Sunday, May 29, 2016

Solutions to Problems on "Friction"-'H C Verma's Concepts of Physics, Part-I, Chapter-6', Questions for Short Answer

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QUESTIONS FOR SHORT ANSWER

1. For most of the surfaces used in daily life, the friction coefficient is less than 1. Is it always necessary that the friction coefficient is less than 1?


Answer: For most of the surfaces in daily life friction coefficient µ is less than 1 but it is always not necessary that µ<1. For example if both the surfaces are glass µ=1 and if both the surfaces are copper it is =1.60 .

µ can be even higher if the surfaces in contact are placed in an evacuated chamber at very low pressure and made smoother and kept free of dust, impurities, oxides etc. 



2. Why is it easier to push a heavy block from behind than to press it on the top and push?

Friction -Questions for Short Answer - 1




Answer: See the picture above. On the left when pressing on the top the block is pushed, the applied force F has a horizontal component Fcosθ which helps in pushing while the vertical component Fsinθ adds to the weight mg. This makes the normal force N=mg+ Fsinθ. If µ is the coefficient of friction


then Fcosθ=µN=µ(mg+ Fsinθ)


Now consider the second case when the heavy block is pushed from behind (On the right in the picture). The applied force F may make an angle θ upwards from the horizontal. In this case the vertical component Fsinθ reduces the force of weight so Normal force becomes N= mg - Fsinθ 


Now pushing force Fcosθ= µ(mg - Fsinθ)


Clearly in this case force required to move the heavy block is less and it is easier to push. 


3.What is the average friction force when a person has a usual 1km walk ?




Answer: We assume that during the walk a person's velocity is uniform, so the distance of 1 km has nothing to do with friction force. But it is the friction force which enables him to walk. This force is equal to the static friction force applied by the road surface on the bottom surface of his shoes. Since during the walk the shoes do not slip on the the surface, the friction force is less than the maximum/limiting friction force. It can be explained as below:-
Let mass of the person is m and static coefficient of friction = µs


Then normal force on the person = weight of the person =mg

Maximum static friction force possible =µs mg

So average friction force when the person in consideration walks ≤ µs mg .

4. Why is it difficult to walk on solid ice ?

Answer: Solid ice is very smooth and the coefficient of friction between it and other material is very small. So the maximum force of friction developed is very small and during normal walking push by walker is more than this maximum force of friction which results in slipping and it gets difficult to walk.



5. Can you accelerate a car on a friction-less horizontal road by putting more petrol in the engine? Can you stop a car going on friction-less horizontal road by applying brakes?
Answer: A car can not be accelerated by putting more petrol in the engine on a friction-less horizontal road because force of friction would not develop which is necessary for acceleration. Same is the case of stopping the car on such surface. Due to lack of the friction no retarding force would develop.

6. Spring fitted doors close by themselves when released. You want to keep the door open for a long time say for an hour. If you put a half kg stone in front of the open door, it does not help. The stone slides with the door and the door gets closed. However,if you sandwich a 20 g piece of wood in the small gap between the door and the floor,the door stays open. Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails.
Answer: It is due to the greater force of friction developed in the second case. The force of friction developed depends upon normal force developed not on the weight of the object. In the first case the normal force developed is only equal to weight of the half kg stone. In the second case the sandwiched piece of wood is pressed down by the door which results in much higher normal force. So the force of friction is much higher too and the small lighter piece of wood is able to stop the door.  





7. A classroom demonstration of Newton's first law is as follows: A glass is covered with a plastic card and a coin is placed on the card. The card is given a quick strike and the coin falls in the glass. (a) Should the friction coefficient between the card and the coin be small or large? (b) Should the coin be light or heavy? (c) Why does the experiment fall if the card is gently pushed?

Answer: (a) The friction coefficient between the card and the coin should be small so that the force of friction on the coin by the card is less and the coin remains almost at the same place.

(b) Friction force on the coin = µmg

Acceleration of the coin =µmg/m =µg

Acceleration is not dependent on mass, so it does not matter whether coin is light or heavy.

(c) If the card is gently pushed the maximum static friction force which opposes the relative motion between card and the coin is not overcome and the coin goes with the card. So the experiment fall.    

8.  Can a tug of war be even won on a friction-less surface?

Answer: A tug of war can not happen on a friction-less surface because the rope will not remain tight as soon as it is pulled. Feet of both party will slip and they will dash against each other.


9. Why do tires have a better grip of the road while going on a level road than while going on an incline?



Answer: On a level road the full weight of the car presses the road and the normal force is maximum and so is the force of friction. While on an incline only a component of weight (mg.cosθ) is available to press the road normally. So normal force is always less than the weight and the force of friction also less than the level road. So the grip of the tires is less.




10.  You are standing with your bag in your hands on the ice in the middle of a pond. The ice is so slippery that it can offer no friction. How can you come out of the ice?



Answer: Since no friction force can develop to move me, I would throw my bag horizontally with a force in the opposite direction to which I want to go. According to Newton's third law of motion the bag too will push me with the same force in the opposite direction. This force will push me and since ice is so slippery that it can offer no friction, that is why me and my bag will come to the edge of the pond sliding though on the opposite directions.




11. When two surfaces are polished, the friction coefficient between them decreases. But the friction coefficient increases and become very large if the surfaces are made highly smooth. Explain.



Answer: When two surfaces are polished the undulations on the surfaces reduce and it offers less resistance to sliding on each other. The actual points of contact where molecular bonds are formed is less. So the friction coefficient between them decreases. 

But when these surfaces are made highly smooth the actual contact points between them increases many folds. At these points molecular bonds are formed. To slide them over each other a large amount of force is needed to break these molecular bonds. So the coefficient of friction becomes very large. 



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Links for the chapter - 

HC Verma's Concepts of Physics, Chapter-7, Circular Motion,

Click here for → Questions for short Answer
Click here for → Objective I
Click here for → Objective II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)

HC Verma's Concepts of Physics, Chapter-6, Friction

Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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HC Verma's Concepts of Physics, Chapter-4, The Forces


"Questions for short Answers"    

Click here for "The Forces" - OBJECTIVE-I

Click here for "The Forces" - OBJECTIVE-II

Click here for "The Forces" - Exercises


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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

Click here for "OBJECTIVE-I"

Click here for EXERCISES (Question number 1 to 10)

Click here for EXERCISES (Question number 11 to 20)

Click here for EXERCISES (Question number 21 to 30)

Click here for EXERCISES (Question number 31 to 40)

Click here for EXERCISES (Question number 41 to 52)

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Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-II"


4 comments:

  1. This comment has been removed by the author.

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  2. Can you plz check firction short answer question 7 b)???

    ReplyDelete
    Replies
    1. Actually in this case only acceleration is not dependent on mass, it is not generalised.

      Delete
  3. In 7 b short ans,I think mass should be heavy so that the coin could maintain inertia of rest

    ReplyDelete