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Understanding Hooke's Law: The Foundation of Elasticity| Chapter 5 Physics 9th

HOOKE'S LAW   Introduction In physics, Hooke's Law is one of the fundamental principles governing how objects deform under external forces . Named after the 17th-century British physicist Robert Hooke, this law provides a crucial understanding of the behavior of elastic materials, such as springs and rubber bands. Whether stretching a rubber band or compressing a spring, Hooke's Law helps explain what happens when forces act on these materials. What is Hooke's Law: Hooke's Law states that the force F needed to extend or compress a spring by some distance x is proportional to that distance. Mathematically, it is expressed as: F= -kx Here k represents the spring constant, which is the measure of the stiffness of the spring, and x is the displacement from the displacement position.  The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.  Understanding the Spring Constant: The spring constant k is a critical co
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Resolution of Force | Chapter 4 Turning Effect of Forces | 9th Physics

 Resolution of Force/ Rectangular Components of Force

The process of splitting of vectors into mutually perpendicular components is called Resolution of Force. 
It is also known as rectangular components of forces. 

Resolution of force
F= Fx+Fy
Now we can find the magnitudes of both of the rectangular components using trigonometric ratios, considering the given right-angled triangle.
Resolution of force
Now in order to solve problems related to the rectangular components of forces we have the following some of the values of the trigonometric ratios at different angles.
Resolution of force
Example 01: A man is pushing a wheelbarrow on a horizontal ground with a force of 300N making an angle of 60 with the ground. Find the horizontal and vertical components of the force.
Data:
F= 300N
θ= 60 Degrees
Fx=?
Fy=?
Solution:
(i)Finding the horizontal component of Force:
Fx= Fcosθ 
Fx=(300).cos(60)
Fx= (300)(0.5)
Fx= 150 N
(ii) Finding the Vertical Component of Force:
Fy= Fsinθ 
Fy=(300).sin(60)
Fy= 300(0.866)
Fy= 259.8N
Result: The horizontal and vertical components of the given force are, 150N and 259.8N.

Example 02: A man is pulling a trolley on a horizontal road with a force of 200N making angle angle 30 with the road. Find the horizontal and the vertical component of force.
Data:
F= 300N
θ= 30 Degrees
Fx=?
Fy=?
Solution:
(i)Finding the horizontal component of Force:
Fx= Fcosθ 
Fx=(200).cos(30)
Fx= (200)(0.866)
Fx= 173.2 N
(ii) Finding the Vertical Component of Force:
Fy= Fsinθ 
Fy=(200).sin(30)
Fy= 200(0.5)
Fy= 100N
Result: The horizontal and the vertical components of the given force are 173.2N and 100N respectively.









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