# SIMPLE CONTACT FRICTION

3.4 SIMPLE CONTACT FRICTION

The problems on simple contact friction consists of

(i)Simple contact friction on a horizontal surface,

(ii)Simple contact friction on an inclined plane, and

(iii)Simple contact friction on Ladder surface

Problem 3.1. A body of weight 100 Newtons is placed on a rough horizontal plan. Determine the co-efficient of friction if a horizontal force of 60 Newtons just causes the body to slide over the horizontal plane.

Sol. Given:

Weight of body, W = 100 N

Horizontal force applied,

P = 60N

∴ Limiting force of friction,

F = P = 60N

Let µ = Co-efficient of friction.

The normal reaction of the body is given as

R = W= 100 N

Using equation (3.1),

F = µR

Problem 3.2. A body of weight 200 N is placed on a rough horizontal plane. If the coefficient of friction between the body and the horizontal plane is 0.3, determine the horizontal force required to just slide the body on the plane.

Sol. Given:

Weight of body, W = 200 N

Coefficient of friction, µ = 0.3

Normal reaction, R = W = 200 N

Let F = horizontal force which causes the body to just slide over the plane.

Using equation (3.1),

F = µR = 0.3 × 200 = 60 N. Ans.

Problem 3.3. The force required to pull a body of weight 50 N on a rough horizontal
plane is 15 N. Determine the co-efficient of friction if the force is applied at an angle of 150 with the horizontal.

Sol. Given

Weight of the body, W = 50 N

Force applied, p = 15 N

Angle made by the force P, with horizontal θ = 15°

Let the co-efficient of friction = µ

Normal reaction = R

When a force equal to 15 N is applied to the body at an angle 15° to the horizontal, the body is on the point of motion in the forward direction. Hence a force of friction equal to µR will be acting in the backward direction. The body is in equilibrium under the action of the forces shown in Fig. 3.6,

Resolving the forces along the plane, µR = 15 cos 15°                 … (i)

Resolving the forces normal to the plane

R + 15 sin 15° = 50

R = 50 – 15 sin 15° = 50 – 15 × 0.2588

= 46.12N

Substituting the value of R in equation (i), we get

µ × 46.12 = 15 cos 15°

Problem 3.4.A body of weight 70 N is placed on a rough horizontal plane. To just move the body on the horizontal plane, a push of 20 N inclined at 20° to the horizontal plane is required. Find the co-efficient of friction.

Sol. Given

Weight of body, W = 70N

Force applied, P = 20N

Inclination of P, θ= 20°.

Let µ = Co-efficient of friction

R = Normal reaction

F = Force of friction = µR.

When a push of 20 N at an angle 20° to the horizontal is applied to the body, the body is just to move towards left. Hence a force of friction F = µR, will be acting towards right as shown in Fig. 3.7.

Resolving forces along the plane, µR = 20 cos 20°                                 … (i)

Resolving forces normal to the plane, R = 70 + 20 sin 20°

= 70 + 20 × 0.342 = 70 + 6.84 = 76.84

Substituting the value of R in equation (i),

µ × 7.6.84 = 20 cos 20°

Problem 3.5. A pull of  20 N, inclined at 25° to the horizontal plane, is required just to move a body placed on a rough horizontal plane. But the push required to move the body is 25 N If the push is inclined at 25° to the horizontal, find the weight of the body and co-efficient of friction.

Sol. Given:

Pull required, P = 20 N

Inclination of pull, θ = 25°

Push required, P’ = 25 N

Inclination of push, θ’ = 25°

Let W = Weight of body

µ = Co-efficient of friction

R = Normal reaction when body is pulled

R’ = Normal reaction when body is pushed.

1st case when body is pulled. The body is in equilibrium under the action of forces shown in Fig. 3.8.

Resolving the forces along the plane,

µR = 20 cos 25° = 20 × 0.9063

= 18.126                                                    … (i)

Resolving forces normal to the plane,

R + 20 sin 25° = W

R = W – 20 sin 25°

=W – 20 × 0.4226

= W – 8.452.

Substituting the value of R in equation (i),

µ(W – 8.452) = 18.126                                                                         … (ii)

2nd case when body is pushed. The body is in equilibrium under the action of forces shown in Fig. 3.9.

Resolving forces along the plane,

µR’ = 25 cos 25° = 25 × 0.9063

= 22.657                                                                   … (iii)

Resolving forces normal to the plane,

R’ = W + 25 sin 25°

= W + 25 × 0.4226

= W+ 10.565

Substituting the value of R’ in equation (iii),

µ(W + 10.565) = 22.657                                    … (iv)

Dividing equation (ii) by equation (iv),

Substituting the value of W in equation (ii),

µ (84.547 – 8.452) = 18.126

Problem 3.6. A block of weight W is placed on a rough horizontal plane surface as shown in Fig. 3.10 and a force P is applied at an angle θ is equal to the angle of friction ɸ.

Sol. Given:

Weight of block = W

Force applied = p

Inclination of force = θ

Let R = Normal reaction

µ = Co-efficient of friction

F = Force of friction = µR.

The forces acting on the block are shown in Fig. 3.10.

Resolving forces vertically, we get

R + P sin θ = W

or R = W – P sin θ                                                      … (i)

Resolving forces horizontally, we get

P cos θ = F = µR                         (  F = µR)        … (ii)

Substituting the value of R from equation (i), the equation (ii) becomes as

P cos θ = µ[W – P sin θ]                                        … (iii)

But from equation (3.2), we know

µ = tan ɸ

where ɸ = angle of friction.

Substituting the value of µ in equation (iii), we get

P cos θ = tan ɸ (W – P sin θ)

The force P will be least, if the denominator i.e., cos (θ – ɸ) is maximum. But cos (θ – ɸ) will be maximum, if

cos (θ – ɸ) = 1

or θ – ɸ = 0

or θ = ɸ

Pleast = W sin ɸ or W sin θ.

Hence the force P will be least if the angle of inclination of P with the horizontal is equal to the angle of friction ɸ.

Problem 3.7. A man wishing to slide a stone block of weight 1000 N over a horizontal concrete floor, ties a rope to the block and pulls it in a direction inclined upward at an angle of 20° to the horizontal. Calculate the minimum pull necessary to slide the block if the co-efficient of friction µ=0.6. Calculate also the pull required if the inclination of the rope with the horizontal is equal to the angle of friction and prove that this is the least force required to slide the block.

(AMIE Winter, 1990)

Sol. Given:

Weight, W = 1000 N

Angle with horizontal, θ = 20°

Co-efficient of friction, µ = 0.6

Let P = Force applied

R = Normal reaction

F = Force of friction = µR

The forces acting on the block are shown in Fig. 3.10 (a).

Resolving forces horizontally,

P cos θ = µR

P cos 20° = 0.6 × R                                                                … (i)

Resolving forces vertically, R + P sin 8 = W

R + P sin 20° = 1000

R = 1000 -P sin 20°                                                                … (ii)

Substituting the value of R in equation (r), we get

P cos 20° = 0.6 (1000 – P sin 20°)

= 600 – 0.6 P sin 20°

or P cos 20° + 0.6 P sin 20° = 600

or P (cos 20° + 0.6 × sin 20°) = 600                                                               … (iii)

Pull required if the inclination of the rope with the horizontal is equal to angle of friction.

Let ɸ = Angle of friction

= the angle made by rope with horizontal (given) = 20°

If In equation (iii), the angle 20° is replaced by angle ɸ, then we get the force required to pull the body as,

P(cos ɸ + 0.6 sin ɸ) = 600

But 0.6 µ. Hence force P will be minimum if tan ɸ = µ = 0.6 (Proved)

Now tan ɸ = 0.6

Substituting this value of ɸ in equation (iv), we get

Problem 3.8. A uniform rod AB of length 50.8 cm weighing 100 N is resting on a rough horizontal surface whose co-efficient of friction is 0.1. It is subjected to a force P applied horizontally at the end A in a direction perpendicular to its length. Determine the point O about which it would commence to turn.

[AMIE (S) 1982, (S) 1986]

Sol. Given:

Length of the rod, AB = 50.8 cm

Weight of the rod, W = 100 N

Co-efficient of friction, µ = 0.1

Force applied horizontally at A = P.

Let the rod commences to turn about point 0, when a horizontal force P is applied at A as shown in Fig. 3.11. Let the distances of O from A is x cm.

Then OB = AB – AO = (50.8 – x) cm

Let w = Weight per unit length of the rod

Weight of length AO fo the rod = w × Length AO

= w × x N

This weight is acting at the middle point of AO. Let G1 is the middle point of AO.

Weight of length BO of the rod = w × Length BO

= w × (50.8 – x) N.

This weight is acting at the middle point of BO. Let G2 is the middle point of OB.

When the force P is applied horizontally at a A, the portion AO will commence to move in the same direction as that of P, while the portion OB will commence of move in the direction opposite to that of P. Hence force of friction, which acts in the opposite direction of the motion of the body, will act for the portion AO in the direction opposite to P whereas for the portion OB will act in the same direction as of P.

Let F1 = Force of friction for portion AO

= µ × weight of AO

= µ × w × x .

This will act at point G1

F2 = Force of friction for portion BO

= µ × weight of BO

= µ × (50.8 – x) × w.

This will act at point G2

The directions of F1 and F2 are shown in Fig 3.11. All the forces i.e., P, F1 andF2 shown in Fig. 3.11 are acting in the horizontal plane (i.e.; in the plane of the paper).

Resolving forces in the direction perpendicular to AB, we get

P + F2 = F1

or P= Fl – F2                                                                     … (i)

Taking moments of all forces about O, we get

Substituting the value of P from equation (i), in the above equation,

x2 = (50.8 – x) (50.8 + x) = (50.8)2 – x2

or x2 + x2 = (50.8)2

or  2x2 = (50.8)2 = 2580.64

Hence the rod will commence to turn about the point O, which is at a distance of 35.92 cm from point A.

3.4.1. Angle of Repose. The angle of repose is defined as the maximum inclination of a plane at which a body remains in equilibrium over the inclined plane by the assistance of friction only.

Consider a body of weight W, resting on a rough inclined plane as shown in Fig. 3.12.

Let R = Normal reaction acting at right angle to the inclined plane.

α = Inclination of the plane with the horizontal

F = Frictional force acting upward along the plane.

Let the angle of inclination (α) be gradually increased, till the body just starts sliding down the plane. This angle of inclined plane, at which a body just begins to
slide down the plane, is called angle of repose.

Resolving the forces along the plane, we get

W sin α = F                                                              … (i)

Resolving the forces normal to the plane, we get .

W cos α = R                                                             … (ii)

Dividing equation, (i) by equation (ii),

3.4.2. Equilibrium of a Body Lying on a Rough Inclined Plane. In Art. 3.4.1 we have studied that if the inclination of the plane, with the horizontal, is less than the angle of friction, the body will remain in equilibrium without any external force. If the body is to be moved upwards or downwards in this condition an external force is required. But if the inclination of the plane is more than the angle of friction, the body will not remain in equilibrium. The body will move downward and an upward external force will be required to keep the body in equilibrium.

Such problems are solved by resolving the forces along the plane and perpendicular to the planes. The force of friction (F), which is always equal to µR is acting opposite to the direction of motion of the body.

Problem 3.9. Prove that the angle of friction (ɸ) is equal to the angle made by an inclined plane with the horizontal when a solid body, placed on the inclined plane, is about to slide down.

Sol. A solid body of weight, W is placed on an inclined plane AC as shown in Fig. 3.13.

Let α = Angle of the inclined plane AC with horizontal plane AB, such that body just, starts moving downward.

The body is in equilibrium under the action of following forces:

1. Weight of the body (W) acting vertically downwards.
2. Normal reaction (R), acting perpendicular to the inclined plane, AC
3. The force of friction, F = µR, acting up the plane as the body is about to slide down the plane.

The weight, W can be resolved in two component one along the plane and other perpendicular to the plane. The components are W sin α and W cos α respectively.

As the body is in equilibrium, the forces along and perpendicular to the inclined plane are:

W sin α = F = µR

W cos α = R

The above relation shows that the angle of friction is equal to angle of the inclined plane when a solid body, placed on the inclined plane is about to slide down.

Problem 3.10. Find the least force required to drag a body of weight W, placed on a rough inclined plane having inclination α to the horizontal. The force is applied to the body in such a way that it makes an angle θ to the inclined plane and the body is (a) on the point of motion up the plane and (b) on the point of motion down the plane.

Sol. Given:

Weight of body = W

Inclination of plane = α

Force applied = P

Angle made by force P with inclined plane = θ

(a)             Least force when the body is on the point of motion up the plane.

When the body is on the point of motion up the plane, the force of friction (F = µR) is acting down the plane. The boy is in equilibrium under the action of following forces as shows in Fig. 3.14.

1. Weight (W) of the body acting vertically downwards,
2. Normal reaction (R), perpendicular to the inclined plane,
3. The force of friction, F = µR acting down the plane, and
4. Force P, inclined at an angle θ to the plane.

Resolving the forces along the plane, we get

Problem 3.11. A body of weight 500 N is pulled up an inclined plane, by a force of 350 N. The inclination of the plane is 30° to the horizontal and the force is applied parallel to the plane. Determine the co-efficient of friction.

Sol. Given:

Weight of body, W = 500 N

Force applied, P = 350 N

Inclination, α = 30°

Let µ = Co-efficient of Q friction

R = Normal reaction

F = Force of friction = µR

The body is in equilibrium under the action of the forces shown in Fig. 3.16.

Resolving the forces along the plane,

500 sin 30° + F = 350

500 sin 30° ‘to µR = 350                                   ( F=μR)

Resolving forces normal to the plane,

R = 500 cos 30° = 500 × .866 = 433 N

Substituting the value of R in equation (i), we get

500 sin 30° + µ × 433= 350

or 500 × 0.5 + 433 µ = 350

or 433µ = 350 – 500 × 0.5 = 350 – 250 = 100

Problem 3.12. A body of weight 450 N is pulled up along an inclined plane having inclination 30° to the horizontal at a steady speed. Find the force required if the co-efficient of friction between the body and the plane is 0.25 and force is applied parallel to the inclined plane. If the distance travelled by the body is 10 m along the plane, find the work done on the body.

Sol. Given:

Weight of body, W = 450 N

Inclination of plane α = 30°

Co-efficient of friction, µ = 10.25

Distance travelled by body = 10 m

Let the force required = P

The body is in equilibrium under the action of forces shown in Fig. 3.17

Fig. 3.17 Body moving up

Resolving forces along the plane,

p = W sin 30° + µR = 450 × 0.5 + 0.25 × R

or P = 225 + 0.25 R                                               … (i)

Resolving forces normal to the plane,

R = W cos 30° = 450 × 0.866 = 389.7 N

Substituting the value of R in equation (i),

P = 225 + 0.25 × 389.7 = 322.425 N. Ans.

Work done on the body = Force × Distance travelled in the direction of force

= 322.525 × 10 Nm = 3224.25 Nm

= 3224.25 J (where J = Joules = Nm). Ans.

Problem 3.13.An effort of 200 N is required just to move a certain body up an inclined plane of angle 15°, the force acting parallel to the plane. If the angle of inclination of the plane is made 20°, the effort, required, again applied parallel to the plane, is found to be 230 N. Find the weight of the body and the co-efficient of friction.

(AMIE Summer, 1980)

Sol. Given:

Effort required, P1 = 200 N ; when inclination, θ1 = 15°

Effort required, P2 = 230 N ; when inclination, θ2 =20°.

In both the cases, the effort is applied parallel to the inclined plane and body is just to move up. Hence the force of friction (F =µr) will be acting downwards.

1st Case

P1=200N, θ1=15°

Let W = Weight of body,

µ = Co-efficient of friction,

R1 = Normal reaction, and

F1 = Force of friction

= µR1

The body is in equilibrium under the action of the forces shown in Fig. 3.18.

Resolving the forces along the plane,

W sin 15° + F1 = 200

or W sin 15° + µR1 = 200

Resolving the forces normal to the plane,

R1 = W cos 15°

Substituting the value of R1 in equation (i),

W sin 15° + µ × W cos 15° = 200

or W (sin 15° + µ cos 15°) = 200                                                      … (ii)

2nd Case

P2 = 230 N and θ2 = 20°

Let R2 = Normal reaction

Problem 3.14.A rough inclined plane co-efficient of friction = µ, rises 1 cm for every 5 cm of its length. Calculate the effort require to drag a body weighing 100 N up the plane:

(i) when the effort is applied horizontally;, and

(ii) when the effort is applied parallel to the plane.

(AMIE Winter 1988)

Sol. Given

Co-efficient of friction = µ

Rise of plane is 1 cm for every 5 cm of its length. See Fig. 3.19(a)

= 0.2

or θ = sin-1 0.2

= 11.53°

Weight of body, W = 100 N.

(i) Find the effort when it is applied horizontally

The body is in equilibrium under the action of the forces shown in Fig. 3.19 (b).

Problem 3.15. (a) Define co-efficient of friction and limiting friction. (b) Block A weighing 15 N is a rectangular prism resting on a rough inclined plane as shown in Fig. ; 3.20. The block is tied up by a horizontal string which has a tension of 5 N. Find:

(i)                           The frictional force on the block,

(ii)                        The normal reaction of the inclined plane, and

(iii)                      The co-efficient of friction between the surfaces of contact.

(AMIE Nov., 1967)

Sol. (a) For definition of co-efficient of friction and limiting’ friction, please refer to Art. 3,2 and Art.. 3.2,1.

(b) Given:

Weight of block, W = 15 N

Tension in string, T = 5 N

Inclination of plane, a = 45°

F = Frictional force,

R = Normal reaction, and

µ = Co-efficient of friction.

Since there is tension in string which means if string is removed, the block will slide down the plane. Hence force of friction will be acting in the upward direction.

The block A is in equilibrium under the action of the forces shown in Fig. 3.21. The forces are:

1. The weight of block, W = 15 N
2. Horizontal tension in the string, T = 5 N
3. Normal reaction, R
4. Force of friction, F = µR, acting upward. Resolving forces along the inclined plane,

15 sin 45° = F + 5 cos 45°

F = 15 sin 45° – 5 cos 45° = 15 × .707 – 5 × .707

= 10 × .707 = 7.07 N

Resolving forces normal to inclined plane,

R = 15 cos 45° + T cos 45° = 15 cos 45° + 5 cos 45°

= 15 × .707 + 5 × .707 = 20 × .707 = 14.14 N

Using equation (3.1), we get

F = µR

(i)                           Frictional force on the block, F = 7.07 N.  Ans.

(ii)                        Normal reaction of the inclined plane, R = 14.14 N.   Ans.

(iii)                      Co-efficient of friction, µ = 0.5   Ans.

Problem 3.16.Find the force required to drag a body of weight W, placed on a rough inclined plane having inclinationα to the horizontal. The force P is applied to the body horizontally and the body is (α) on the point of motion up the plane, and (b) on the point of motion down the plane.

Sol. Given:

Weight of the body = W

Inclination of the plans = α

Force applied = P.

(a) Force required to drag the body when it is on the point of motion up the plane.

When the body is on the point of motion up the plane, the force of friction F = µR is acting down the plane. The body is in equilibrium under the action of the forces shown in Fig. 3.22.

(b) Force required to drag the body when it is on the point of motion down the plane.

In this case the force of friction F = µR will be acting up the plane. The body is in equilibrium under the action of the forces shown in Fig. 3.23.

Fig. 3.23. Body moving down

Problem 3.17. Find the force required to move a load of 30 N up a rough inclined plane, the force being applied parallel to the plane. The inclination of the plane is such that when the same body is kept on a perfectly smooth plane inclined at that angle, a force of 6N applied at an inclination of 30° to the plane keeps the same in equilibrium. Assume co-efficient of friction between the rough plane and the load is equal to 0.3.

(AMIE Winter, 1983)

Sol. Given:

Load, W = 30 N

Co-efficient of friction between the rough plane and load,

µ = 0.3

Let α= Inclination of the plane with horizontal

P1 = Force required to move the load up a rough inclined plane, when the force is applied parallel to the plane.

The force applied when same body is kept on a smooth inclined plane, P2 = 6 N.

Inclination of the force with the inclined plane, θ = 30°.

1st Case. Consider the body of weight 30 N placed on a smooth inclined plane as shown in Fig. 3.24.

The forces acting on the body are:

i.  The weight (W = 30 N) vertically downward.
ii. The force P2 (= 6 N) at an angle of 30° with the inclined plane.
iii. The normal reaction R.

Resolving forces normal to the inclined plane.

R + P2 sin 30° = W cos α

The forces acting on the body are:

i. The weight W (= 30 N) vertically downward.
ii. The force P1 parallel to the plane.
iii. The normal reaction R*.
iv. Force of friction F = µR*.

Resolving the forces along the inclined plane,

W sin α + F = P1

or 30 sin 9.974°+µR = P1            ( F = µR* and α =9.974°)    … (ii)

Resolving the forces normal to the inclined plane

R* = W cos α

=30 × cos 9.974°                          (  0 = 9.974°)

= 30 × 9.9848 = 29.544.

Substituting the value of R* in equation (ii), we get

30 × sin 9.974 + µ × 29.544 = P1

or 30 × .1732 + 0.3 × 29.544 =P1

or 5.196 + 8.8632 = P1

or P1 = 14.059 N. Ans.

Problem 3.18. A cord connects two bodies of weights 300 N and 800 N. The two bodies are placed on an inclined plane and cord is parallel to inclined plane. The coefficient of friction for the weight of 400 N is 0.15 and that for 800 N is 0.4. Determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place, down the inclined plane. The body weighing 400 N is below the body weighing 800 N.

Sol. Weight of first body, W1 = 400 N

Weight of second body, W2 = 800 N

Co-efficient of friction for first body, µ1 = 0.15

Co-efficient of friction for second body, µ2 = 0.40

Let α = Inclination of plane

T = Tension in cord

R1 = Normal reaction for 1st body

R2 = Normal reaction for 2nd body

F1 = Force of friction between 1st body and plane = µ1 R1

F2 = Force of friction between 2nd body and plane = µ2R2

As the motion of the two bodies is about to take place down the inclined plane, the force of friction AF1 and F2 will be acting upward. The two bodies are in equilibrium under the action of forces shown in Fig. 3.26.

Forces on the first body

Resolving forces along the plane,

400 sin α= T + F1 =T + µ1R1 = T + 0.15 R1                                        … (i)

Resolving forces normal to the plane,

400 cos α = R1

Substituting the value of R1 in equation (i),

400 sin α = T + 0.15 × 400 cos α

= T + 60 cos α

T = 400 sin α – 60 cos α                                                               … (ii)

Forces on the second body

Resolving forces along the plane,

800 sin α + T = F2= µ2R2 = 0.40 R2                                               … (iii)

Resolving forces normal to the plane,

R2 = 800 cos α

Substituting the values of R2 in equation (iii),

800 sin α + T = 0.40 µ 800 cos α = 320 cos α

T = 320 cos α – 800 sin α

Equating the values of T, given by equations (ii) and (iv),

400 sin α – 60 cos α == 320 cos α – 800 sin a

400 sin α + 800 sin α = 320 cos α + 60 cos α

1200 sin α = 380 cos α

Problem 3.19. Two blocks A and B are connected by a horizontal rod and are supported on two rough planes as shown in Fig. 3.27. If the weight of block B is 1500 N and co-efficient of friction of block A and B are 0.25 and 0.35 respectively, find the smallest weight of block A for which equilibrium can exist.

Sol. Given: Weight of block B,

WB = 1500N

Co-efficient of friction for block A,

µA = 0.25

Co-efficient of friction for block B,

µB = 0.35

Let the smallest weight of block A for equilibrium

=WA·

If the weight of block A is less than the value required for equilibrium, the block B will be slide downwards. But the block A and B are connected by a horizontal rod of fixed length. Now when blocks B starts moving in the downward direction, the block A starts moving towards left. Hence a force of friction FA equal to µA RA will be acting on block A towards right as shown in Fig. 3.27. On block B, the force of friction FB, equal to µBRB will be acting in the upward direction.

For block A

Resolving force normal to plane, RA = WA.

Force of friction, FA = µA RA = µ 0.25 × WA = 0.25 WA·

This force will be transmitted to block B through rod AB.

For block B

The block B will be in equilibrium under the action of the forces, shown in Fig. 3.28. The forces are:

i. The weight of block B = 1500 N acting vertically downwards.
ii. The normal reaction RB of the plane.
iii. The horizontal force = 0.25 WA, transmitted to block B through rod AB.
iv.  Force of friction FB = µB RB = 0.35 RB acting up the inclined plane.

In this case, the forces are resolved horizontally and vertically instead of along the
inclined plane and normal to the plane. For an equilibrium state, the forces acting in any direction must be zero.

Resolving forces horizontally,

0.25 WA + FB cos 60° = RB cos 30°

0.25 WA + 0.35 RB cos 60° = RB cos 30°                        (FB=0.35 RB)

0.25 WA + 0.35 × 0.5 RB = RB × .866

0.25 WA + 0.175 RB = 0.866 RB

or 0.25 WB = 0.866 RB – 0.175 RB = 0.691 RB                     … (i)

Problem 3.20. A block A weighing 100 N rests on a rough inclined plane whose inclination to the horizontal is 45°. This block is connected to another block B weighing 300 N resting on a rough horizontal plane, by a weightless rigid bar inclined at 30° to tile horizontal. Find the horizontal force required to be applied to the block B to just move the block A in upward direction. Assume angle of limiting friction as 15° at all surfaces where there is a sliding.

(AMIE Summer, 1984)

Sol. Given:

Weight of block A, WA = 100 N

Inclination of plane,α= 45°

Weight of block B, WB = 300 N

Inclination of rigid bar with horizontal = 30°

Angle of limiting fraction, ɸ = 15°

Let P = Horizontal force to be applied to block B to just move the block A in the upward direction

T = Thrust in the rigid bar.

Consider the equilibrium of block A

When the block A is at the point of moving up the plane, the forces acting on block A are:

i. Weight (= 100 N) acting vertically downward
ii. Normal reaction RA
iii. Force of friction (FA = µRA) acting down the plane
iv.  Thrust T in the rod at an angle of 30  to the horizontal and 15  with the inclined plane.

Resolving the forces along the plane,

100 sin 45  + FA = T cos 15

100 × 0.707 + µ RA = T × 0.9659

70.7+ 0.2679 RA = 0.9659 T                   (μ=0.2679)

Resolving the forces normal to the plane

100 cos 45  + T sin 15  = RA

100 × 0.707 + T × 0.2588 = RA

Substituting the value of RA in equation (i), we get

70.7 + 0.2679 (100 × 0.707 + T × 0.2588) = 0.9659 T

70.7 + 0.2679 × 100 × 0.707 + T × 0.2588 × 0.2679 = 0.9659 T

70.7 + 18.94 + 0.0693 T = 0.9659 T

89.64 = 0.9659 T – 0.0693 T = 0.8966 T

Consider the equilibrium of block B

The forces acting on the block D are

i. Weight equal to 300 N acting vertically downward
ii. Normal reaction RB
iii. Horizontal force P
iv. Thrust T at an angle of 30  with the horizontal
v. Force of friction, FB = µRB acting towards lift.

Resolving the forces normal to the plane,

300 + T sin 30  = RB

300 + 49.985 =RB

RB = 300 + 49.985 = 349.985 N.

Resolving the forces along to the plane,

P = T cos 30  + µRB

= 99.97 × 0.866 + 0.2679 × 349.985

( µ = 0.2679, RB = 349.985)

= 86.574 + 93.76 = 180.334 N. Ans.

Problem 3.21. Refering to the Fig. 3.29 (a) given below, determine the least value of the force P to cause motion to impend rightwards. Assume the co-efficient of friction under the blocks to be 0.2 and pulley to be frictionless.

(AMIE Winter, 1988)

Sol. Given:

Co-efficient of friction under both blocks, µ = 0.2

Pulley is frictionless. Motion of block of weight 100 N is towards right. Find least value of P.

Consider the equilibrium of block of weight 150 N

As the block of weight 100 N tends to move rightwards, the block of weight 150 N will tend to move upwards. Hence force of friction will act downwards as shown in Fig. 3.29 (b).

Fig. 3.29 (b)

The weight 150 N is acting vertically downwards. The body is in equilibrium under the action of forces shown in Fig. 3.29 (b). Resolving the forces along the plane,

T = 150 sin 60° + µR                                       … (i)

Resolving the forces normal to inclined plane,

Substituting the value of R in equation (i),

T = 150 sin 60° + 0.2 × 75                          (μ=0.2)

= 144.9N

2nd Case

Now consider the equilibrium of block of weight 100 N

The block of weight 100 N tends to move rightwards, hence force of friction will be acting towards left as shown in Fig. 3.29 (c). Also the pulley is frictionless hence the tension in the string which is attached to the block of weight 100 N will be 144.9 N. The body will be in equilibrium under the action of forces shown in Fig. 3.29 (c).

Resolving forces along the plane, i.e., horizontally),

p cos θ = T + 0.2 R*

= 144.9 + 0.2 R*                                                      … (iii)

Resolving the forces normal to the plane (i.e., vertically)

R* + P sin θ = 100

R* = 100 – P sin θ

Substituting the value of R* in equation (iii),

P cos θ = 144.9 + 0.2 × (100 – P sin θ) = 144.9 + 20 – 0.2 P sin θ

P cos θ + 0.2 P sin θ = 164.9

P (cos θ + 0.2 sin θ) = 164.9

Problem 3.22. What should be the value of the angle θ in Fig. 3.29 (d) so that the motion of the 90 N block impends down the plane? The co-efficient of friction µ for all the surfaces is 1/3.

(AMIE Summer, 1990)

Sol. Given:

Co-efficient of friction for all surfaces, µ = 1/3

Motion of weight 90 N impends down the plane. Find the value of θ.

First Consider the equilibrium of weight 30 N.

As the weight 90 N tends to move downwards, there will be a rubbing action between the surfaces of weight 90 N and 30 N. Hence a force of friction will be acting between these two forces.

The weight 30 N is tied to a string, the other end of

The string is fixed to the plane. When the weight 90 N tends to move downwards, the weight 30 N with respect to weight 90 N will move upwards. Hence the force of friction on the lower surface of the weight 30 N will act downward as shown in Fig.3.29 (e). The weight 30 N will be in equilibrium under the action of the forces shown in Fig. 3.29 (e) in which

T = Tension in the string

R1 = Normal reaction on the

Lower surface of weight 30 N

F1 = Force of friction = µR1

Resolving the forces along the plane,

T = 30 sin θ + µR1

3.4.3. Analysis of Ladder Friction. Fig. 3.30 shows a ladder AC resting on the ground and leaning against a wall.

Let RA = Reaction at A

RC = Reaction at C

FA = Force of friction at A

=µRA

FC = Force of friction at C

=µRC

Due to the self weight of the ladder or when some man stands on the ladder, the upper end A of the ladder tends to slip downwards, and hence the force of friction between the ladder and the vertical wall FA = µRA will be acting upwards as shown in Fig. 3.30. Similarly, the lower end C of the ladder will tend to move towards right and hence a force of friction between ladder and floor FC = µRC will be acting towards left.

For the equilibrium of the system, the algebraic sum of the horizontal and vertical components of the forces must be zero. Also the moments of all the forces about any point must be zero.

Note. If the vertical wall is smooth, there will be no force of friction between the ladder and the vertical wall.

Problem 3.23. A uniform ladder of length 10 m and weighing 20 N is placed against a smooth vertical wall with its lower end 8 m from the wall. In this position the ladder is just to slip. Determine :

(i)                           the co-efficient of friction between the ladder and the floor, and

(ii)                        frictional force acting on the ladder at the point of contact between ladder and floor.

Sol. Given:

Weight of ladder, W = 20 N

Length of ladder, AC = 10 m

Distance of lower end of ladder from wall, i.e.,

In right-angled triangle ABC,

As vertical wall is smooth, hence there will be no force of friction between ladder and wall.

In the position of the ladder shown in Fig. 3.31, the ladder is just to slip. Hence the lower end C of the ladder will tend to move towards right and hence a force of friction between ladder and floor (i.e., FA = µRC) will be acting towards left.

Let RA = Reaction at A

RC = Reaction at C

µ = Co-efficient of friction between ladder and floor at C

FC = Force of friction at C =µRC

Ladder is uniform and the weight of ladder (W = 20 N) is acting at the middle point of AC to G. The line of action of W will pass through the middle point of BC. Hence distances

Taking the moments of all the forces about point C,

Cloakwise moments = Anti-clockwise moments

RA × AB = 20 × CD

(The moments of RC and FC about point C is zero)

(ii) Frictional force acting at C is given as

FC = µ × RC = 0.67 × 20 = 13.40 N.   Ans.

Problem 3.24. A uniform ladder of length 13 m and weighing 25 N is placed against a smooth vertical wall with its lower end 5 m from the wall. The co-efficient of friction between the ladder and the floor is 0.3. Show that the ladder will remain in equilibrium in this position. What is the frictional force acting on the ladder at the point of contact between the ladder and floor ?

(AMIE Winter, 1981)

Sol. Given:

Length of ladder, L = AB = 13 m

Weight of ladder, W = 25 N

Distance of lower end of ladder from wall,

AC = 5 m

Co-efficient of friction between ladder and floor,

µ = 0.3

Vertical wall in smooth and hence there will be no force of friction between ladder and wall

Let FA = Limiting frictional force acting at A = µRA = 0.3 RA

RA = Normal reaction at A

RB = Normal reaction at B

The ladder AB is placed as shown in Fig. 3.32. The weight of 25 N is acting at the middle point of AB (i.e., at G) vertically downwards. If the ladder is not in equilibrium, it will start moving at A towards right and force of friction (FA) will act towards left as shown in Fig. 3.32. The forces acting on the ladder are shown in Fig. 3.32.

Equating the vertical forces,

RA = 25 N

Equating the horizontal forces, we have

RB = FA = µRA = 0.3 RA = 0.3 × 25 = 7.5 N

Maximum amount of force of friction available at A is

FA = 7.5 N

To prove that ladder is in equilibrium for the given position

Let. FA’ = Force of friction when ladder is in equilibrium

RB’ = Normal reaction at B when ladder is in equilibrium

RA’ = Normal reaction at A when ladder is in equilibrium

The ladder will be in equilibrium if FA’ is less that FA.

Hence the equilibrium, the force of friction required as 5.21 N. But maximum amount of force of friction available is 7.5 N which is more than required amount. Hence the ladder will remain in equilibrium in the given position.

Problem 3.25. The weight of 14 m long bar as shown in Fig. 3.33 (a) is 600 N and it may be considered to be concentrated at a point 6 m from the bottom. It rests against a smooth vertical wall at A and on a rough horizontal floor at B. The co-efficient of static friction between the bar and the floor is 1/3. Establish by calculations if the bar stand in the 600 position as shown.

Sol. Given:

Length, AB = 14 m

Weight, W = 600 N

The weight is acting at D, which is 6 m from point B

Length BD = 6 m and length AD = 8 m

Vertical wall is smooth and hence FA = 0

Co-efficient wall of friction, µ = 1/3

Angle CBA = 600

Let us find:

(i)                           actual force of friction available at point B, and

(ii)                        force of friction required for equilibrium at point B

The force of friction required for equilibrium will be obtained from moment equation i.e., M=0. The actual force of friction available will be obtained from summation of forces in horizontal and vertical directions i.e., Fx = 0 and Fy = 0.

Let, RA = Normal reaction at A

RB = Normal reaction at B

FB = Force of friction at B = µRB

The forces acting on the ladder are shown in Fig. 3.33 (a)

Resolving forces horizontally, RA = FB                                           … (i)

Resolving forces vertically, RB = 600 N                                         … (ii)

As the actual force of friction available at B is 200 N, which is more than the force of friction required for equilibrium, the ladder will stand stable in the 600 position as shown in the given figure.   Ans.

Problem 3.26. (S.I.Units). A uniform ladder of weight 850 N and length 6 m rests on a horizontal ground and leans against a smooth vertical wall. The angle made by the ladder with the horizontal is 650. When a man of weight 750 N stands on the ladder at a distance 4 m from the top of the ladder, the ladder is at the point of sliding. Determine the co-efficient of friction between the ladder and floor.

Sol. Given:

Weight of ladder, W = 850 N

Length of ladder, L = AB = 6 m

Angle made by ladder with horizontal, α = 650

Weight of man, W1 = 750 N

Distance of man from top of the ladder, L1 = 4 m

Let µ = Co-efficient of friction between the ladder and floor. Vertical wall is smooth and hence there will be no force of friction between the ladder and vertical wall.

Let AB is the ladder and G is the middle point of the ladder at which the weight 850 N is acting. The man of weight 750 N is standing at E. At this position, the ladder is at the point of sliding. This means that ladder at A will be start moving towards right. Hence a force of friction FA = µRA will be acting towards left as shown in Fig. 3.34.

Now RA = Normal reaction at A

RB = Normal reaction at B

The forces acting on the ladder are shown in Fig. 3.34.

Resolving forces vertically,

RA = 850 + 750 = 1600 N                                                   … (i)

Resolving forces horizontally,

RB = FA = µRA = µ × 1600 = 1600 µN                                                … (ii)

In right-angled ΔABC, BC = AB sin 650 = 6 sin 650

= 6 × .9063 = 5.437 m

AC = AB cos 650 = 6 cos 650 = 6 × .4226 = 2.5357 m

As G is the middle point of AB and GD is normal to AC.

Problem 3.27. (S.I.Units). A uniform ladder of weight 200 N of length 4.5 m rests on a horizontal ground and leans against a rough vertical wall. The co-efficient of friction between the ladder and floor is 0.4 and between ladder and vertical wall is 0.2. When a weight of 900 N is placed on the ladder at a distance of 1.2 m from the top of the ladder, the ladder is at the point of sliding. Find:

(i)                           The angle made by the ladder with horizontal,

(ii)                        Reaction at the foot of the ladder, and

(iii)                      Reaction at the top of the ladder.

Sol. Given:

Weight of ladder, W = 200 N

Length of ladder, L = AB = 4.5 m

Co-efficient of friction between ladder and floor,   µA = 0.4

Co-efficient of friction between ladder and wall, µB = 0.2

Weight of ladder, W1 = 900 N

Distance BE = 1.2 m

Let α = Normal reaction at A

RA = Normal reaction at A

FA = Force of friction at A = µA RA

RB = Normal reaction at B

FB = Force of friction at B = µB RB

When the ladder AB is on the point of sliding, the foot of the ladder will move towards right and hence a force of friction FA = µA RA will be acting towards left. The top of the ladder will be moving downwards on the vertical wall and hence a force of friction FB = µB RB will be acting upwards as shown in Fig. 3.35.

Problem 3.28. A ladder 5 m long and 250 N weight is placed against a vertical wall in a position where its inclination to the vertical is 300. A man weighing 800 N climbs the ladder. At what position will he induce slipping? The co-efficient of friction for both the contact surfaces of the ladder i.e., with the wall and the floor is 0.2.

(AMIE Summer, 1983)

Sol. Given:

AB is the ladder and G is the middle point of the ladder at which the weight 250 N is acting. The man of weight 800 N is standing at E, at a distance x from A, when the ladder is on the point of slipping. This means the ladder at A will start moving towards right and a force of friction FA = µRA will be acting towards left at A. The ladder at B will be moving downwards and hence a force of friction FB = µRB will be acting upwards at B.

Now RA = Normal reaction at A

RB = Normal reaction at B

The forces acting on the ladder are shown in Fig.3.36.