1. UNIT - V
FRICTION

2. Friction Dry friction (Coulomb friction)
Friction forces oppose the tendency of contacting surfaces to slip one relative to the other.
Dry friction- the tangential component of the contact force.

3. Friction
Forces have to be concurrent in order
to be in equilibrium.

4. Friction forces as a function of P
Maximum value of friction forces is called the limiting value of static friction. This condition is also called the impending motion.

5. Tip vs. slip
P2>P1 F1 F2
Body Tips

6. The magnitude of the friction forces
Fmax=msN
Where ms is the coefficient of static friction.
It is independent of normal forces and area of contact.
F<msN
The general case for equilibrium condition
Once the body starts to slip
F=mkN
mk is the coefficient of dynamic friction
mk < ms

7. The Resultant of the friction and normal forces
R= N2+F2
tan q= F/N
At the point of impending motion:
R= N2+F2max = N2+(msN)2 = N 1+ ms2
tan fs= F/N= msN/N= ms
fs is the angle of static friction
q < fs

8. Gravity forces in inclined surfaces
For equilibrium:
q < fs

9. Example
Two blocks with masses mA = 20 kg and mB = 80 kg are connected with a flexible cable that passes over a frictionless pulley as shown in Fig. P9-4. The coefficient of friction between the blocks is If motion of the blocks is impending, determine the coefficient of friction between block B and the inclined surface and the tension in the cable between the two blocks.

10. Solution Free body diagram for block A Fy = An - 196.14 cos(35) = 0
An = N
For impending motion of the block A: (Af = ms.An)
+  Fx = -T (160.67) sin 35 = 0
T = N

11. Block B weight:
WB = mB.g = 80 (9.807) = N
Free body diagram for block B:
Fy = Bn cos 35 = 0
Bn = N
Fx = - Bf sin 35 = 0
Bf = N
= Bf/Bn= / = 0.320

12. Determine the horizontal force P required to start moving the 250-lb block shown in Fig. P9-1 up the inclined surface. The coefficient of friction between the inclined surface and the block is s = 0.30.
P = lb  265.3b

13. Solution For impending motion: Ff = Fn = 0.30Fn
Ff = Fn = 0.30Fn
Free body diagram for the block:
Fy = Fn cos 30 - Ff sin 30 -W
= Fn cos 30 Fn sin 30 = 0
Fn = lb
Fx = P - Fn sin 30 - Ff cos 30
= P sin 30 (349.15) cos 30 = 0
P = lb  265.3b

14. The masses of blocks A and B of Fig
The masses of blocks A and B of Fig. P9-36 are mA = 40 kg and mB = 85 kg. If the coefficient of friction is 0.25 for both surfaces, determine the force P required to cause impending motion of block B.
P  935 N

15. Solution
WA = mAg = 40 (9.807) = N
From a free-body diagram for the block A when motion is impending:
Af = Af (max) = A An = 0.25 An
Fy = An cos 45 + Af sin 45 - WA
= An cos 45 An sin 45 = 0
An= N
Af= 0.25 An = N

16. WB = mBg = 85 (9.807) = N
From a free-body diagram for the
block B when motion is impending:
Bf = Bf (max) = Bn = 0.25Bn
Fx = P cos 20-WB sin 45-Af -Bf
= P cos 20 sin 45 Bn = 0
Fy = P sin 20 - WB cos 45 - An + Bn
= P sin 20 cos 45 Bn = 0
P  935 N

17. Workers are pulling a 400 lb crate up an incline as shown in Fig. P9-3
Workers are pulling a 400 lb crate up an incline as shown in Fig. P9-3. The coefficient of friction between the crate and the surface is 0.20, and the rope on which the workers are pulling is horizontal.
(a) Determine the force P that the workers must exert to start sliding the crate up the incline.
(b) If one of the workers lets go of the rope for a moment, determine the minimum force the other workers must exert to keep the crate from sliding back down the incline.
P=197.8 lb
P=25.8lb

18. A 120 lb girl is walking up a 48-lb uniform beam as shown in Fig. P9-21. Determine how far up the beam the girl can walk before the beam starts to slip if
(a)The coefficient of friction is 0.20 at all surfaces
(b)The coefficient of friction at the bottom end of the beam is
increased to 0.40 by placing a piece of rubber between the
beam and the floor.
x = 3.21 ft
x = 6.26 ft

19. Solution (a) For a free-body diagram for the
beam when motion is impending:
Ar = An = 0.2An
Bf= Bn = 0.2Bn
f = tan-16/8 = 36.87
Fh = Bn sin 36.87 - 0.2Bn cos 36.87 - 0.2An = 0
+  Fy = Bn cos 36.87 + 0.2Bn sin 36.87 + An = 0
Solving yields: An = lb; Bn = lb
+ MA = 48(6) cos 36.87 + 120(x) cos 36.87 (10) = 0
x = ft  3.21 ft

20. For a free-body diagram for the beam when motion is impending:
Ar = An = 0.4An
Bf= Bn = 0.2Bn
+  Fh = Bn sin 36.87 - 0.2Bn cos 36.87 - 0.4An = 0
+  Fy = Bn cos 36.87 + 0.2Bn sin 36.87 + An = 0
Solving yields:An = lb; Bn = lb
+ MA = 48(6) cos 36.87 + 120(x) cos 36.87 (10) = 0
x = ft  6.26 ft