Friction And Wedges

1. INTRODUCTION TO
TECHNOLOGY B
FRICTION AND WEDGES

2. FRICTION
A block of weight W is placed on
a horizontal plane surface. The
force acting on the block are its
Weight and the reaction force of
the table. Since the weight force
has no horizontal component, the
reaction of the surface also has
no horizontal component. The
reaction is therefore, normal to
the surface and is represented by
the normal force N

3. FRICTION
Suppose now that a horizontal
force P is applied to the block. If
P is small the block will not
move, some other horizontal
force therefore must exist which
balances P. this other force is
known as the STATIC FRICTION
FORCE.

4. FRICTION
If the force P is increased, the
friction force F also increases,
continuing to oppose P, until
its magnitude reaches a
certain maximum value Fm. If
P is increased the friction
force cannot balance it any
more and the block starts to
move. As soon as the block
starts to move the value of F
drops to a lower value of Fk
and is known as the kinetic
force of friction

5. FRICTION
Fm of the static friction force is
proportional to the normal
component of N of the reaction
of the surface and so we have:
Fmax = μ
s
N
Fkinetic = μ
k
N
Where μ
s
and μ
k
are called the
coefficient of static and kinetic
friction respectively.

6. ANGLES OF FRICTION
Tan Φs = μs
Tan Φk = μk
If the applied force P has a horizontal
component Px which tens to move the
block, the force R will have a
horizontal component F and thus will
form an angle Φ with the normal to the
surface. This value is called the angle
of static friction and is denoted Φs.
Then from the geometry
Tan Φs = Fm / N

7. FRICTIONAL FORCES ON A
INCLNED PLANE

8. SAMPLE PROBLEM
A sand sled of mass 10kg is used on a
sand dune having a lope of 15°. If the
coefficient of friction is 0.35, what force
must the boy exert on the tow rope to
drag the sand sled up the dune?

9. SAMPLE PROBLEM
FBD
P – W sin 15° - Fr = 0
Since Fr = μ N
= 0.35 mg cos 15°
= 0.35 (10)(9.8) cos 15°
= 33.1N
P – 9.8 sin15° - 33.1 = 0
P = 58.5N

10. Sample Problem A support block is acted upon the by two forces as
shown. Knowing that coefficients of friction between the block and
the incline plane are μs = 0.35 and μk = 0.25, determine the force
P required to (a) to start the block moving up the inclined plane
(b) to keep it moving up
(c) to prevent it from sliding down.
Solution
For each part we draw the FBD of the block
and a force triangle including the 800N
vertical force, the horizontal force P and the
force R exerted on the block by the inclined
plane. The direction of R must be
determined in each separate case.

11. Sample Problem A support block is acted upon the by two forces as
shown. Knowing that coefficients of friction between the block and
the incline plane are μs = 0.35 and μk = 0.25, determine the force
P required to (a) to start the block moving up the inclined plane
FBD
Tan Φs = μs
= 0.35
Φs = 19.29°
25° + 19.29° = 44.29°
Force P to start block moving up
P = (800N) tan 44.29°
= 780N

12. Sample Problem A support block is acted upon the by two forces as
shown. Knowing that coefficients of friction between the block and
the incline plane are μs = 0.35 and μk = 0.25, determine the force
P required to (b) to keep it moving up
FBD
Tan Φk = μk
= 0.25
Φk = 14.04°
25° + 14.04° = 39.04°
Force P to keep block moving up
P = (800N) tan 39.04°
= 649N

13. Sample Problem A support block is acted upon the by two forces as
shown. Knowing that coefficients of friction between the block and
the incline plane are μs = 0.35 and μk = 0.25, determine the force
P required to (c) to prevent it from sliding down.
FBD
Φs = 19.29°
25° - 19.29° = 5.71°
Force P to prevent block sliding down
P = (800N) tan 5.71°
= 80.0N

14. The Wedge
A wedge is one of the simplest and
most useful machine. A wedge is used
to produce small adjustments in the
position of the body or to apply large
forces. Wedges largely depend on
friction to function.

15. Simple application example
A wedge used to position or lift a
large mass m
The FBD showing the force
triangle for the mass and
the wedge.

16. Sample Problem
A heavy block of mass 500kg is to be lifted
vertically by the system of two wedges as
shown. If the coefficient of the static friction
for the all the surfaces is 0.3, determine the
force P which will cause the system to be on
the point of movement.

17. Sample Problem
Forces on the blockTriangle of forces for the block
For the block:
2 = R2 sin
56.6° sin106.7°
R2 = 5667N

18. Sample Problem
Forces on the top wedge.
Force triangle for the wedge
P =5667
Sin 39.4° sin 67.3°
P = 3899N

19. Tutorial Questions
A Piece of flat steel is clamped in the jaws of an
engineers vice with the clamping force of 600N. If
the coefficient of friction (μ) = 0.3, what minimum
pull on the steel would just cause it to move?
The horizontal position of the 500kg
rectangular block of concrete is adjusted by the
5° wedge under the action of the force P. is the
coefficient of static friction for both pairs of
wedge surfaces is 0.30 and if the coefficient of
static friction between the block and the
horizontal surface is 0.60, determine the least
force P required to move the block.

20. Tutorial Questions
If the coefficient of the friction between
the steel wedge an the moist fibers of
the newly cut stump is 0.20, determine
the maximum angle which the wedge
may have and not pop out of the wood
after being driven by the sledge.
Determine the range of values which
the mass mo may have so that the
100kg block neither start moving up the
plane nor slip down. μs = 0.30

21. Tutorial Questions
A 100N force acts on a 300N block placed
on an inclined plane. Determine whether
the block is equilibrium and find the
value of the friction force. μs = 0.25 and
μk = 0.20