Friction
Consider a block resting on a rough surface and then subjected to load P;
F
θ
N
Friction – is the contact resistance exerted by a body upon another body when one of the bodies
tends to move past another body.
In symbol:
F = μN where: F = Frictional force (N, lb)
μ = Coefficient of static friction
N = Normal force (N, lb)
Factors affecting the value of friction:
1. Types of material
2. Roughness of the surface
3. Normal force developed between contacting surface
W P
R
F N
θ R F
N
F
tanθ=
N
F
μ=
N
∴ tanθ=μ
Determine the range of values of P so that the 250 N block will tend to move (a) up the incline and
(b) down the incline, use μ = 0.25.
30°
°
Given:
μ = 0.25
W = 250 N
Req’d:
Pmax
Pmin
Solution:
a. Up to the incline
P
ΣFy = 0 ΣFx = 0
N – 250cos30° = 0 P – F – 250sin30° = 0
F N = 216.51 N P = 179.13 N
Direction of movement from F = μN
F = 0.25(216.51 N)
N
b. Down the incline
ΣFy = 0 ΣFx = 0
250 N
P N – 250cos30° = 0 P + F – 250sin30° = 0
N = 216.51 N P = 70.81 N
F F = 54.13 N
N
Bodies A & B are joined by a cord parallel to the inclined plane as shown. Determine the angle θ at
which motion impends. What is the tension in the cord?
μ = 0.50
A
μ = 0.20
θ
WA = 200 lb
WB = 300 lb
Given:
WA = 200 lb
WB = 300 lb
Solution:
Consider Block A
ΣFy = 0 ΣFx = 0
200 lb
T NA – 200cosθ = 0 T – FA – 20sinθ = 0
NA = 200cosθ T = 200sinθ – 40cosθ 1
From F = μN
FA = 0.20(200cosθ)
NA FA = 40cosθ
FA
Consider Block B
300 lb ΣFy = 0 ΣFx = 0
NB – 300cosθ = 0 -T + FB – 300sinθ = 0
NB = 300cosθ T = 150cosθ – 300sinθ 2
T From F = μN
FB = 0.50(300cosθ)
NB
FB FB = 150cosθ
Solve equations 1 & 2:
200sinθ – 40cosθ = 150cosθ – 300sinθ
500sinθ = 190cosθ
190
tanθ=
500
θ = 20.81°
T = 33.63 lb
A horizontal bar, 10 ft long and of negligible weight, rests on rough inclined planes as shown. If the
angle of friction is 15, how close to B may the 200 lb force be applied before motion impends?
Given:
θ = 15°
Req’d:
x=? FA
Solution: 25°
tanθ = μ NB
μ = 15° NA 25° 60°
FA = tan15°NA FB
FB = tan15°NB
ΣFH = 0
FAcos25° - NAsin25° + FBcos60° - NBsin60° = 0
tan15°NAcos25° + NAsin25°+ tan15°NBcos60° - NBsin60° = 0 1
ΣFV = 0
-FAsin25° + NAcos25° + FBsin60° + NBcos60° - 300 lb = 0
-tan15°NAsin25° + NAcos25°+ tan15°NBsin60° + NBcos60° = 300 2
NA = 205.69 lb ; FA = 55.11 lb
NB = 186.98 lb ; FB = 50.10 lb
ΣMB = 0
-200 lb(x) – 100(8) + NAcos25°(10) – FAsin25°(10) = 0
x = 4.8’
Using Triangle Law:
R
RA θ = 15°
40° N
95° 300 lb
45°
RB F
RA RB 300 lb tanθ =F/N cos15° = NA/RA
= =
sin 45 ° sin 40° sin 95 ° F = Μn NA = 205.69 lb
μ = F/N sin15° = FA/RA
RA = 212.94 lb sin15° = FB/RB FA = 55.11 lb
RB = 193.57 lb FB = 50.10 cos15° = NB/RB
NB = 186.98 lb
Find the least value of P that will just start the system of blocks shown moving to the right. The
coefficient of friction under each block is 0.30.
Given:
WA = 200 lb μ = 0.30
WB = 300 lb
Req’d: P = ?
Consider Block A: Consider the Pulley:
ΣFy = 0 T
200 lb
NA – 200 (4/5) = 0 3
T
NA = 160 lb 4
From F = μN FA
T
F = 0.30(160 lb)
F = 48
ΣFx = 0 NA
T – FA – 200(3/5) = 0
T = 168 lb
Consider block B:
300 lb
ΣFV = 0
P
T = 168 lb Pcosα – 90 lb + 0.30Psinα = 168 lb
α
P(cosα + 0.30sinα) = 258 lb
FB
P = 258 lb / (cosα + 0.30sinα)
ΣFV = 0 NB *get the derivative of P*
dP 258(−sinα +0.30 cos α )
NB + Psinα – 300 lb = 0 =
dα (cos α + 0.30 sin α )2
NB = 300 lb – Psinα sinα =0.30 cos α
ΣFH = 0 tanα =0.80
-168 lb + Pcosα – FB = 0, but FB = 0.30; α =16.70 °
FB = 0.30(300 lb – Psinα) P = 247.12 lb
Block A supports a load W = 1000 lb and is to be raised by forcing the wedge B under it. The angle of
friction for all surfaces in contact is 15°. Determine the force P which is necessary to start the wedge
under the block. The block and wedge have negligible weight.
Given: θ = 15°
Req’d: P
Solution:
Consider block A:
W = 1000 lb
ΣFH= 0
FA
FABcos20° + NAsin20° - NA = 0
but μ = tan15°
NA ∴ FA = tan15°NA
FAB = tan15°NAB
FAB
20°
tan15°NABcos20° + NABsin20° - NA = 0 ––Eq. 1
ΣFV= 0
20°
-FABsin20° + NABcos20° - FA = 1000 lb
NAB
-tan15°NABsin20° + NABcos20° - tan15°NA = 1000 lb ––Eq. 2
NA = 861.92 lb FA = 230.95 lb
NAB = 1451.51 lb FAB = 388.93 lb
Consider the wedge:
NAB ΣFV= 0
20°
-NABcos20° + FABsin20° + NB = 0
20°
NB = 1230.90 lb
FAB but FB = tan15°NB
P FB = 329.83 lb
ΣFH= 0
P – FB – FABcos20° - NABsin20° = 0
FB
NB
P = 1191.75 lb
A homogeneous cylinder, 3 ft in diameter and weighing 400 lb is resting on two rough inclined
surfaces as shown. If the angle of friction is 16°, find the moment M applied to the cylinder that will
start it rotating clockwise.
M
1.
5’
4
2° 5
5°
Given: W = 400 lb
Req’d: M
Solution:
tan16° = μ
FA = μNA
FA = tan16°NA
FB = tan16°NB
ΣFH = 0
FAcos42° + NAsin42° + FBcos55° - NBsin55° = 0
tan16°NAcos42° + NAsin42° + tan16°NB(cos55° - NAsin55°) = 0 ––Eq. 1
ΣFV = 0
-FAsin42° + NAcos42° + FBsin55° + NBcos55° - 400 lb = 0
-tan16°NAsin42° + NAcos42° + tan16°NBsin55° + NBcos55° = 400 lb ––Eq. 2
Solve equations 1 & 2:
NA = 243.79 lb FA = 69.91 lb
NB = 978.53 lb FB = 94.20 lb
ΣM O = 0
M – FB(1.5’) – FA(1.5’) = 0
M = 243.17 lb-ft
R
Using triangle law 16° N
400 lb RA RB
= =
sin 83 sin 39 sin 58 F
RA
58 RA = 253.62 lb cos16° = NA / RA
83 ° 400 RB = 341.77 lb NA = 243.80 lb
° 39 lb cos16° = NB / RB = 328.53 lb
R
°
B
Block A weighs 120 lb, block B weighs 200 lb and the cord is parallel to the incline. If the coefficient of
friction for all surfaces in contact is 0.25, determine the angle θ of the incline for which motion of B
impends.
Given:
WA = 120 lb
WB = 200 lb
μ = 0.25
Req’d: θ = ?
Solution:
Consider block B:
ΣFy = 0 ΣFx = 0
NA
NB – NA – 200cosθ = 0 -FA – FB + 200sinθ = 0
FA
NB = 320cosθ 200sinθ = 30cosθ + 80cosθ
FB = 0.25 NB tanθ = 110/200
FB = 0.25(320cosθ) θ = 28.81
NB
FB = 80cosθ
FB
Consider block A:
ΣFy = 0 ΣFx = 0
NA – 120cosθ = 0 -T + FA + 120sinθ = 0
NA = 120cosθ T = 30cosθ + 120sinθ
but FA = μNA T = 30cos(28.81) + 120sin(28.81)
FA = 0.25(120cosθ) T = 84.12 lb
FA FA = 30cosθ
NA
