Solving Friction Problems with Forces and Coefficients
ESC-S201
27-SEP
Friction problem
1.
A pull of 180 N applied upwards at 30
0
to a rough horizontal plane was required to just
move a body resting on the plane while a push of 220 N applied along the same line of
action was required to just move the same body downwards. Determine the weight of the
body and the coefficient of friction.
W
180 N
30
0
W
220 N
30
0
Case-2
Case-1
Case-A
W
180 Sin (30 )
W
1.
First we resolve the 180 N force in Horizontal and
vertical components.
2.
Due to horizontal force 180 Cos (30) block tend to
move in direction of positive x –axis.
3.
Since block is kept at floor therefore floor will try
to oppose this motion. So floor will apply
a friction force on block towards left.
180 Cos (30 )
F
s
N
a)
Force equilibrium in x-direction
:
180 * Cos (30) – F
s
= 0
F
s
= 180 *0.866.
F
s
= 155.9 N
b)
Force equilibrium in y-direction :
N – W + 180 sin (30)= 0
N = W - 90
c)
F
s
= µ
s
N
155.9 =
µ
s
(
W - 90)
Case-B
220 sin(30)
220 cos(30)
W
N
F
s
FBD
Force acts on the block
1. N = Normal contact force.
2. F
s
= static friction force.
3. W = weight of the block.
4. 220 N applied force act at an
angle of 30
0
.
Force equilibrium in x- direction.
F
s
= 220 Cos (30
0
)
Force equilibrium in y-direction.
N = W + 220 Sin( 30
0
).
Since block is about to move therefore, friction
force act on the block is equal to maximum static
friction force.
Max static friction force =
Coefficient of static friction x Normal
Contact force.
F
s
= ( F
s
)
max
= µ
s
N.
Friction problem
1. F
s
= 220 Cos (30
0
)
2. N = W + 220 Sin( 30
0
).
3.
F
s
= ( F
s
)
max
= µ
s
N.
By (1),
µ
s
N = 220 Cos(30
0
)
µ
s
(W + 220 Sin( 30
0
) ) = 220 Cos(30
0
)
µ
s
(W-180sin(30
0
)) = 180 * cos (30
0
).
Case-1
Case-2
Solve these two equation and get µ
s
and W.
Answer
W = 990 N
µ
s
= 0.1722
Problem-2
Friction problem
2. Find whether block A is moving up or down the plane. Weight of
block B = 600 N and Weight of block A = 300 N. Coefficient of limiting
friction between plane AB and block A is 0.2. Coefficient of limiting
friction between plane BC and block B is 0.25. Assume pulley as
smooth.
60
0
40
0
B
A
A
B
C
Friction problem
60
0
40
0
B
A
B
C
A
Total three cases possible:
1.
Either both block is stationary.
2.
Block A moves in downwards direction
3.
Block A moves in upwards direction.
Friction problem
60
0
40
0
B
A
B
C
A
CASE-1 : Either both block is stationary.
1. Weight of block A = 300 N
2. Weight of block B = 600 N
A
W
A
N
T
60
0
W
A
Cos( 60
0
)
W
A
Sin ( 60
0
)
1.
Take component of weight parallel to surface AB:
W
A
Sin ( 60
0
) = 300 * 1.732 * 0.5
= 259.8 N
2. Take component of weight normal to surface AB:
W
A
Cos ( 60
0
) = 300 *0.5
= 150 N
3. Since block A
Friction problem
A
N
T
150 N
259.8 N
F
s
If T < 259.8 N
A
N
T
150 N
259.8 N
If T > 259.8 N
F
s
Condition for static friction force F
s1
<=
μ
A
N
= 0.2*150
= 30 N
Friction problem
A
N
T
150 N
259.8 N
F
s1
If T < 259.8 N
Condition for static friction force F
s 1
<= 75N
Assume T < 259.8 N
1)
Force equilibrium along y-axis :
T + F
s1
= 259.8 N
B
T
If T < 259.8 N
385.7 N
459.6 N
N
’
F
s2
Condition for static friction force F
s2
<= 114.9 N
Assume T < 259.8 N
1)
Force equilibrium along y-axis :
T + F
s2
= 385.7 N
Friction problem
T + F
s1
= 259.8 N
T + F
s2
= 385.7 N
- - -
-
F
s2
+ F
s1
= - 385.7 N + 259.8 N
F
s2
– F
s1
= 125.9 N
F
s2
<= 114.9 N
F
s 1
<= 30N
If my direction is correct then F
Maximum value of F
s2
– F
s1
= 114.9 N.
So our direction of static friction force is incorrect.
Problem-3
Ladder problem
3. A ladder of length 4 m weighing 200 N is placed against a vertical
wall. The coefficient of friction between the wall and the ladder is 0.2.
and that between the floor and the ladder is 0.3.The ladder in addition to
its own weight has to support a man weighing 600 N. at a distance of 3m
from A. Calculate the minimum horizontal force to be applied at A to
prevent slipping.
60
0
A
B
200 N
C
600 N
AB = 4 m
AG = 2 m
AC = 3 m
G
P
AIM:
P
min
= ??
Ladder problem
Step-1
o
DRAW FREE BODY DIAGRAM OF LADDER
Ladder problem
Step-1
o
DRAW FREE BODY DIAGRAM OF LADDER
60
0
200 N
600 N
P
N
B
V
B
V
A
A
B
At the time of slipping direction of
velocity of point A and B are shown in
above figure.
o
Velocity direction of point A and B.
F
s
2
N
A
F
s
1
Ladder problem
Step-2
o
Moment equilibrium about point A
Ladder problem
60
0
200 N
600 N
P
N
B
F
s
2
N
A
F
s
1
o
Moment equilibrium about point A
1.
Moment due to Normal force N
B
about point A.
= Force x perpendicular distance from A to line
of action of force N
B
= N
B
x AX
= N
B
x AB * Cos( 30
0
)
= N
B
x 4 * Cos( 30
0
)
Direction : Anticlockwise moment
2. Moment due to Friction force F
s2
about point A.
= Force x perpendicular distance from A to line
of action of force F
s2
= F
s2
x AY
= F
s2
x AB Cos ( 60
0
)
A
B
C
AB = 4 m
AG = 2 m
AC = 3 m
G
X
30
0
Y
= F
s2
x 4 x 0.5.
Direction: Anticlockwise direction.
Ladder problem
60
0
200 N
600 N
P
N
B
F
s
2
N
A
F
s
1
o
Moment equilibrium about point A
3. Moment due to 600 N force act at point C about
point A.
= Force x perpendicular distance of line of action
of force 600 N from A.
= 600 N x AZ
= 600 N x 3 * Cos ( 60
0
) m
= 900 N m
Direction : Clockwise moment
4. Moment due to 200 N force act at point G about
point A.
= Force x perpendicular distance of line of action
of force 200N from A.
A
B
C
AB = 4 m
AG = 2 m
AC = 3 m
G
X
30
0
Y
Z
= 200 N x AM
= 200 N x AG * Cos( 60
0
)
= 200 N x 2 * 0.5 m
= 200 N m
Direction : Clockwise moment
M
Friction problem
o
Moment equilibrium about point A
4 * N
B
* Cos( 30
0
) (ACW) + 2 * F
s2
( ACW) + 900 (CW) + 200 (CW) = 0
Take anti clockwise moment as positive.
4 * N
B
* Cos( 30
0
) + 2 * F
s2
- 900 - 200 = 0
4 * N
B
* Cos( 30
0
) + 2 * F
s2
= 1100
At the time of slipping friction force F
s2
= µ
B
x N
B
4 * N
B
* Cos( 30
0
) + 2 * µ
B
* N
B
= 1100
60
0
A
B
µ
B
µ
B =
Coefficient of friction between vertical wall and rod
Problem-4
Friction problem
1.
Two identical blocks A and B are connected by a rod and they rest
against vertical and horizontal planes respectively. If sliding
impends when Ɵ = 45
0
, determine the coefficient of friction,
assuming it to be the same for both floor and wall.
A
B
Ɵ
Answer
µ = 0.414
In these physics problems, forces, friction, and coefficients are utilized to determine weight and movement of objects on inclined planes. The solution involves resolving forces, applying equilibrium principles, and calculating friction coefficients. Through detailed steps and calculations, the weight of the body, coefficient of friction, and the direction of block movement are found.
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Presentation Transcript
ESC-S201 27-SEP
Friction problem 1. A pull of 180 N applied upwards at 300to a rough horizontal plane was required to just move a body resting on the plane while a push of 220 N applied along the same line of action was required to just move the same body downwards. Determine the weight of the body and the coefficient of friction. 220 N 180 N 300 300 W W Case-2 Case-1
Case-A 180 Sin (30 ) 1. First we resolve the 180 N force in Horizontal and vertical components. 2. Due to horizontal force 180 Cos (30) block tend to to oppose this motion. So floor will apply N 180 Cos (30 ) W move in direction of positive x axis. Fs 3. Since block is kept at floor therefore floor will try W a friction force on block towards left. c) Fs = s N a) Force equilibrium in x-direction: 180 * Cos (30) Fs = 0 Fs = 180 *0.866. Fs = 155.9 N b) Force equilibrium in y-direction : 155.9 = s (W - 90) N W + 180 sin (30)= 0 N = W - 90
Case-B Force equilibrium in x- direction. F s = 220 Cos (300) FBD 220 cos(30) Force equilibrium in y-direction. W N = W + 220 Sin( 300). Fs 220 sin(30) N Since block is about to move therefore, friction force act on the block is equal to maximum static Force acts on the block friction force. 1. N = Normal contact force. Max static friction force = 2. Fs = static friction force. 3. W = weight of the block. Coefficient of static friction x Normal Contact force. 4. 220 N applied force act at an angle of 300 . Fs = ( Fs )max = s N.
Friction problem 1. F s = 220 Cos (300) 2. N = W + 220 Sin( 300). 3. Fs = ( Fs )max = s N. By (1), s N = 220 Cos(300) s (W + 220 Sin( 300) ) = 220 Cos(300) Case-2 s (W-180sin(300)) = 180 * cos (300). Case-1 Answer W = 990 N s = 0.1722 Solve these two equation and get s and W.
Friction problem 2. Find whether block A is moving up or down the plane. Weight of block B = 600 N and Weight of block A = 300 N. Coefficient of limiting friction between plane AB and block A is 0.2. Coefficient of limiting friction between plane BC and block B is 0.25. Assume pulley as smooth. B 600 400 A C
Friction problem B 600 400 C A Total three cases possible: 1. Either both block is stationary. 2. Block A moves in downwards direction 3. Block A moves in upwards direction.
Friction problem CASE-1 : Either both block is stationary. B T 600 400 C A 600 N 1. Weight of block A = 300 N 2. Weight of block B = 600 N WA 1. Take component of weight parallel to surface AB: WA Sin ( 600 ) = 300 * 1.732 * 0.5 = 259.8 N 2. Take component of weight normal to surface AB: WA Cos ( 600 ) = 300 *0.5 = 150 N 3. Since block A
Friction problem If T < 259.8 N If T > 259.8 N T T N N Fs Fs Condition for static friction force Fs1 <= A N = 0.2*150 = 30 N
Friction problem If T < 259.8 N Condition for static friction force Fs 1 <= 75N T Assume T < 259.8 N N Fs1 1) Force equilibrium along y-axis : T + Fs1 = 259.8 N If T < 259.8 N Condition for static friction force Fs2 <= 114.9 N N T Assume T < 259.8 N Fs2 1) Force equilibrium along y-axis : T + Fs2 = 385.7 N
Friction problem T + Fs1 = 259.8 N T + Fs2 = 385.7 N - - - - Fs2 + Fs1 = - 385.7 N + 259.8 N Fs 1 <= 30N Fs2 Fs1 = 125.9 N Fs2 <= 114.9 N If my direction is correct then F Maximum value of Fs2 Fs1 = 114.9 N. So our direction of static friction force is incorrect.
Ladder problem 3. A ladder of length 4 m weighing 200 N is placed against a vertical wall. The coefficient of friction between the wall and the ladder is 0.2. and that between the floor and the ladder is 0.3.The ladder in addition to its own weight has to support a man weighing 600 N. at a distance of 3m from A. Calculate the minimum horizontal force to be applied at A to prevent slipping. 600 N B AB = 4 m C G AG = 2 m AC = 3 m 600 200 N P A AIM: Pmin = ??
Ladder problem Step-1 o DRAW FREE BODY DIAGRAM OF LADDER
Ladder problem Step-1 o DRAW FREE BODY DIAGRAM OF LADDER o Velocity direction of point A and B. 600 N B NB VB Fs2 600 200 N P A Fs1 VA NA At the time of slipping direction of velocity of point A and B are shown in above figure.
Ladder problem Step-2 o Moment equilibrium about point A
Ladder problem o Moment equilibrium about point A 1. Moment due to Normal force NB about point A. = Force x perpendicular distance from A to line of action of force NB 600 N B X NB 200 N C G Fs2 300 600 = NB x AX P Y A Fs1 AB = 4 m AG = 2 m AC = 3 m = NB x AB * Cos( 300 ) NA = NB x 4 * Cos( 300 ) = Fs2 x 4 x 0.5. Direction: Anticlockwise direction. Direction : Anticlockwise moment 2. Moment due to Friction force Fs2 about point A. = Force x perpendicular distance from A to line of action of force Fs2 = Fs2 x AY = Fs2 x AB Cos ( 600)
Ladder problem o Moment equilibrium about point A 3. Moment due to 600 N force act at point C about point A. 600 N B X NB 200 N C G Fs2 300 = Force x perpendicular distance of line of action of force 600 N from A. 600 P Y Z M A Fs1 AB = 4 m AG = 2 m AC = 3 m = 600 N x AZ NA = 600 N x 3 * Cos ( 600 ) m = 200 N x AM = 900 N m = 200 N x AG * Cos( 600) Direction : Clockwise moment = 200 N x 2 * 0.5 m 4. Moment due to 200 N force act at point G about point A. = 200 N m = Force x perpendicular distance of line of action of force 200N from A. Direction : Clockwise moment
Friction problem o Moment equilibrium about point A 4 * NB * Cos( 300 ) (ACW) + 2 * Fs2 ( ACW) + 900 (CW) + 200 (CW) = 0 Take anti clockwise moment as positive. 4 * NB * Cos( 300 ) + 2 * Fs2 - 900 - 200 = 0 4 * NB * Cos( 300 ) + 2 * Fs2 = 1100 At the time of slipping friction force Fs2 = B x NB 4 * NB * Cos( 300 ) + 2 * B * NB = 1100 B B 600 A B = Coefficient of friction between vertical wall and rod
Friction problem 1. Two identical blocks A and B are connected by a rod and they rest against vertical and horizontal planes respectively. If sliding impends when = 450 , determine the coefficient of friction, assuming it to be the same for both floor and wall. A B Answer = 0.414
