Friction MCQ Quiz in मल्याळम - Objective Question with Answer for Friction - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Friction:

When one surface slides or tends to slide over another surface, resistance offered to the motion is known as friction. It is also called as a force of friction or frictional resistance.

Coefficient of friction (μ):

It is the ratio of the magnitude of limiting force of friction to the normal reaction between surfaces in contact

\({\rm{\mu }} = \frac{{{\rm{Limiting\;force\;of\;friction}}}}{{{\rm{Normal\;reaction}}}} = \frac{F}{R}\)

∴ F = μ R

The angle of friction (ϕ):

It is defined as the angle made by resultant reaction (S) with normal reaction (R).

We can see that

\(\tan \phi = \frac{{\rm{F}}}{{\rm{R}}} = \frac{{{\rm{\mu R}}}}{{\rm{R}}} = {\rm{\mu }}\)

∴ ϕ = tan^-1 μ

The angle of repose (θ):

It is the maximum inclination at which a block placed on the inclined plane just begins to slide due to its own weight only.

Concept:

Using equation of motion:

V = u + at

where V = Final velocity, u = Initial velocity, a = acceleration, t = time

Calculation:

Given:

u = 60 km/hour, = \(\frac{{60\; \times 1000}}{{60\; \times \;60}}\)  = 16.67 m/sec                                

t = 6 sec

When brake is applied then final velocity V = 0

So,

V = u + at

0 = 16.67 + a × 6 

a = \(\frac{{ - \;16.67\;}}{{6\;}}\) = - 2.77 m/sec²

Acceleration is negative that means retardation will occur,

a = 2.77 m/sec² (retardation)

From figure

F = f = µN

By newton’s law F = ma

Comparing above equations we get,

ma = mgµ

µ = \(\frac{a}{g} = \frac{{2.77}}{{9.81}} = 0.3\;\)

So, the minimum coefficient of friction between the wheels and the road would be 0.3.                                                 

Concept:

Frictional force resists the motion of the object upon a surface, however, this resisting force is limited, and once the applied force on the body is more than the resisting force the body will start moving.

The value of limiting friction is given by

Limiting friction F = μN

where μ is static friction coefficient between the body and the surface and is the N is the normal reaction of the body to the surface.

Calculation:

Given:

W = 1000 N, μ = 0.5, P = 100 N

Limiting friction force

F = μN

F = 0.5 × 1000 = 500 N

Since the applied force (100 N) is less than limiting friction so the body will be at rest.

 As long as the body is at rest, the friction force between the surface and the body will be equal to the force applied on the body

∴ Friction force = P = 100 N

Explanation:

Static friction F_s: 

When the external force is applied to move a body the frictional force resists the body's motion and the object remains at rest until the external force exceeds the static friction force.

• If there is no external force applied to move the body, the frictional force also becomes zero.
• Range of static friction is:  0 < F_s < F_max
• The maximum frictional force (Fmax ) is also called limiting friction. It is the maximum resistance to the motion that a body produces.
• When F < Fmax the body remains at static equilibrium.

If, F = F_max the frictional force is equal to the maximum frictional force and the body is at a limiting equilibrium and the body will start its motion if the external force is increased slightly.

Additional InformationDynamic friction (F_k): 

When two bodies are in relative motion the friction force experienced by them is known as dynamic or kinetic friction.

Kinetic friction is slightly less than the limiting friction force.

Calculation

 Block weight as 200 N, Subjected to a load P inclined at 30° with horizontal

 Surface is not smooth , 

Normal reaction N = 200 + P sin 30° = 200 + 0.5 P

 Frictional force (f) = μN = 0.25 (200 + 0.5 P) = 50 + 0.125 P

For block to slide,

 P cos30° > 50 + 0.125 P

 0.866 P – 0.125 P > 50

\(P > \frac{{50}}{{0.741}} > 67.5N\)

∴ Value of P, so that block starts sliding, is 67.5 N

Concept:

• Friction: The property of a surface that opposes the relative motion between two surfaces is called friction.

Friction (f) = μ N

where μ is the coefficient of friction and N is the normal force

There are three types of friction:

1. Static friction: The friction that exists between an object at rest and a rough surface on which it is placed, such friction is termed as static friction.
   1. The coefficient of friction due to this friction is called a static friction coefficient.
2. Kinetic friction/sliding friction: The friction between the two surfaces when there is a relative motion between the surfaces then it is called kinetic friction.
3. Rolling friction: When a block is rolling (usually a circular body like a ball, tire) on a surface then the friction acting is called rolling friction.

• In the case of rolling motion, there is no relative motion between the two contact surfaces. So there is no work done by the friction in a rolling motion. That's why the rolling friction is less than the sliding/kinetic friction. 
• μRolling < μSliding < μStatic
• μr < μk < μS

1. Friction opposes the relative motion between two surfaces. 
2. When the body is moving then the friction acting is called sliding friction. 
3. Friction in machines wastes energy and also causes wear and tear.

Concept:

The angle at which the resultant of force of limiting friction f and normal reaction R makes with the direction of normal reaction R is called the angle of friction.

Coefficient of static friction, μ = tan θ

In \(\Delta ROC\;tan\;\theta = \frac{{RC}}{{OR}} = \frac{{OB}}{{OR}} = \frac{F}{R} = \mu \)

Hence μ = tan θ

Or

F = W sin θ

R = W cos θ

\(\mu = \frac{F}{R} = \tan \theta \)

So angle of static friction is:

θ = tan^-1μ 

Concept:

For solving this problem, we need to understand the types of friction problems.

Type – I: When the condition of impending motion is known to exist.

Here, the body is in equilibrium and on the verge of slipping.

Friction force is given by;

F = Fmax = μSN …1)

Where μS: coefficient of static friction

N: Normal reaction

Type – II: Neither condition of impending motion nor condition of motion is known to exist. In such cases, steps to be followed are:

i) Assume static equilibrium and solve for frictional force (F), necessary for equilibrium.

ii) If F < Fmax; Fmax = μSN

⇒ Body is in static equilibrium as assumed and F is determined by equations of equilibrium.

iii) If F = Fmax = μSN, motion impends but assumption of equilibrium is still valid.

iv) If F > Fmax; Fmax = μSN

⇒ This is not possible because surfaces can’t support more friction force than μSN. Here assumption of equilibrium is not valid and motion occurs.

Friction force is calculated by,

Fk = μkN …2)

Type – III: When relative motion is known to exist between contacting surfaces, then friction force is calculated by;

F = Fk = μkN

Calculation:

Given data, μS = 0.2, m = 10 kg

Draw free body diagram (FBD);

This problem belongs to type II. Assume body to be in equilibrium.

ΣFx = 0, ΣFy = 0

⇒ F = 10 N; N = mg …3)

Now, calculate maximum value of friction force i.e. limiting friction

Fmax = μSN = (0.2)(10)(9.81) = 19.62 N …4)

∵ F < Fmax ⇒ Body is in static equilibrium as assumed.

The body is in static friction regime and frictional force (F) will be equal to the applied external force.

⇒ F = 10 N  

Key Points

Concept:

Rolling Resistance: It is the minimum force required at center of roller parallel to the contacting surface so that rolling starts. In rolling resistance there will not be point or line contact there is surface contact.

The length of contact is known as coefficient of rolling resistance.

The force necessary to roll the wheel is given by \({{F}_{R}}=\frac{{{\mu }_{R}}W}{R}\)

Where μR is coefficient of rolling resistance

W = Load acting on the wheel, R = Radius of the wheel

Calculation:

Given that D = 600 mm, \(R=\frac{D}{2}=300~\text{mm}=0.3~m\), μR = 0.3, W = 500 N

\({{F}_{R}}=\frac{{{\mu }_{R}}W}{R}=\frac{0.3 \times 500}{300}=0.5~N\)

Concept:

The angle of limiting friction:

Limiting Angle of Friction is defined as the angle which the resultant reaction (R) makes with the normal reaction (RN).

In the limiting case, when the body just begins to move, it is in equilibrium under the action of the following three forces:

1. Weight of the body (W),

2. Applied horizontal force (P), and

3. Reaction (R) between the body A and the plane B.

The reaction R must, therefore, be equal and opposite to the resultant of W and P and will be inclined at an angle (ϕ) to the normal reaction (RN). This angle ϕ is called the limiting angle of friction or simply angle of friction.

\(\phi = {\tan ^{ - 1}}\left( {\frac{F}{{{R_N}}}} \right) = {\tan ^{ - 1}}\left( \mu \right)\)

Important Points

Angle of Repose (α)

The angle of repose or angle of sliding α is defined as the minimum angle of inclination of a plane with the horizontal such that a body placed on the plane just begins to slide down.

Angle of Friction (θ) 

The angle of friction between any two surfaces in contact is defined as the angle which the resultant of the force of limiting friction Flim and normal reaction N makes with the direction of normal reaction N.