# Tag Info

## New answers tagged friction

0

The friction should be optimum. The cars surely run better on roads than in sand. The cars have tires so the friction is less and the friction of cars mainly depend on normal force than roads. More rough roads will increase grip during acceleration but as the car starts moving in a constant velocity the roads will grip the tire and drag. Cars will not be ...

0

Normal force $F_N$ is just the force between two surfaces. It's called "normal" because it acts perpendicular (normal) to the surfaces. Gravitational force is completely unrelated. Gravity always acts with $F_g = -mg$. The minus sign indicates that the force points down. These two forces often oppose each other, which is why $F_N$ OFTEN, BUT NOT ALWAYS, ...

0

Normal Force arises due to the Newton's Third law. Normal Force will be always acting opposite to the force falling on the surface. Normal Force is a reaction force. Remember Normal force is equal to mg only when the object is placed horizontally, and the force is acting in the direction of the gravitational field. Now your second question Here you ...

0

Briefly, the normal force is $F_N=mg$ when the surface that mass $m$ is resting on is horizontal (when the surface is inclined by an angle $\theta$ to the horizontal, then it's just $F_N=mg\cos\theta$). Friction has nothing to do with $F_N$, per se. But the frictional force experienced by $m$ sliding down an inclined plane is the coefficient of (kinetic) ...

0

There is a maximum fricction force, and is F=uN. u is the coefficient of friction and N the paralel component of the weight to the normal. If the car is at rest, a higher coefficient makes a better accelration. But if the car has a velocity v, only will be a diference between coefficient at a range. If U is the maximu coefficient at that range, ...

0

You arrived at almost correct answer. I get net horizontal accelerating force as: $$F_x(\theta) = F_{pull}(\cos\theta+\mu\sin\theta) - \mu m g$$ You are just missing the $\mu$ before the sin and it also has a different sign. Double check your diagram, especially the orientation of forces. But nevertheless you still seem to arrive at almost correct ...

0

Drawing a free body diagram, the following equations can be found: $F_y = F_\text{pull}\sin(\theta) + F_\text N - mg$ $F_x = F_\text{pull}\cos(\theta) - F_\text R$ Stipulations on variables: $F_\text N$ (normal force) must be positive; the ground cannot exert an attractive force. $F_\text R$ (frictional force) must be less than or equal to $\mu$ ...

1

To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0.$$ But if you insert the actual values from your ...

0

For an object to move in a constant velocity the zero acceleration then the Forces should be zero. The $$\mu=\frac{65*g*\sin 30}{65*g*\cos30}=0.57735$$ Your answer is perfectly true. $$\mu<1$$ if your question is why does the object still move even when force is zero. Example is Terminal Velocity.in air. The reason the object comes down is that there is ...

0

Let $\theta$ is the slope angle. The normal force is $F_{N} = mg\cos{\theta}$, which you have calculated. Now there are three important points to consider. The object is in constant speed, so $F_{net} = 0$. There are three forces acting on the object $F_{friction}, ~F_{gravity} ~and ~ -F_{normal }$ whose net sum is zero to justify (1). $F_{friction}$ by ...

0

In simplest terms and using Newton's mathematics: F = m * a. or Force F = mass m * acceleration a. Example #1 - A body on Earth. Now on the planet Earth, the gravitational acceleration "g" is about equal to 9.8 meters/second^2. So let's substitute a=g in the above equation. Then the force required to keep an object of mass m AT REST near the surface of ...

0

It's important to consider exactly where the forces of friction are being applied in a typical rising elevator. In most cases (I would hope), the elevator car isn't just scraping along the walls of the elevator shaft on its way up. That would make for a very noisy and expensive ride to the top, I would think. Elevators are more complex from a kinematics ...

0

there is chemical reason behind it. interlocking is the reason for the friction b/w rough surfaces(either) but in case of super smooth surfaces, interlocking can't take place. chemically, when two super smooth surfaces comes in contact with each other , their chemical structures disturbs. then their outet shells get merged into one another so they oppose ...

0

Yes, friction force is $F_r=\mu N$, where $N$ is the normal force exerted by the floor on the object, Here $N=m(g + a)$ So yes friction force also increases

2

I gather that the large source of error you are worried about is the ability of the experimenter to accurately hit the start/stop button on the stopwatch at the start/stop of the ball's journey down the ramp. What is the approximate magnitude of error we'd expect? Before I directly answer your question, let's estimate how bad the experimental error will be ...

1

The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both $m$ and the spring is characterized by constant $k$ the angular frequency of ...

0

Yes it will. The spring behaves like friction here. When the first block is pushed, it transfers the energy to the spring which converts the kinetic energy to its potential energy. Once the second block overcomes its inertia, it will also start to move. Think of two blocks on a surface with friction without a spring in between. Pushing the first block ...

1

Let's look at this problem from the point of view of equations of motion, see diagram below: Firstly let's make a few assumptions. Ball and cube are of same weight ($mg$) and same size. Simple friction model $F_f=\mu F_n$ holds and $\mu$ is independent of speed. Both objects are completely stationary (no sliding, rolling or tumbling) at $t=0$, at which ...

0

This is a very common question asked by students like me in mechanical engineering. The friction is irrespective of area means the friction generated, the vector F (The letter with which we denote) will form whether area in contact is less or more But the ability to stop is determined by the Number of friction vectors developed per area of contact. On the ...

0

Whenever one applies a sideways force trough the center of gravity of an object, that force has two components: 1) a direct force that tries to overcome friction and slide the object, and 2) a torque that uses friction to produce a rotation of the object by lifting its center of gravity over the leading edge. A short, flat object will tend to slide ...

0

Your situation is dynamic in that the ball accelerates under the force of gravity, initially at rest but then reaching a steady state, terminal velocity, similar to what a skydiver experiences when jumping from an aircraft. The terminal velocity is reached when the force of gravity is in balance (equilibrated) with the viscous drag force imposed by the flow ...

6

The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1) $$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$ rather than just $$\tag{2} {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. ... 0 In your equation (1), T = 8.02 m and in your equation (4) -T\sin{21^{\circ}} + \mu mg - \mu T \cos{21^{\circ}} = 0. Rearranging (4) and then substituting the value of T from (1) gives$$\mu = \frac{T\sin{21^\circ}}{mg - T\cos{21^\circ}}= \frac{8.02(0.358)}{9.8-8.02(0.934)}=1.24$$There is a minus sign in the denominator. You must have added instead of ... 0 I did not quite understand your problem in solving a simple question. But i think i get your doubt. The friction when ever taken in the work energy theorem is always taken as negative and as the work is done against the direction of the gravity the work is negative. here do not worry about the potential energy just calculate the work done when you equate ... 0 As the total work on the system equals the change in kinetic energy:$$W_{Friction}+W_{Potential} = \Delta K$$Taking:$$v_{i}=v_{f}=0$$we can write$$W_{Friction}+W_{Potential} = 0W_{Friction}+\Delta V = 0W_{Friction}-V_{i} + V_{f} = 0W_{Friction} + V_{f}= V_{i}E_{i} = W_{Friction} + E_{f}$$obs: Work by definition is$$\int ...

0

Yes and no, but I think in your context the answer is yes: only account for friction work. This is a good place to consider that energy accounting depends on the choice of system. In introductory courses, potential energy is usually introduced as gravitational potential energy, $mgh$. IMO, this is dangerous because it can lead to confusion, as we see in ...

0

When you have a non conserving force in the system(such as friction) an energy only approach only gives you an estimate of the results. It's not a particularly good way to look at the situation. The initial energy is going to be the kinetic energy. The resulting energy will be the gravitational potential less the friction losses. The frictional losses ...

2

Consider the diagram below of a ball on a horizontal surface: Newton's Laws tell us that if no net force acts on the ball it will remain in a constant state of motion ($v=0$ or $v=\text{constant}$). Consequently, if no net torque acts on the ball its state of rotation will also remain constant ($\omega=0$ or $\omega=\text{constant}$). Where friction does ...

0

From the linked Wikipedia page: The term has a related usage in mechanics, where it refers to the maximum angle at which an object can rest on an inclined plane without sliding down. This angle is equal to the arctangent of the coefficient of static friction μs between the surfaces. Given that you are asking about solid objects rather than granular ...

2

Your equation for the damped solution is wrong. In order to match the boundary conditions (initial velocity = 0) you have to add either a phase, or a $\sin$ term. I prefer the phase. If initial velocity is zero, the derivative must be zero: $$A_t = A_0 e^{-t/\tau} \cos(\omega t + \phi)\\ v_t = A_0\left(-\frac{e^{-t/\tau}}{\tau}\cos(\omega t + \phi) - ... 0 If the two surfaces of the same material are very smooth, the molecules and atoms of one surface get closer to the molecules and atoms of the other surface. As they get closer the molecular cohesive forces begin to take effect and the two surfaces will actually stick to each other. If the surfaces are rough there are far fewer molecules that are close but ... 1 It depends on the surfaces how high the friction becomes. If they are smooth and clean enough how do they "know" that they are separate surfaces? It is possible for them to actually weld into a a single part. Although this is difficult to achieve in practice - it is annoying when you don't want it to happen. 0 HINT: Torque on the rod about point on the ground, Since the rod is not rotating the \tau_{net}=0$$ Tl\cos69-mg\frac l2\cos42=0\\ $$Forces acting on the rod, Since the rod is not slipping F_{net}=0$$ N=mg-T\sin69\\ T\cos69-F_{friction}=0\\ F_{friction}=\mu N\\  F_{friction}=\mu(mg-T\sin69)\\ T\cos69-\mu(mg-T\sin69)=0 $$There you go now I ... 0 The answer you have found is in fact the correct one, including the range of values; the question does not give enough information to be any more exact. Consider that you are pulling this block up the slope, and decide that you need to take a break. You know that gravity is exerting 170\text{ N} down the slope at all times, so you lower the tension ... 0 The above explanation is generally correct: usually there only exists positive friction and frictional force is always in the direction opposite of the direction of motion. However, that explanation considers only positive coefficients of friction. There has been recent (2013) research with graphite that demonstrates a negative coefficient of friction. ... 0 I think you're failing to distinguish between static friction and kinetic friction (the latter also called dynamic friction), e.g., http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/ben_townsend/StaticandKineticFriction.htm Briefly, once moving, it takes less force to keep an object moving against friction than it took to get that object moving in the first ... 2 Forces are only affecting acceleration. Not any other parts of motion. Think of Newton's 2nd law:$$\sum F=ma If the sum of forces is zero as in your case, nothing accelerates. If it was standing still, then it continues standing still. If it was moving then it continues moving! It only takes a force to change a motion. Not to keep it up.

1

Answering your Title question. NO, the objects do not change their state of motion if all the forces cancel each other. I will assume that the friction on the object is maximum. That means $200N$ is the maximum friction on the surface. Now the force needed is $200N$ to overcome the static frictional force. So if initially you apply $200N$ force, there will ...

1

Because the work done by friction is converted into rotational kinetic energy of the cylinder, since friction provides the torque to roll down the cylinder.

1

I assume that friction is an external velocity dependent force in your simulation code. Since you have such external forces, your total energy, total angular momentum, total momentum are likely not to be conserved. In your case, the friction is a phenomenological external force, but similar behavior could also be simulated with a large particle, moving in a ...

3

According to conservation of momentum, the center of mass of a system cannot accelerate without external forces. In other words, if the center of mass starts out at rest (which is generally a good procedure in simulations), then it should always stay at rest. It is normal for numerical errors to introduce deviations, but the motion you are seeing looks ...

0

I believe that firstly we need to get one thing straight and that is, on infinitesimal level, where is the acceleration of a body moving on some circular path pointing? The answer is, it is always pointing away from the center of curvature of your path, road, or whatever. So then, there is always a force in action, force which is trying to move you ...

1

If you are sliding across the surface, then "static friction" is not applicable. Consider first your motion on a merry go round without sliding. At any instant, your tangential velocity is the same as the tangential velocity of the surface under your feet. Since the two velocities are the same, no instantaneous frictional force is required to keep you moving ...

4

Friction causes the chalk to stay on the chalkboard. While the chalkboard appears smooth, under a microscope its surface is rough. Chalk is a much weaker material than the chalk board. When it is forced across the chalk board, small parts of chalk ('dust') are broken and remain trapped by friction in the surface asperities of the chalk board. The rougher ...

0

When two objects are pressed together and then slid against each other there is a force which opposes the 'sliding force' called friction. This force is parallel to, and located at, the contact plane of the surfaces. It is also opposite in direction to the 'sliding force'. The part of the "moving force" which contributes to the frictional force is ...

1

The idea is that, roughly speaking, the block moves down the gradient of the slope. Because the slope changes its direction of motion, it pushes the block left and right in roughly equal measure, and because it has a short period, the block never moves with significant horizontal velocity relative to the fixed axes. Thus, the horizontal speed of the block ...

2

When talking theoretically, an ideal flywheel rotating clockwise or anticlockwise would face the same magnitude of frictional forces in opposite directions. But if the flywheel is made such so as to rotate in one direction only or has worn out and/or has been damaged might produce different magnitudes of frictional forces. The surfaces which is rotating on ...

0

Firstly, a more general advice(something that was told to my class by my professor): The "opposite to the motion" direction of friction is not the best way to see it. In fact, nothing in physics should be viewed as being an absolute rule except from the very basic foundations of physics, which are its laws. One case in which friction is not opposite of ...

Top 50 recent answers are included