Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction:
Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact. Dry friction is subdivided into static friction ("stiction") between non-moving surfaces, and kinetic friction between moving surfaces. With the exception of atomic or molecular friction, dry friction generally arises from the interaction of surface features, known as asperities (see Figure 1).
Fluid friction describes the friction between layers of a viscous fluid that are moving relative to each other.Lubricated friction is a case of fluid friction where a lubricant fluid separates two solid surfaces.Skin friction is a component of drag, the force resisting the motion of a fluid across the surface of a body.
Internal friction is the force resisting motion between the elements making up a solid material while it undergoes deformation.When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire. Kinetic energy is converted to thermal energy whenever motion with friction occurs, for example when a viscous fluid is stirred. Another important consequence of many types of friction can be wear, which may lead to performance degradation or damage to components. Friction is a component of the science of tribology.
Friction is desirable and important in supplying traction to facilitate motion on land. Most land vehicles rely on friction for acceleration, deceleration and changing direction. Sudden reductions in traction can cause loss of control and accidents.
Friction is not itself a fundamental force. Dry friction arises from a combination of inter-surface adhesion, surface roughness, surface deformation, and surface contamination. The complexity of these interactions makes the calculation of friction from first principles impractical and necessitates the use of empirical methods for analysis and the development of theory.
Friction is a non-conservative force – work done against friction is path dependent. In the presence of friction, some kinetic energy is always transformed to thermal energy, so mechanical energy is not conserved.
So, ignore the -0.72, I was just trying to see if I had a sign error (I then remembered magnitude is absolute value) but basically:
Since the weight is 12.0N, theta is 53.1, and the coefficient of kinetic friction is 0.100, I just plugged those values into the equations above...
I don't undertand the equation. It is Newtons's second law of motion, so it decribes a force that acts on a single disc relative to the ground. So when the force is proportional to velocity, shouldn't it be ##-bv##? Because the dics's velocity is ##v## relative to the ground. Relative to the...
First i show the sketch of the setup:
My first attempt was just to balance out the forces on the box. On the sketch below i have shown the situation where the spring is stretched distance L.
In this situation we get the equations:
Which when solved leads to
All good. I then looked at the...
For this,
I don't understand why they don't have a negative sign as the torque to the friction should be negative. To my understanding, I think the equation 5.27 should be ##I\frac{d \omega}{dt} = -F_{friction}R## from the right hand rule assuming out of the page is positive.
Noting that ##f_k...
Simple question. Let's say a solid cylinder has an initial speed ##v_o## and it's rotating on infinitely hard ground without air resistance.
The cylinder will come to a stop eventually. There are two sources of friction.
Since the wheel/cylinder is deformed at the contact patch, there is some...
(mentor note: moved from Classical Physics forum hence no template)
Hello, I am having trouble with this question: Imagine in real life there was a coefficient of kinetic friction of 0.4 between the plastic wheels of the cart and the wooden ramp. If there is only friction on the flat part of...
Consider the standard pendulum with a weightless rod of length b and a mass point m and mg is applied. In the hinge there is a torque of viscous friction which is proportional ##\omega^2##.
Now release the pendulum from the horizontal position. What biggest height does the point m attain after...
Let's say that the mass of the objest is suddenly bigger, so when I want to maintain the constant movement, my force must increase as well. But will the velocity have the same magnitude? I think that the velocity will be smaller, so if I want to have the same initial velocity, I must apply an...
Homework Statement: A door opened at 75 degrees from the latch remains at rest. When moved to a position 60 degrees from the latch, it slowly closes on its own. Is there another possible cause besides a sloping floor?
Relevant Equations: \tau = r x F; v = \omega x r; a_T = \alpha x r, where...
We have 2 objects, m1 and m[SUPlB]2[/SUB]
Friction is present between the two objects but not between m1 and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of...
Topography of both the object and the surface.
Mass/inertia.
Moisture, but that can probably fall under topography.
I suppose atmospheric pressure, maybe. Or wind.
Magnetism.
Any others?
Question picture:
My solution:
Where:
S is the lineforce
Ff is the force as a result of friction
a is the resulting acceleration
F is the acting force
The answear is supposed to be a=(F-2mg(mu))/(m+M)
Any idea what i could have missed?
Thanks for your help on beforehand!
The question was this:
My calculations show that the answer should be equal to work done on crate to make it reach the same velocity which is equal to 216 J but the answer given is 432 J
It is believed that extra energy is needed to overcome friction but friction is an internal force and...
My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt:
Let block 1 be the vertically moving block and let block 2 be the horizontally moving one.
Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0...
The answer is (D), but I don't understand why.
Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts:
1) motion on snowy surface
Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction...
So basically I need to find the coefficient of friction given the listed information.
What bothers me is that I am getting two different accelerations for two different approaches. When I calculate acceleration using Fg=mgsin60 I do it this way: Fg=mgsin60 -> ma=mgsin60 ->a=gsin60 -> a=8.66. But...
So I have difficulty understanding this https://physics.stackexchange.com/questions/331173/why-is-friction-only-on-the-back-wheel
With constant speed cycling uphill the front wheel has no friction. But if that would be the case how does the wheel even spin?
First of all, the pulling force is
300N cos(30) = 260 N
At this point, I try to find the friction force
Fn = mg = 20kg * 9.81 m/s^2 = 196.2 N
Then,
Ff = μ * Fn = 0.5 * 196.2 N = 98.1 N
So after canceling the horizontal forces,
260N - 98.1N = 161.9N
And the acceleration will be 161.9N / 20kg =...
There is a well-known Abraham-Lorentz equation describing radiative friction. Suppose a particle moves in an electromagnetic field.
ma(t)=q(E+vxB) + m(tau)a’(t)
By solving this equation numerically, I get non-physical solutions(runaway solutions) Although, it would seem that an electron in an...
We want to figure out how much work friction does on a block as it slides down an inclined plane with a rough surface.
we find the force due to gravity that pulls the block down the ramp, that's found by M * g * sin(θ),
The normal force on the block is given by M * g * cos(θ).
The force of...
So the only problem I am having is determining the direction of static friction. I did the same problem but while they were going in a vertical circular motion instead, where the static friction force was in the direction of centripetal force (pointing to the center of the circle).
Would it be...
The block starts to slide if friction can no longer hold the block.
F=u*n and F=(m1+m2)a
so: (m1+m2)a=uN=>am1+am2=uN=>am2=(uN)/(am1)
So:am2=(uN)/(am1) is the force.
The answer is F=(u*m1g(m1+m2))/m2
I do not see how the acceleration terms are canceled. Is my answer equivalent to this?
Are we allowed to use free body diagrams in dynamic systems where there is non-negligible friction in say the rotational pivot between two linkages? If so, how to incorporate friction into the equations of motion? If not, what method allows friction to be accounted for?
I reason the frictional force on the plate from the ice is doing work first 3 meters (while the motorbike is moving on top) and then an "x" distance after the motorbike has left it. Does anybody have an idea of how one might solve this problem?
I’m writing a presentation for younger students explaining how science can help them satiate their curiosity in a fun way. This presentation’s goal is to apply as much science as possible to a ridiculous question in hopes of promoting STEM fields. My example is, how many capybaras --the world’s...
I have a flat planar part made of crystalline sapphire (about ~2k weight, and polished to a mirror finish) that rests on three ball bearings, and I want to minimize the static friction at these 3 interfaces. The ball bearings are fixed so they cannot roll, and the sapphire part can only slip...
I'm unsure on where to begin with this question, i've tried many different formulas that aren't giving me the right answer. I believe to start I need to convert the kilo newtons to newtons.
I tried w = fs + mgh
w = 10500 x 8.9/sin(13.9)+(1845.69 x 9.8 x8.9) = 549986.46 J
and then convert to...
First, I assumed that the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N##
It's the first problem...
Hello Everyone
I want to model forces affecting on syringe plunger , but I do not know how to calculate terms like friction and damping coefficient.
What I imagine is that : F_driving = ma + cv + f ----------------(1)
where:
f: friction
c: coefficient of viscous damping
m: mass of plunger (is...
Hi everyone,
i have been trying to find an answer to this problem I have but couldnt find any good answers...
(I dont know much about this stuff, but need a formula for a Project I am currently working on).
So The problem goes as follows:
Assuming we have a ball with a mass of m which is...
Ok, logically, it must be that the static friction force of block A equal to the force of gravity on block B, so mass of block A is:
m_A * 9.8 * 0.30 = m_B * 9.8
m_A * 2.94 = 2 * 9.8
m_A * 2.94 = 19.6
m_A \approx 6.7 kg.
However, when I look at block A individually, there is one thing...
Young & Freedman 13th ed, Exercise 7.81
Starting with the crate, here is its free-body diagram:
In accordance with Newton's First Law:
$$ \Sigma F_y = 0 = n+(-w_c \cos{\alpha}) $$
Thus ## n = w_c \cos{\alpha} ##.
And according to Newton's Second Law:
$$ \Sigma F_x = m_c a_x = w_c...
I don't understand part (b)
In part (a), I need to calculate the coefficient of the static friction:
mg * \mu_static = 35
58.8 * \mu_static = 35
\mu_static = 35 / 58.8 \approx 0.6
So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot...
Since there is no friction : $$ m \ddot{x} = 0 $$ (no x motion).
For the kinetic energy , I've tried: $$ K = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m v^2_cm = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m \dot{z}^2$$ . Giving me a weird expression , shouldn't the kinetic energy just be half the the...
Initially I tried to explain it via kinetic energy of the object and how the frictional force can only do as much work on the object as the object has kinetic energy but I got confused. Could someone here please explain why if I have an object with a net force due to friction (and no other...
The answer is .32m. I set the elastic potential energy as equal to the work, but at first I put the force in the work equation as (F elastic - F kinetic friction) times distance and rearranged.
1/2kx^2 = (kx-Ff) d
(0.5) (22) (0.035)^2 = (22 x 0.035-0.042) d
0.013475= 0.728 d
0.013475/0.728 = d...
Dear all,
Me and some colleagues (non-physicists) are discussing how force works when passing a cylinder (which we are holding) into a narrow tube. As we insert more of the cylinder into the tube, the force we are exerting is increasing. My theory is that the normal force is increasing and his...
Hello,
In reviewing friction, I realized a couple of things: the coefficient of static friction can be larger than 1 (always thought it would be smaller than 1), that the surface area does not matter for static and kinetic friction ONLY for simple solid, rigid objects ( for materials like...
TL;DR Summary: Distance traveled by a car considering only air friction?
How much distance would a 3-ton car travel if its initial speed was 17 km/h and we only take into account air's friction? (Assume that the car has an airfoil-like shape, so that the resistance against the air is very low)...
When I exert enough force that overcome the static friction. The object start moving and surface create kinetic friction on object if I exert harder overcome the maximum of friction it start accelerate. When I release the object will the kinetic friction disappear immediately proportional to...
When we push an object on the surface, if I push hard enough such that it reach the maxium static friction of the object then it start moving with a constant speed and F_push = F_friction. But when I realease the object F_push immediately become zero remain only F_friction. Does the friction...
Friction happens because of adhesion between high points of the pertubrances of the two surfaces. The pertubrances deform. More the force between surfaces more deformation. Is the deformation elastic or inelastic? Will the surface of one body become smoother if pressed with hydraulic press?
Hello,
When we consider a block sitting on a surface, the gravitational force ##W## and the normal force ##F_N## are applied to the block. Both equal i magnitude and opposite in direction. We call the normal force the reaction force exerted by the surface on the block.
Now we consider the...