time window it is integrated. If object was already traveling with initial speed $v_o$, until you started to push it, then you will boost it's ...
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The body will be moving with a constant velocity which is the velocity the body had at the moment the force was removed and it will be 10 m/sec.
The moment the force is removed , the body continues with the speed gained at the last moments. If the force was not removed and acted continuously the body would have been accelerating all the time.
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He will accelerate only till the force is acting on it in accordance to Newton's Second Law. It will reach a velocity due to that acceleration and then continue with that constant velocity according to the Newton's First Law.
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I had the same question exactly when I was learning dynamics (I think).
The question at hand is that if a constrained is formed in terms of velocities, how do we transform the constraint in terms of accelerations so that it can be used in ODE system, of equations.
The simplest example is that of a slipping wheel on a plane.
Velocity Kinematics
It is ...
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Velocity along the common normal direction must be the same as the distance between the bodies must remain constant, but the same may not be true for acceleration.
Consider a rod rotating with constant angular velocity. Different points on the rod have different accelerations along the normal direction but the velocities in that direction are same for all ...
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On your sketch I'm going to choose +x to the left and +y down. Similarly for x' and y' moving with the slant block. I will use m to indicate $M_2$. The slant block will move to the left (+x) with an acceleration A. Then m will have two components of acceleration in the inertial frame (assuming A is small): $a_x$ = A + a' cos(θ) and $a_y$ = a' sin(θ). ...
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One reason for this is we dont know the direction where friction will act
For example in this case due to different inclinations we cant say which block will move down , we can only say the system will move downwards but from which side we cant tell
So we just assume cases taking different direction of friction in different cases and estimate the minimum ...
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Ever wonder why you cannot hold up an offset weight by just a simple equal and opposite force? Even you you supply an upwards force $W$, to counter-act the weight $W$, the system isn't stable.
The reason is the you need an additional downwards force on the back to counter the torque generated by the offset weight.
The upwards force needs to be greater ...
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Note that, when we speak about Newton's laws at a basic level, we treat all masses as point masses, hence we do not consider the rotational effects caused on the body.
In this case, it is an extended object, hence we have to consider the rotational aspect of the body when force is applied. What happens here is as the weight is taken away from the point where ...
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If the weight is at the end of the stick you are holding, a vertical distance traveled by the weight ($y$) will be the greater the longer the stick. So if the weight travels down a bit, this bit gets bigger the longer the stick ($y=l\sin{\theta}$, where $y$ is the vertical displacement, $l$ the length of the stick and $\theta$ the angle between the ...
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The bag is applying a torque (denoted by $N$). Which has the expression $\vec N=\vec r\times \vec F$,
Now as $r$ increases $N$ also increases. So you will have to apply a higher torque for the same amount of force $F=mg$ ($m$ being mass of the bag).
So the force acting on the bag is same but it is the torque which is increasing. So no violation is coming.
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If I have understood your question correctly, you're asking why it's easier to hold up an arrangement like (a) than it is (b)
That's because when you're holding it up, it's never perfectly vertical. You're counteracting not only the force due to gravity, but also the moment because the center of mass is not at the support.
In case (a), the center of mass (...
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The net force that the person holding the rod is exerting on the bag is the same. In the second case he unfortunately is exerting this force through a torque on the rod, which requires large opposing forces applied to close points on the rod.
Newton's laws apply.
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What you are feeling is not an additional force, but rather an additional torque. The net force is the same in both circumstances, but the torque that you have to exert is substantially higher in the second case.
When you exert a torque you are essentially fighting yourself, one force counteracts another force in an opposite direction at a different location....
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Why didn't he consider the centripetal force as a force acting on that object.
Because Centripetal force is not a force acting on the body. The Centripetal force is not necessarily a unique force in its own right. As an answer pointed out, The "centripetal force" is simply defined as the net force acting towards the centre.
If there were some more ...
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First of all let me clear you that whenever a body moves in a circular path there is always an acceleration radially inwards which we refer to as Centripetal acceleration. And then there must be a force associated with this acceleration again radially inward. We call this force as Centripetal force.
Here is the proof of the presence of a centripetal ...
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According to Newton’s second law, NET FORCE = mass times acceleration. Your weight or the gravitational force on you is cancelled by the normal force (net force = 0) and hence you do not accelerate. Friction only comes into play when objects slide ACROSS one another or are attempting to do so, not when they are moving perpendicular to one another. There CAN ...
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Your English is little hard for me to understand, but let me still try to answer.
According to what I think, gravity is accelerating myself no matter what, and what it's actually preventing me from going down it's not the normal force, but the friction my body experiences against the surface which it's total, this is the reason then why I'm at the bottom of ...
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He is doing right .
Centripetal force is not any particular force ,any force acting in the radial direction behaves as a centripetal force,so your thinking that we should add mv²/R as centripetal force in LHS always is wrong.
The net of all the forces acting in the radial direction is the centripetal force.The role of centripetal force can be played by any ...
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When you draw a free-body diagram, you actually don't need to assume any mathematical relationships about the forces. Just identify the forces and their directions. Now, a rule of thumb is that, unless there are electric or magnetic forces in your problem, gravity is the only one that acts without touching the object. All other forces are contact forces.
...
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If you are saying that the tension in the string at the point of contact with an object hanging freely from it is equal to its gravitational weight in equilibrium, then yes.
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If it is hanging on a string, then normal reaction would be zero and tension in string would be equal gravitational force.In case the block is touching the ground then tension would be zero and what you said would be correct since the rope becomes slack in this case.Hope you got it.
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But the point of the wheel that touches the ground is not moving
relative to the ground; therefore, neither the reaction force nor the
action force does any work; therefore, this force cannot be what
causes the bike to accelerate from a standstill.
That is not correct. If the wheel is not slipping on the ground, then the static friction force will cause the ...
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These are the steps through which i believe the bicycle accelerates from rest.
You push the pedal .
2.The back wheel moves to rotate (clockwise) to stop that relative motion $f_1$ acts forward , now the whole bicycle will accelerate forward due to $f1$.
3 . Now come to rear wheel . Since the whole bike along with rear wheel moves forward , the point of ...
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I completely agree with the wonderful answer of @Bob D sir.
However I would like to add a few things to help your understanding even better.
First of all,what a lot of people seem to believe is that friction always opposes motion.This is not true.Friction may also help in motion. Eg We can walk only because friction pushes us forward when we push our feet ...
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The car takes the corner because its wheels are turned (by the steering mechanism) at an angle to its direction of motion. There is still rolling resistance that is trying to slow the wheels down, but this is overcome by the car's engine which exerts a rotational torque on the wheels via the gearbox and transmission (just as it does when the car is going in ...
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Suppose you have a mass connected to the end of a string. You swing the mass in a circular horizontal path at constant speed. You feel a force pulling on your hand. Per Newton's third law that force is equal and opposite to the force that the string is applying to the mass. That force is called the centripetal (center seeking) force acting on the mass and is ...
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If block $C$ moves a distance $x$ (let's say) then for the string to remain taut the block $B$ has to move the same distance i.e. $x$. Which makes the block $A$ to move a distance $x$. Since this is happening in same time each block has same velocity and acceleration.
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He didn't considered normal force since the rope and pulley are massless . So there is no weight of rope (mg pulling the rope or the pulley).
In real life you are correct , there is a normal force between the rope and pulley.
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Overview
First, identify the degrees of freedom in order to account for all unknowns. In this case, it is $x$ and $\alpha$ and together with the unknown normal $N$ are the 3 unknowns to be solved for by the 3 equations of motion.
First, you need the kinematics of the center of mass, as a function of $x$ and $\alpha$. I designate time derivatives with a dot $\...
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The object would rotate about the point of contact, you can easily find the angular acceleration of the bar, about the point of contact , about this point of contact the only torque would be due to $\cfrac{Mgx}{\cos \theta}$,
if you find the angular acceleration, you can find the vertical acceleration of center of mass of the bar by multiplying angular ...
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You have to include the torque acting on the system. A normal force that is off-center will cause rotation in your drawing.
$\mathbf{N} \times \ldots$ is the torque due to the normal force.
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but the condition of tyres are same as in previous case and friction must be in forward direction
Not the same. Friction acts in the opposite direction of relative motion. Imagine the car is driving on a frictionless surface instead of a road. It's moving at a slowish speed, and although there is no friction, the wheels are rolling without a slip.
If you ...
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Frictional Force always acts in the opposite direction of Applied force ( acceleration). As we know, Frictional force is a dissipative force, so a system (here is the car) loses energy (kinetic energy) and eventually it comes to rest.
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Plenty of energy is dissipated to make a car stop.
Air rolling over the the surface of the car creates a small friction skin where energy is translated, the perpendicular airflow is an opposing force, the gearbox has plenty of energy losses,the brake pads on a car clasping the brake discs dissipate a lot of energy in the form of heat and opposes the ...
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It is not the same friction, but rather the friction between the wheels and the breaking pads (and in general between the rotating parts of the car and their supports.)
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If the marble rolls without slipping, then there is a (static) friction force which provides the torque required for angular acceleration. At point Q the marble is losing linear and angular velocity, so the friction force is directed upward. The centripetal force comes from the normal force at the surface. Don't forget that the radius of motion for the ...
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