Matter & Interactions 2nd ed. Practice Problems
Aaron Titus | High Point University
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1310002     The spring force of parallel springs in an exercise apparatus.     1310002
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Question

 

The piece of exercise equipment shown below includes two handles that are on each end of two parallel springs, each with a stiffness 500 N/m. To build strength, a person holds the two handles and pulls them apart with equal magnitude forces applied in opposite directions to the handles, thus stretching the springs. Neglect the mass of the springs and assume that they are pulled horizontally.


Figure: Two parallel springs between two handles that can be pulled apart.

  1. If a person stretches the springs 5 cm from their relaxed lengths and holds it at rest, what is the net force applied to the system and what magnitude force must he apply to each handle?
  2. What magnitude force must he apply to each handle to stretch it twice as far (10 cm) and hold it at rest?
  3. If the two parallel springs were replaced with a single spring such that the same force would be required to stretch it 10 cm as for the two parallel springs, what would be the stiffness of the single spring?

 



Solution

 

(a) Define the system as the entire apparatus of springs and handles. Because equal magnitude, oppositely directed forces are applied to each handle, the net force on the system is zero. As a result the system's momentum (which is zero in this case) will remain zero. This can be seen in the free-body diagram.


Figure: Two parallel springs between two handles that can be pulled apart.

To calculate the magnitude of the force applied to each handle, define a single handle as the system. Let's begin with the right handle at an instant when the springs are stretched 5 cm. The forces on the handle are shown below.


Figure: Free-body diagram for the right handle.

Apply the momentum principle and note that the handle's momentum is not changing (after all, it's at rest).

Therefore, the net force on the handle is zero. Now, apply the definition of net force as the sum of the forces acting on the handle to calculate the net force. Note that the top and bottom springs have identical stiffness and are each stretched 5 cm. Therefore, they are equal forces and can be combined. Use Hooke's law for the spring force.

The force by the right hand on the right handle is to the right, exactly as expected from the picture. Since the net force on entire apparatus is zero, then the left hand pulls with the same magnitude force as the right hand, but in the leftward direction. Thus,

(b) According to Hooke's law, if the spring is stretched twice as far, then it will exert twice the force on the handle. Therefore, the magnitude of the force applied to each handle in this case will be .

(c) Now, replace the two parallel springs with a single spring that exerts the same force, as shown below.


Figure: An identical apparatus with a single spring.

If the force by two springs on a handle is 100 N when stretching them 10 cm, then the force by the substituted single spring on the handle also must be 100 N when stretching it 10 cm. According to Hooke's law

One can derive that the effective spring stiffness of two springs in parallel is

For the identical springs in this case, the effective spring stiffness is . Not surprisingly, this is in agreement with our answer found from Hooke's law.