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Wize High School Grade 12 Physics Textbook > Dynamics

Friction

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Friction


Friction is the rubbing force between two surfaces is caused by the atomic forces between objects. Friction force is proportional to the normal force, and is usually written as f⃗\vec{f} or F⃗f\vec{F}_f . The "roughness" which depends on the surface is captured by the unit-less coefficient of friction, denoted by μ\mu .


Ff=μN\boxed{F_f = \mu N }



Friction force is always parallel to the surface of contact. So, it is perpendicular to the normal force acting on that surface.

Static Friction

  • Static friction is the frictional force when the objects are not moving relative to each other. It is a force that opposes any applied force that tries to move the object from rest.
  • The maximum static friction is proportional to the normal force and has the following equation:

fstaticmax⁡=μsN\boxed{f_{static}^{\max}=\mu_sN}

  • μs\mu_sis the coefficient of static friction and it is unit-less
  • The actual value of the static friction can be anything up to this value, depending on how strong the applied force is:
fstatic≤fstaticmax⁡=μsNf_{static}\le f_{static}^{\max}=\mu_sN


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  • If you increase the force applied to the mass a little beyond the maximum of static friction, the mass starts to move, and the friction force decreases instantly when the body begins to move. This point is known as threshold of motion or verge of motion.






Watch Out!
A common mistake students make is that they always put static friction equal to μsN\mu_s N. But this is ONLY true when the object is about to move or slide. So, make sure you read the question carefully!


Kinetic Friction

  • Kinetic friction is the frictional force when the objects are in motion relative to each other. It is a force that opposes this motion.
  • The magnitude of this force is obtained through:
fkinetic=μkN\boxed{f_{kinetic}=\mu_kN}

  • where μk\mu_kis the coefficient of kinetic friction and it is unit-less
  • fkineticf_{kinetic} is less than fstaticmax⁡f_{static}^{\max}, so, μk<μs\mu_k<\mu_s



Wize Concept
You always need more force to start an object moving than when it's already moving.





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Rolling Friction


Friction force for a rolling object can be found using:

frolling=μrN\boxed{f_{rolling}=\mu_rN}

  • where μr\mu_ris the coefficient of rolling friction, and it is unit-less. μr≪μk<μs\mu_r\ll\mu_k<\mu_s

  • Kinetic friction is the resistance of an object to move and is related to the two surfaces.
  • Rolling friction is the resistance of an object to roll and is related to deformation.








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Example: Motion of an Object on a Rough Surface


An object with a mass of 5 kg5\ kg is moving on a horizontal surface with a coefficient of kinetic friction equal to 0.40.4. If at t=0t=0 the speed of the object is 20 m/s20 \ m/s,

a) How long later is the object going to stop?
b) What is the distance travelled by the object before it stops?

Solution:

First. let's look at the free body diagram of the object:


Part a)


vi=20 m/sv_i=20\ m/s, F=maF=ma

acceleration can be found by looking at the second law of Newton:. Note that since the object is moving we are talking about kinetic friction.

x:ma=−fk=−μkNy:0=N−mg→N=mg}ma=−μkmg→a=−μkg\left.\begin{array}{l}x:ma=-f_k=-\mu_kN\\y:0=N-mg\to N=mg\end{array}\right\}ma=-\mu_kmg\to a=-\mu_kg


=−0.4×9.81=−3.924 m/s2=-0.4\times9.81=-3.924\ m/s^2

Now we can use kinematic equations to find the time of this motion:

vf=vi+atv_f=v_i+at→t=vf−via=0−20−3.924=5.1 s\to t=\frac{v_f-v_i}{a}=\frac{0-20}{-3.924}=5.1\ s

Part b)


Again using kinematic equations we have:


vf2−vi2=2a(Δx)→Δx=vf2−vi22a=0−4002×(−3.924)v_f^2-v_i^2=2a\left(\Delta x\right)\to\Delta x=\frac{v_f^2-v_i^2}{2a}=\frac{0-400}{2\times\left(-3.924\right)}→Δx=51 m\to\Delta x=51\ m


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Example: When Will the Bass Drop?


A fisher is driving back from the shore after a day of fishing. He's loaded up a box of bass fish onto the back of a flatbed truck. The light in front of you is turning yellow already... so you accelerate (just a little bit to get through the light) at 4.0m/s2 .

What is the coefficient of static friction so that the fisher doesn't drop the bass?




If you decelerate, you want the box to decelerate with you, or else it will fly through your back window!

There is just one force acting in the horizontal direction, and that is friction. As long as the coefficient of friction is high enough, the box will stay stuck to the bed of the truck.

This is the simplest case for the normal force, giving N = mg. We find fs,max=μsNf_{s,max}=\mu_s N.

By Newton's Second Law,

ΣF=mafs,max=matruckμsN=matruckμsmg=matruckμsg=atruck\begin{array} {c} \Sigma F = ma \\ f_{s,max} = ma_{truck} \\ \mu_s N = ma_{truck} \\ \mu_s mg= ma_{truck} \\ \mu_s g= a_{truck} \end{array}
So, as long as μs>atruck/g\mu_s > a_{truck}/g, the maximum force of static friction will not be exceeded.

Practice: Block on a Slope with Friction

A mass of 5.0 kg5.0\ kg is stationary on the surface of an inclined surface with angle of θ\thetaand coefficient of kinetic friction of 0.300.30 and coefficient of static friction of 0.500.50. You start increasing the angle of the inclines surface.

a) At what angle does the mass start to slide down? (Round to the nearest whole number)
b) If you set the system at an angle which is 5 degrees more than the angle you found in part a), how much time does it take for the speed of the mass to reach 2.0 m/s2.0\ m/s? (Units are in seconds, write the answer so that it has one digit)


A 10-kg box is initially at rest on a flat surface. A horizontal force of 28 Newtons is applied on it. The static and kinetic coefficient of friction between the box and the floor are 0.32 and 0.18 respectively. What is the magnitude of the friction force?


Practice: Tugboat

A tugboat of mass 13,300 kg has an engine that exerts a force of 112 kN forward while it pulls a much larger boat behind it. This system experiences 16.1 kN of forces resisting motion (think of this as the "friction" between the boat and the water, although there would be more than friction going on in this case).

a) If the acceleration of the system is 0.700 m/s2m/s^2, what is the mass of the larger boat?
b) Calculate the force exerted by the tugboat on the larger boat, assuming that 14.7 kN of the resistive forces are experienced by the larger boat.

Practice: Pulley and Friction

You set up a pulley system as shown. There is kinetic friction acting on the top block (μk=0.175\mu_k=0.175) which has mass 4.50 kg. The vertical block has mass 10.0 kg.

Answer with three significant figures.