Mechanical Engineering Principles

FRICTION 173

15.4 Friction on an inclined plane

Angle of repose

Consider a massmlying on an inclined plane, as
shown in Figure 15.3. If the direction of motion of
this mass is down the plane, then the frictional force
F will act up the plane, as shown in Figure 15.3,
whereF=μmg.

q

q

F

mg

N

Direction of motion

Figure 15.3

Now the weight of the mass is mg and this will
cause two other forces to act on the mass, namely
N, and the component of the weight down the plane,
namelymgsinθ, as shown by the vector diagram of
Figure 15.4.

q

q

W= mg

mg cos q

mg sin q

Plane

Figure 15.4 Components of mg

It should be noted that N acts normal to the surface.
Resolving forces parallel to the plane gives:

Forces up the plane = forces down the plane

i.e. F=mgsinθ( 15. 1 )

Resolving force perpendicular to the plane gives:

Forces ‘up’ = forces ‘down’

i.e. N=mgcosθ( 15. 2 )

Dividing equation (15.1) by (15.2) gives:

F

N

=

mgsinθ

mgcosθ

=

sinθ

cosθ

=tanθ

But

F

N

=μ, hence,tanθ=μ

whereμ=the coefficient of friction, andθ=the

angle of repose.

Ifθ is gradually increased until the body starts

motion down the plane, then this value ofθ is

called thelimiting angle of repose. A laboratory

experiment based on the theory is a useful method of

obtaining the maximum value ofμfor static friction.

15.5 Motion up a plane with the

pulling forceP parallel to the

plane

In this case the frictional forceF acts down the

plane, opposite to the direction of motion of the

body, as shown in Figure 15.5.

Motion P

F

q

q

N

mg

Figure 15.5

The components of the weight mg will be the same

as that shown in Figure 15.4.

Resolving forces parallel to the plane gives:

P=mgsinθ+F( 15. 3 )

Resolving forces perpendicular to the plane gives:

N=mgcosθ( 15. 4 )