New answers tagged free-body-diagram
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if i assume that the "supporting force" is actually the contact force i.e. there is no friction than force acting on the object in the direction of gravity is simply the component of horizontal thrust F in the direction of gravity which is zero and the component of normal contact force i.e.fcos0.if friction is taken into consideration then component of ...
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Static friction is keeping the block from sliding down the wedge. If the angle is steep enough, or the static friction coefficient is low enough, the block will start sliding, then it is kinetic friction that slows it's descent.
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Let's try one more! Let's attach a rope to the ceiling and suspend a 10 kg weight.
The tension force on the rope = ma = 10 kg × 10 m/s^2 = 100 Newtons.
With a 20 kg weight: 20 kg × 10 m/s^2 = 200 Newtons
Now, attach a simple pulley to the ceiling and put 10 kg on either end:
Tension force on the rope is (10kg + 10kg) × 10m/s^2 = 200 Newtons
We've ...
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This is a lovely question, as it poses an obvious common sense interpretation that is at odds with our physical models, specifically do to with the difference between a passive reaction to a force and an active one; intuitively one expects there to be a quantitative difference between their magnitudes, when in fact they are the same.
Consider three ...
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Consider a variation on the Atwood machine where instead of hanging straight down, the two masses $m_1$ and $m_2$ each sit on a separate frictionless plane at an angle of $\theta_1$ and $\theta_2$ to the vertical. The planes are joined along a ridge and the string connecting the two masses runs over the ridge, again without friction. The simple Atwood ...
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I think you are on the right track. Instead of looking at a net force on the string, which would not be possible as @Marco Ocram points out, consider that the net force is acting on a system consisting of the string and masses connected to each end. Then the system, including the string, can be considered accelerating with no net force on the string itself.
...
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The arrangement you describe is impossible. The tension of the string will be 70N. Whatever was trying to restrain the end of the string with a force of 60N will be subject to a force of 70N by the string. As a result it will accelerate subject to a net force of 10N. The reaction on the string will be 70N.
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...is the value of tension always fixed for it's length never changes?
No. The tension in ideal strings is not at all related to the extension, because their length is fixed. To get a feel for this, imagine a normal elastic string which is quite stretchable. Now slowly increase the "stiffness" i.e. rigidity of this string. So, initially if a force of $5N$ ...
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The key to this problem, is that all the boxes have the same acceleration, $a$. And in each case, you have $F=ma$. In the first part, the force $F$ needed to accelerate all the boxes at $a$ is $F=ma=(M_1+M_2+M_3)a$.
In the second part, the force needed to accelerate $M_3$ at $a$ will be $F=ma=M_3a$, and in the third part you'll be accelerating both $M_2$ ...
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Objects without mass can accelerate without an external force.
$$F = ma$$
If $m = 0$ , $F$ is always equal to zero but $a$ may or may not be zero.
There is no restriction on the acceleration of a string, only that the net force acting at any point in the string is zero as $m = 0$. This is why at every point on the string tension acts in opposite ...
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I think that the correct system is: massless spring + mass m at one end + force F at the other end.
If the system is hanging from a hook, F is a reaction force, balancing the weight mg.
If the system is free from a gravitational field, the effect of the force is an acceleration: a = F / m and there is a tension F in the spring.
The deflection of the ...
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Going faster definitely leaves shallower tracks. I know this because as a young man I rode dirt bikes (motocross) a lot, when going through sand or soft silt if you went slower you would sink up in it, and usually lose speed and control. If you went very fast through it, you would stay more on top of it, leaving noticeably shallower tracks. For the tire to ...
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This cancelling of tensions at a point is only possible in equilibrium. Your reasoning shows this. If every point is in equilibrium then each point can't accelerate and there is no way for the rest of the string to know that one end is accelerating. The problem is that we think of strings as percectly inelastic and while that is often almost the case, there ...
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In principle they do, although I do not know how measurable is the effect at the range of speeds a car can go. In order to depress the soil the wheel makes a force, let us call it the normal. This normal force is independent of speed.Once the soil feel the force, it will move down a distance $d$ at some rate, for instance $d=1/2N/mt^2$ is the simplest ...
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I think what is confusing you here is the nature of a massless string, which is a bit unrealistic (but very useful for demonstrating concepts).
Consider Newton's second law, it is the mass that resists the acceleration. A massless string has no resistance to acceleration. The required force to accelerate it from Newton's second law is: $$F = m a$$
If $m =...
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The net force is acting on whatever masses are connected to the string, not the string itself which is massless. Think of the string and attached masses as a system. The masses and the string accelerate as a unit due to a net external force on the system. The tension in the string is an internal force.
To further illustrate, see the free body diagrams (FBD)...
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Gravity!!!
I know you might be surprised with this answer, but it is not the good old classic Newton's gravitation. This is general relativity, which basically states that all laws of physics must be the same irrespective of the reference frame, this symmetry also known as equivalence principle, leads to what we think of as centrifugal forces.
The ...
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Assuming block A is not moving and block B is hanging from it, (if it is accelerating, is it pulling block B or vice versa, but the concept would remain the same) we cut horizontally at the plane where the adhesive layer contacts the block B to spare ourselves worrying about the weight of the adhesive. And consider the FBD of the block a with the adhesive at ...
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I believe you'll want to keep the entities (Block A, Adhesive, Block B) conceptually separate if you want to think about the free body diagrams. In particular, the free body diagram indicates the forces acting on the object. As you point out, Block A is in equilibrium (not accelerating), so the only force that can cancel gravity, which is pulling down, is ...
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It is static friction that prevents your foot from slipping. But static friction has a limit:
$$f_s\leq \mu_sn$$
By taking a step, you exert a backwards force. Static friction then appears as the equal but opposite reaction force and pushes forward to avoid you slipping. As you walk faster, you exert more force. Since static friction must equal your force, ...
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look at this diagram $F=F(t)$
where:
$F_i$ is your foot force
$F_{\mu\,k}=\mu_k\,m\,g$ is the kinematic friction force between your foot and the surface.
As long as your foot force is less then the kinematic friction force you don't move
if you run then your foot force gradient is greater then as if you walk, thus:
$\frac{dF_1}{dt} > \frac{dF_2}{...
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While walking the static friction acts on your feet to hold it in a place and when you apply a force via pushing the ground backward it in turn pushes you forward (Newton's third law). Here you would see that your bones push your feet backward and hence are pushed forward and in a similar manner your whole upper body is pushed forward. This generates a ...
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Water on a surface can lower the co-efficient of static friction, which will make it take less force to break static friction (shoe not slipping), and become kinetic friction (shoe slipping). Moving slowly on a wet surface usually causes less horizontal forces which can break the static friction between the shoe and the floor. Hydroplaning can also be a ...
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See.. This is a question of rolling motion. If you are given all the data you may solve it. Assume the direction of friction in any arbitrary direction whether in left or right. Solve the equations and if you come out with the friction as positive value your assumed direction is correct else opposite. This method is very very useful in mechanics section as I ...
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First of all imagine what would happen without friction: the cylinder will rotate clockwise. Now put friction back. Can you see which direction it will act?
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Nothing is moving it backward
The train is accelerating, but the bob is not. In order for the bob to be accelerated at the same rate as the train, there needs be a force acting sideways on it. This is effected as the displacement of $\theta$ on $\vec T$- which combined with gravity $m \vec g$ produces a resultant acceleration $\vec a$
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I would believe the weight does change
A running watch has more mass than a stopped watch due to the energy created by the motion of its parts. This would be somewhat equivalent to your example. The moving sand grains have kinetic energy and there would be some thermal energy created by friction too. The extra energy means extra mass. E=mc2
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This is the problem of accelerating reference frames. Newton's first law:
In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force
can be seen as a condition for the other laws to hold true. The frame that is moving with the train is obviously an accelerating reference ...
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The answer by rghome is correct when Earth's surface is chosen as the reference frame.
Now consider this with the train car as the reference frame.
According to the Equivalence Principle, the effect of an accelerating frame of reference is identical to a gravitational field.
This means that if you are in the train car (without windows), you have no way of ...
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The bob doesn't move backwards at all. The train is moving forwards (according to your reference frame) and if the bob wasn't attached to the train it would remain stationary. Since it is attached to the train there is a tension in the wire (as the top anchor point moves with train and the bottom is attached to the weight, so stretching the wire slightly). ...
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The assumption that the force $R$ is in the direction of the rod is false. In fact, the direction of $R$ is exactly what you need to compensate the forces $T_1$ and $T_2$ with, which will produce some nonzero torque on the rod for general angles.
Simple experiment with your hand and some rope (and with a pulley if you have one) can give you an idea what ...
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Whenever an object follows a circular path then a centripetal force is required to explain such(the change in direction). The force is given by $F_{centripetal } = m \frac {v^2}{l}$. Therefore here rather than being $T - m g \cos \theta =0$ it is $T - m g \cos \theta =m \frac {v^2}{l}$
[Note: Here $v$ is the instantaneous velocity and varies from one moment ...
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The end mass rotates, thus you have a centripetal force added.
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I found explanations of this unsatisfactory as well, including in answers to related questions like this one. Here is a direct explanation in terms of equilibrium of forces.
The green rectangle is the lever, with thickness $h$. (It's upside-down from a typical picture, but that doesn't matter.)
Because the system is in equilibrium, we have $F_x = G_x$ (...
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I don't see how the power delivered to the wheels gets to be kinetic energy of the car without friction doing work.
Forces don't have to do work.
Imagine a compressed spring with a mass. The compressed spring has a bit of energy in it. When we release the spring, it can accelerate (do work) on the mass. This seems pretty simple.
But for the mass to ...
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The wheels exert a backwards force on the ground, so the ground exerts a forward force on the wheels. That's what actio = reactio tells us.
Since the center of the wheel is (largely) stationary relative to the car, the impulse (force times time) delivered from the ground must be forwarded at the axis of the wheel. Here we have again the axis of the wheel ...
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Work has two different but equivalent definitions. One is that it is a force applied over a distance, and the other is that work is a transfer of energy. I prefer the second definition, as it is a bit clearer in circumstances like this.
The car accelerates, and therefore there is clearly an increase in KE. But was work done by the road? The road is ...
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The friction just provides the grip. The torque of the engine does the work. Its the same when you start running- your legs exert the force and friction just prevents your feet slipping backwards.
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Well, I would compare the velocity of each block relative to the other blocks, for the bottom block, I would compare its velocity to the block above it, and if it is moving with respect to the bottom block, then there must be kinetic friction there. The max static friction applied for the top and bottom of block 1 would be different, though, because there is ...
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There are two forces acting on the mass are gravity and tension from the rope. This is seen in your free body diagram on the right side of your image. The mass does not fall (or rise) because these forces are of equal magnitude and opposite direction, thus the net force acting on the mass is $0$.
Yes, if we look at the rope the force pulling it down at $A$ ...
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It is the normal force ( https://en.wikipedia.org/wiki/Normal_force ) of the Earth against the supports that hold the platform up. Normal force is what keeps our feet from sinking into the Earth, due to our weight, it keeps a book on a table from falling through the table. It will keep the supports from falling into the Earth, the supports hold the platform, ...
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The first diagram (a) shows 3 forces acting on 3 different objects: $\vec{F}_P$ is the force of the person pushing on the rope, $\vec{F}_{BR}$ is the force on the boulder from the rope, and $\vec{F}_{CR}$ is the force on the car from the rope. To see how these forces relate we look at the free-body-diagram for the rope (b).
Newton's 3rd law says $\vec{F}_{...
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There's a tension in the rope that exerts a force in both directions.
Think about it this way: suppose you're participating in a tug-of-war with your friend. You pull him, and he pulls you. Although you're exerting a force on the rope, the rope is certainly also exerting a force on you (that's why you feel you are pulled forward). Similarly, your friend ...
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When in free fall, neglecting air drag, the only force acting on your body is gravity, which is a non contact force, and you feel weightless. In other words the feeling of “weightlessness” is that of not experiencing the sensation of any contact forces. Stand on the ground and you feel the upward contact force of the ground on your feet that opposes and ...
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It's because what you feel is distortions to your body's shape. When in free fall, when air resistance is low, every part of your body is accelerating very nearly the same amount. No difference in force between any parts of your body means your body doesn't have to exert any forces to keep its shape, so you don't feel anything.
If the gravitational field ...
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@Marco Ocram answer is essentially correct. And I think you should accept it. I would just offer an additional perspective since it appears the question is not being put on hold as a "homework and exercise" question.
Since the two blocks are constrained to have the same acceleration in response to the external force $F_{applied}$ we can then apply Newton's ...
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Mass A has to accelerate at the same rate as mass B, because they are connected by a string of fixed length. That means that the relative size of the net force on each object has to be in proportion to their masses, otherwise they would accelerate at different rates.
So T/(F-T) (ie the force on block A as a fraction of the force on block B) has to equal ...
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