FRICTION - GEN PHYSICS, Lecture notes of Physics

Download FRICTION - GEN PHYSICS and more Lecture notes Physics in PDF only on Docsity!

MODULE 6: friction

FRICTION

Friction – a force that always acts to resist the motion of one object sliding on another.

Principles of Friction

1. Frictional force is independent of the contact area.

Frictional force is unchanged whether the object lies on its side or bottom.

2. For a given pair of surfaces, frictional force is proportional to normal force.

F (^) s max  N

Normal force (N) – the force exerted by the surface to the object placed to it. This force is always perpendicular to the surface.

3. The number relating maximum static friction (fsmax) and normal force (N) is called the coefficient of static friction (s) is defined by fsmax = s N

s = fs max N where :

 - coefficient of friction ff – frictional force fs – static friction fk – kinetic friction N – normal force

Solution:

Since the speed is constant, the total force acting on the body is equal to zero.

 F = 0

horizontal forces vertical forces

 Fx = 0  Fy = 0

Fa + (-)ff = 0 N + (-)w = 0

Fa = ff N = w

w = mg = (40kg) (9.8 m/s^2 = 392 kg m/s^2 or N

A. k = ff = fs ; deriving the formula of static friction, we obtain N N

fs = k N

= (0.5) (392 N)

= 196 N since fs = Fa, the force needed to start the object moving is 196 N.

B. Fk = k N

= (0.3) (392 N)

= 117.6 N ; therefore to keep the object moving a force of 117.6N is needed at constant speed.

Problem #2: A block is pulled to the right at constant velocity by a force of 10 N acting 30^0 above the horizontal. The coefficient of sliding friction between the block and the surface is 0.5. What is the weight of the block?

N Fa = 10 N

Fk 3  = 30^0

k = 0.

w =?

Since weight is unknown, all forces acting in the system vertically must be known. First, the must calculate components of applied force.

Fax = cos  Fay = sin  Fa Fa Fax = Fa cos  Fay = Fa sin 

Fx = 0 Fy = 0

Fax + (-)fk = 0 N + (-)w + Fay = 0

Fax = fk w = N +Fay

Fa cos  = fk w = N + Fa sin 

N

Fa = 300 N

m = 50kg

ff θ

θ =30^0 w

Horizontal component of weight

wx = w sin 

Vertical component of weight

wy = w cos 

Fx = 0 Fy = 0

F + (-)ff +(-) wx = 0 N + (-)wy = 0

F - wx = ff wy = N

F - w sin  = ff w cos  = N k = ff N

 (^) f = F - w sin w cos = 300N – ( 490N) (sin 30^0 ) (490N) (cos 30^0 )

= 300N – (245N)

(424N)

Activity: Perform the following:

1. A block weighs 2000 N is being dragged on a horizontal surface by a force to keep it in constant motion. If the coefficient of friction between the block and the horizontal surface is 0.06. What is the magnitude of the force applied?
2. A 120 lb block is placed on a plane inclined at an angle of 35^0. The coefficient of static friction is 0.25; the coefficient of kinetic friction is 0.15. What force parallel to the plane is necessary (a) to hold the block from sliding down the plane; (b) to pull the block up the plane with uniform speed?

Note: If Yes, go to the quiz section, If No,

review our lessons and ask for an additional

activities before taking the quiz.

Ready for the QUIZ guys?