APT ACADEMIC SOLUTIONS : APT ACADEMIC SOLUTIONS
Slide 2 : www.aptacads.com
Session : www.aptacads.com Session Particle Dynamics - 2
Session Objectives : www.aptacads.com Friction & frictional force
Kinetic friction
Static friction
Examples Session Objectives
Friction : www.aptacads.com Friction What does it do?
It opposes relative motion!
How do we characterize this in terms we have learned?
Friction is a force in a direction opposite to the direction of relative motion or the tendency of relative motion between the two objects which are in contact with each other
Causes of Friction : www.aptacads.com Causes of Friction Friction is caused by the “microscopic” interactions between the two surfaces:
Friction... : www.aptacads.com Friction... Force of friction acts to oppose relative motion:
Parallel to surface.
Perpendicular to Normal force.
How to find direction of friction ? : www.aptacads.com How to find direction of friction ? First assume that there is no friction between the two surfaces between which we want to find the friction.
Identify the object in which the external force is acting.
The object will move along the direction of this external force.
Now friction will oppose the relative motion of this object with respect to which it is in contact.
Static Friction... : www.aptacads.com Static Friction... So far we have considered friction acting when something moves.
We also know that it acts in un-moving “static” systems:
In these cases, the force provided by friction will depend on the forces applied on the system.
Static Friction : www.aptacads.com Static Friction Just like in the sliding case except a = 0.
i : F fS = 0
j : N = mg While the block is static: fS = F (unlike kinetic friction)
Static Friction : www.aptacads.com Static Friction S is discovered by increasing F until the block starts to slide:
i : Fmax – msN = 0
j : N = mg S FMAX / mg
Friction: important points : Friction: important points There are two different friction forces: the static friction force (no motion) and the kinetic friction force (motion).
The static friction force increases with the applied force but has a maximum value.
The kinetic friction force is independent of the applied force, and has a magnitude that is less than the maximum static friction force.
Friction and Normal Forces : Friction and Normal Forces N N The maximum static friction force and the kinetic friction force are proportional to the normal force.
Changes in the normal force will thus result in changes in the friction forces.
NOTE:
The normal force will be always perpendicular to the surface.
The friction force will be always opposite to the direction of (potential) motion.
Friction and Normal Forces : Does friction depend on area of contact ? A F1 > F2 B F1 = F2 C F1 < F2 fmax = S N Friction force does not depend on area! fmax = S mg Friction and Normal Forces
Angle of repose : www.aptacads.com At qcrit
F = f
mg sin qcrit = f = mS N Independent of m, or g. Property of surfaces only mS = tan qcrit = mS mg cosqcrit Angle of repose qcrit is called angle of repose
Which is harder-pushing or pulling ? : Which is harder-pushing or pulling ? N N
Solution : www.aptacads.com Solution (Ffri)push > (Ffri)pull
What force drives/stops the car? : www.aptacads.com What force drives/stops the car?
Stopping Distance depends on friction : www.aptacads.com Braking force
Friction road/tyres v2 =vo2 + 2a(x-xo)
0 = vo2 + 2ad amax = - sg Stopping Distance depends on friction
Stopping Distance depends on friction : www.aptacads.com dmin depends on v2!! Take care!! If v0 = 90 kph (24 m s-1) and m = 0.6 ==> d = 50 m!! Stopping Distance depends on friction
Example : www.aptacads.com Example A 0.2 kg hockey puck sliding over ice stops because of the frictional force of ice after covering a distance of 14 m.
a. If the initial speed is 7 m/s, what is the magnitude of the frictional force?
b. What is the coefficient of kinetic friction between puck and ice?
Solution : www.aptacads.com Solution Applying v2 – u2 = 2aS , we get
0 – 7 × 7 = 2 a × 14 N =mg u = 7 m/s, v = 0, s = 14 m
Example : www.aptacads.com Example A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is m. The force required to initiate horizontal motion of the block is maximum when
(a) Lift is moving up with constant acceleration
(b) Lift is moving down with constant acceleration
(c) Lift is stationary
(d) Lift is in free fall
Solution : www.aptacads.com Solution F is largest when N is maximum N is maximum when lift moves up with constant acceleration a [N = m(g + a)]. So F is maximum at that condition.
Example : www.aptacads.com Example A 70 Kg box is pulled along a horizontal surface by a 400 N force at an angle of 30° above horizontal. If the coefficient of friction is 0.50, what is the acceleration of the box? (g = 10 m/s2)
Solution : www.aptacads.com Solution Vertical: N = mg – F sin30 Horizontal: F cos 30 – f = ma Solving for a: a = 1.37 m/s2
Slide 27 : www.aptacads.com