Hooke’s law

An overview of hooke's law

1. HOOKE’S LAW

2. Hooke’s Law for relatively small deformations  of an object, the displacement   or size of the deformation is directly proportional to the deforming force or load

3. Hooke’s Law Elastic behaviour of solids according to Hooke’s law can be explained by the fact that small displacements of their constituent molecules, atoms, or ions from normal positions is also proportional to the force that causes the displacement

4. Hooke’s Law The concept of Hooke’s Law is that the amount of force applied to a spring or elastic object is proportional to the amount of deformation (length of stretch or compression). The greater the force applied to an elastic object, the more deformation (stretch or compression) there is. With less force applied, there will be less deformation in the spring.

5. the force on unit areas within a material that develops as a result of the externally applied force. Stress Strain the relative deformation produced by stress. For relatively small stresses, stress is proportional to strain.

6. Formula F= kx Where: F = force applied k = force constant x = amount of deformation   To determine whether there is a stretch or compression in the spring, we use a method of signs. If the spring is compressed, both “x” and “F” are negative(-) and when stretched both are positive(+).

7. Sample problems 1. What is the force required to stretch a spring whose constant value is 100 N/m by an amount of 0.50 m?

8. Solution: What is the force required to stretch a spring whose constant value is 100 N/m by an amount of 0.50 m ? Given : F=? k=100 N/m x= 0.50 m   F= kx F=(100N/m)(0.50m) F=50 N

9. Sample problems 2. A force of 100 N is stretching a spring by 0.2 m. Calculate the force constant.

10. Solution: 2 . A force of 100 N is stretching a spring by 0.2 m. Calculate the force constant . Given: Force F = 100 N,            Extension x = 0.2 m, Force constant k = ? k = F/x   k = (100N)/( 0.2m)  k = 500 N/m.

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