Chapter Four Class 9th

CHP # 4 Q.1 A.

Turning effect 0f forces

State and explain addition of forces (vectors)? Addition of vectors Two or more than two vectors are graphically added by “Head to tail rule”. This rule states that “draw the first vector, connect the first vector head and the second vector tail. Connect the second vector head and the third vector tail and so on. At last connect the first vector tail and the last vector tail. This is the resultant vector”R”. For example A B

 R

 B

 A

   R  A B Q.2 A.

What is trigonometry? Trigonometry The branch of mathematics with the help of which we can calculate the sides and angles of a right angle triangle B

Hypotenuse

Perpendicular

 O

A Base

In triangle OAB, m  OAB = 90 0 and m AOB =  , Then

Sin 

Per AB  Hyp OB

Cos 

base OA  Hyp OB

Tan 

Per AB  base OA

Trigonometric functions and their values are;  00 30 0 450 60 0

Q.3 A.

90 0

Sin

0

0.5

0.707

0.866

1

Cos

1

0.866

0.707

0.5

0

Tan

0

0.577

1

1.732



State and explain resolution of forces (vectors)? Resolution of vectors The process in which one vector is decomposes to two or more than two vectors. The vectors obtain are called components of the original vector. For example Let a force “F” is acting along x-y plane, if we draw a normal on x-axis and also on y-axis such as shown in the figure. Y

F

Fy



X

Fx    The components of the force F are Fx and F y .

We know from trigonometry that

 Fy Sin   F   Fy  FSin -------------- (1)  Fx Cos   F

  Fx  FCos

------------- (2) Squaring and adding equation (1) and (2)

Fx  Fy  F 2 Cos 2  F 2 Sin 2 2

2

Fx  Fy  F 2 ( Sin 2  Cos 2 ) 2

2

Sin 2  Cos 2  1 Fx  Fy  F 2 (1) 2

F 

2

Fx  Fy 2

2

For direction of “F” we have

Tan 

Fy Fx

  Tan 1 Q.4 A.

Fy Fx

What is meant by rectangular components? The components of a vector which are mutually perpendicular to each other are called Rectangular components. For example F

Fy

Fx

 The rectangular components of the force F are

  Fx  FCos   F y  FSin

Q.5 A.

Define moment of a force (Torque). On what factors does it depends? Moment of force The turning effect produced by the force is called moment of force (Torque). It is the product of force and moment arm of the force. It is denoted by “ ”. Mathematically Torque = Force  moment arm of force



    F r 



  FrSin

The SI unit of torque is Newton-meter (N-m). Clockwise torque is considered negative and anti-clockwise torque is positive. Torque depends on the following factors; (1) Force Greater the applied force, greater will be the torque and vice-versa. (2) Moment arm of force The perpendicular distance between the force and the axis of rotation is known as moment arm of force. Greater the moment arm of force, greater will be the torque and vice-versa.

Q.6 A.

Define equilibrium of a body. State the two conditions of equilibrium. Equilibrium The state of rest or uniform motion of an object is called its state of equilibrium. For examples A boy holding a vase in his hand, the vase is at rest under the action of two forces. The weight of the vase acting downward and normal reaction of the hand acting upward. Conditions of equilibrium are as under. First condition of equilibrium According to this condition “If the sum of all the forces acting on a body is zero, then the body is in equilibrium state.” Mathematically Sum of all forces = 0

 F  0

Fx  0

Fy  0 Second condition of equilibrium According to this condition “If the sum of all the torques acting on a body is zero, then the body is in equilibrium state.” Mathematically Sum of all torques = 0

   0

Clockwise torques = anti-clockwise torque.

Q.7 A.

Q.8 1.

2.

Explain the stability of an object with reference to position of center of mass. The stability of a body in equilibrium depends upon the position of the center of mass of the body. When an external force is applied on the body, it will disturb the position of center of mass. Consider a circular cone which when lying from its base on the horizontal surface, it is in equilibrium. The cone will remain its stability forever (stable equilibrium). But if this cone is kept on its end, it falls and can not keep its equilibrium position due to change of the center of mass (unstable equilibrium). If the cone is kept on the side way, it will roll on the surface without changing the center of mass (neutral equilibrium). Define the terms; Parallel forces Those forces whose point of action is different but the line of action are same are called parallel forces. Like parallel forces Those parallel forces which are act in same direction are called like parallel forces. For example

Q 3.

P

unlike parallel forces Those parallel forces which are act in opposite direction are called unlike parallel forces. For example

Q

P

4.

Center of mass That specific point of a body about which mass is equally distributed in all directions. A B c.m

C

5.

Couple Those two forces which are equal in magnitude, opposite in direction but does not cancel each other are called couple.

Q

P 6.

Center of gravity It may be define as “that fixed point of a body at which the whole weight of the body appears to act is known as center of gravity.” Table of center of gravity of regular shape objects; S.NO Object Centre of gravity 1

Sphere

Centre of the sphere

2

Uniform rod

Centre of the rod

3

Rectangular plate

Intersection of diagonals

4

Triangular plate

Intersection of medians

Conceptual questions 1. A.

2. A.

In which of the following case or cases, there is no horizontal force on the suitcase and why? In figure (a) the force on the suitcase when resolved it will have only vertical component, because it is acting along y-axis. in which case (A, B or C) the x and y components of the force equal to one another? It is clear that the value of Sin45 and Cos45 are equal to 0.707. Therefore in B the x and y components of the force are equal.

3. A.

Why is it more difficult to lean backward, explain? The center of gravity of foot is half way along the foot from head to toe. When we keep the foot on ground, it is almost in contact. If we lean backward it is not possible to exert a backward force on the foot and hence we will loss our balance and will fall.

4. A.

Explain why door handles are not put near hanger? The door rotates about the hinges, due to torque. The torque depends upon the force and moment arm of the force. If the door handles are put near the hanger, then the moment arm of force become zero. The torque also becomes zero and the door will not open.

5. A.

Why does a helicopter has a second rotor on its tail? The main rotor of the helicopter provide a torque, in order to provide an anti toque to the helicopter, the second small rotor is mounted at the tail of the helicopter.

6. A.

Why it is better to use a long spanner rather than short one to tighten a nut on the bolt? In the long spanner the moment arm for the rotation of the nut on bolt increases which decreases the required force to tighten the nut on the bolt. Therefore it is better to use the long spanner than short one.

7. A.

Which of the glasses is the most stable? Explain your answer. The glass in figure (c) is the most stable due to plane surface area of stability.

8.

A girl is seated safely and steadily in a canoe but when she tries to stand up, the canoe capsizes. Explain this in terms of center of mass and stability. When girl is seated in canoe, the center of mass is uniformly distributed due to which center of gravity in this position can not be changed easily and hence she is in stable position. But when she tries to stand up, the center of gravity of her body changes and then she loss her stability and fall.

A.

9. A.

How you would determine the center of gravity of irregular shape body? The center of gravity of irregular object can be determined experimentally. Take a card board of irregular shape. Drill a few holes A, B and C near the edges of the card board. A B C

Now suspended the card board from point A with the help of a nail on the wall and draw a plumb line from the suspended nail. A B C Similarly draw two plumb lines from point B and C. The three lines meet at the same point. This point is called center of gravity of the object. A B C 10. A.

The gravitational force acting on the satellite is always directed towards the center of the earth. Does this force exert torque on satellite? The gravitational force acting on the satellite is always directed towards the center of the earth, so this is a central force. For the central force the moment arm is zero and hence torque is zero.

Numerical questions 1.

A force of 100N is applied perpendicularly at a distance of 0.50m to turn a nut of the wheel of a bus. Find the torque acting on the nut? Given data

F  100N r  0  50m  ?

We know that

  rF   0  50  100   50Nm

2.

A mobile crane lifting some material of the building of 10000N. This weight produces turning effect on the arm of the crane which is 15m, calculate moment of the force? Given data

F  10,000N r  15m  ?

We know that

  rF   15  10,000   150000Nm

3.

Two kids of weighing 300N and 350N are sitting at the ends of 6m long sea-saw. The see-saw is pivoted at its center. Where would a third kid sit so that the see-saw is in equilibrium in the horizontal position? The weight of the third kid is 250N (ignore the weight of see-saw). Given data Weight of 1st kid = W 1= 300N Weight of 2nd kid = W 2= 350N Weight of 3rd kid = W 3= 250N Length of see-saw = AB = 6m AC = CB = 3m CD =?

A

D

W1

W3

?

C

B

W2

We know that Clockwise torques = anti clockwise torques

W1  AC  W3  CD  W2  CB 300  3  250  CD  350  3 900  250CD  1050 250CD  1050  900 250CD  150 150 CD  250 CD  0  6m

4.

A student carries out an experiment to balance a regular 4m long plank at its mid point. How much weight is placed at 80cm to the left of the pivot so that it balance a mass of 3 2 Kg placed at 100cm to the right of the pivot? 100cm

1m

W1

80cm

1 2m

W2

Given data

m1  3  2 Kg

W1  m1 g  3  2  10  32N r1  1m r2  1  2m W2  ? We know that Clockwise toques = Anti-clockwise torques

W1  r1  W2  r2 W2 

W1  r1 r2

32  1 1 2 W2  26  6 N W2 