Saturday, September 5, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 4, THE FORCES, Questions for Short Answers

"Questions for short Answers"


The Forces

1. A body of mass m is placed on a table. The earth is pulling the body with a force mg. Taking this force to be the action, what is the reaction?


Answer: This action force mg is being applied by the body on the table. To hold the body in position the table also exerts equal and opposite force on the body which is the reaction. So in this case, downward force mg by the body on the table is action and the upward force mg by the table on the body is the reaction. 



2. A boy is sitting on a chair placed on the floor of a room. Write as many action-reaction pairs of forces as you can.

Answer: Let the weight of the boy be W. If this force W be taken as action (downward) on the chair. Equal and opposite force W (Upward) is applied to the boy by the chair which is a reaction. So this is the first action-reaction pair W-W.

Now this weight W plus the weight of the chair W' applies force on the floor through the four legs of the chair. Let P=W+W'. The force P is divided into four forces P1, P2, P3 and P4 (according to the sitting posture of the boy) which act downwards through the legs of the chair. If these be taken as actions, equal and opposite (upwards) forces by the floor on the legs will be reactions. So other pairs of action-reaction are P1-P1, P2-P2, P3-P3, and P4-P4. 


  

3. A lawyer alleges in court that the police has forced his client to issue a statement of confession. What kind of force is this? 


Answer: The lawyer means that the police has threatened his client to issue a statement of confession. So this force is not a physical force but an emotional one. 



4. When you hold a pen and write on your notebook, what kind of force is exerted by you on the pen? By the pen on the notebook? By you on the notebook. 


Answer: When two bodies are in contact and forces of push, pull or friction come into the picture, these forces are categorized in electromagnetic forces. So the push and friction forces exerted by us on the pen, the push force exerted by the pen on the notebook are all kind of electromagnetic force. When we write, the root of the palm exerts a push on the notebook which again is a kind of electromagnetic force. If we do not consider this then we do not directly exert forces on the notebook, it is through the pen. 



5. Is it true that the reaction of a gravitational force is always gravitational, of an electromagnetic force is always electromagnetic and so on?

 

Answer: Yes.



6. Suppose the magnitude of nuclear forces between two protons varies with the distance between them as shown in figure (4-Q1). Estimate the ratio "Nuclear force/Coulomb force" for (a) x=8 fm (b) x=4 fm, (c) x=2 fm and (d) x=1 fm (1 fm =10-15 m).
         

Given figure for Q No - 6

Answer: In the given figure force on the Y-axis is plotted on a logarithmic scale. Magnitude of Nuclear forces from the given figure by logarithmic interpolations are,

For x=8 fm, → 0.18 N  

For x=4 fm, → 1.00 N   

For x=2 fm, → 17.7  N 

For x=1 fm, → 177 N 

Now Coulomb force is given by,

F=q²/4π ε0 r²  = 9x109 .e²/r² = 9x109 .(1.6x10-19 )²/r². (10-15)²  

= 144/r² N    (Here r is in fm)  

So this force,
for x=8 fm, → 2.25 N 
for x=4 fm, → 9.00 N 
for x=2 fm, → 36.00 N   
for x=1 fm, → 144.0 N   
So the ratio "Nuclear force/Coulomb force"   
(a) for x=8 fm is  0.18 N/2.25 N = 0.08 
(b) for x=4 fm is  1.0 N/9.0 N = 0.11  
(c) for x=2 fm is  17.7 N/36.0 N = 0.491 
(d) for x=1 fm is  177 N/144 N = 1.23  
It is clear that in the range of r< 2 fm Nuclear forces are quite comparable to Coulomb forces. 
   

7.  List all the forces acting on the block B in figure (4-Q2).

Given figure for Q No - 7

Answer: Forces acting on Block B are shown below,
Forces acting on block B
 

        Let us take the weight of block B as W. This force of weight W acts on block B. Since the man exerts force (say F) above the level of block B there may be movement tendency between both blocks and a force of friction (P) will be applied by block A on block B downwards. There will be a force R acting upwards on block B by the floor such that R=W+P. The push force F by the man on block A which is transferred to block B through A is counterbalanced by a force of friction between block B and the floor. It is equal but opposite in direction to F.



8.  List all the forces acting on (a) pulley A, (b) the boy and (c) the block C in the figure (4-Q3).
The figure for Question number 8

Answer: (a) For forces acting on pulley A, see the figure below:-

Forces on pulley A


Let us take the weight of block C as W. This force W is transmitted as tension in the string throughout. So in a balanced state, the strings on both sides of the pulley A will apply forces W horizontally and W vertically downward. Taking the pulley weightless and frictionless, the hinge of the pulley will exert a force on the pulley that is equal in magnitude but opposite in direction to the resultant of two perpendicular forces W and W. As shown in the figure it will be equal to √2W and 45° from horizontal/vertical.

(b) The forces on the boy are shown in the figure below,
Forces on the boy
   The string exerts a force W on the boy upwards, the weight F of the boy acts downwards and a force R is exerted by the floor on the boy upwards such that F=W+R. Since the string exerts the force away from the body (or C.G,) of the boy resulting a slipping tendency at the feet touching the floor, it is counterbalanced by a force of friction P (Horizontal) by the floor at the feet of the boy.  

(c) The forces acting on the block C are shown in the figure below,
Forces on the block C
   Since the block C hangs with the string, the forces acting on it are weight W of the block C downwards and the counterbalancing force W by the string upwards.  

9. Figure (4-Q4) shows a boy pulling a wagon on a road. List as many forces as you can which are relevant to this figure. Find the pairs of forces connected by Newton's third law of motion. 
The figure for question number 9

Answer: The various forces acting in the figure are shown below,
Different Forces relevant to the figure
        Weight of wagon W is distributed on wheels as W1 and W2. P1 and P2 are the forces exerted by the floor on the wagon wheels. Forces of friction R1 and R2 are exerted by the wheels on the floor and equal and opposite forces are exerted by the floors on the wheels. Tension force F in the string is exerted by the hand of boy and equal but opposite force is exerted by the string on the hand of the boy. This force F is also exerted by the boy through string on the wagon and equal but opposite force F is exerted by the wagon on the boy. Resultant of forces in the string at the shoulder of the boy is counterbalanced by force R on the string by the shoulder of the boy. Weight W3 of the boy acts downwards while an upward force P3 is applied by the floor on the boy. The backward push on the floor by the boy results in forces of friction R3 and equal but opposite force R3 is acted by the floor on the boy. 

           Pairs of forces connected by Newton's third law of motion are F-F, R1-R1, R2-R2, R3-R3, and R-R. Though other forces are also connected by Newton's third law of motion they do not show as pairs in the figure. Weight W1 and W2 are distributed on the wheels according to loading patterns and C.G. of the wagon and the floor reactions P1 and P2 are according to the net force downwards on the wheels.   



10. Figure (4-Q5) shows a cart. Complete the table shown below.
The figure for question number 10

Force on
Force by
Name of the force
Direction
 Cart
1.
2.
3. 


 Horse
1.
2.
3. 


 Driver
1.
2.
3. 


    

Answer: The table is completed below according to the following figure:-  
Different forces

Force on
Force by
Name of the force
Direction
 Cart

1.Road
2.Road                                        
3.Horse 

 4.Driver
5.Driver
6.Gravity
  R
  F
  P

 W
  F1
  W3
  Upwards
  Horizontally right
Along the rod on the horse, left
  Downward
  Horizontally left
  Downwards
 Horse
1.Cart 

 2.Driver
3.Gravity 
4.Road
5.Road
  P

 H
 W1
 W2
 F1
Along the rod on the horse, right
Along the rope, right
downwards
 Upwards
 Horizontally left
 Driver
1.Horse
2.Cart
3. Cart
4.Gravity 
 H
 R1
 F1
 W
Along the rope, Left
 Upwards
 Right
 Downwards

     

===<<<O>>>===

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For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com 

My Channel on YouTube  →  SimplePhysics with KK

Links to the Chapters



CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

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)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

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)

CHAPTER- 6 - Friction

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".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

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


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


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)


CHAPTER- 2 - "Vector related Problems"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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11 comments:

  1. Tanmay AgarwalSeptember 23, 2018 at 9:48 AM

    thanku so much sir, for providing awesome explanation for questions for short answers of HC verma

    ReplyDelete
  2. GunuappaFebruary 8, 2019 at 12:45 AM

    Thank you Sir for posting answers. Have one doubt in Q10 isn't frictional force exerted by road on horse in right direction.

    ReplyDelete
    Replies
    1. K K MISHRAFebruary 8, 2019 at 6:34 AM

      Dear student, The horizontal force on the road by the horse is towards right but the horizontal force on the horse by the road is towards left. Equal and opposite forces.

      Delete
  3. GunuappaFebruary 10, 2019 at 9:08 PM

    Thank you Sir. I understood this point, but getting confused by the direction of frictional force.
    Horizontal force on the cart by the road is towards right and opposes the motion.
    Push by the horse results in opposite push by the road, but what about the frictional force between them. I think frictional force between them is helping the horse not to slip on the road and helping in the motion of the horse and hence the cart. Could you please tell me the direction of frictional force exerted by the horse on the road.

    ReplyDelete
    Replies
    1. K K MISHRAFebruary 11, 2019 at 6:37 AM

      Dear student! Here do not think that push by the horse is different than friction. In fact the horse is pushing the road through the friction force.

      Delete
    2. GunuappaFebruary 11, 2019 at 10:46 PM

      Ok, Thanks Sir.

      Delete
  4. GunuappaFebruary 11, 2019 at 10:44 PM

    This comment has been removed by the author.

    ReplyDelete
  5. UnknownAugust 5, 2019 at 8:47 AM

    Thank u sir ....but i am having doubt in q 7 ...how we will know that the frictional force exerted by block a on block is in downward direction and not in upward direction ?

    ReplyDelete
    Replies
    1. K K MISHRAAugust 11, 2019 at 10:48 AM

      If due to the friction, the block A has an overturning tendency, then the block B will tend to slide up. Thus the friction by A on the B will be downward.

      Delete
  6. UnknownAugust 5, 2019 at 8:58 AM

    1 more doubt sir ...the push force F will decrease on reaching to bock B ...according to me the push force on B will be F'' which will be less than F ....also sir if the friction force is equal to push force the block B will not move (how we know that it will not move )....please correct me if i am wrong

    ReplyDelete

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