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Wize High School Grade 12 Physics Textbook > Dynamics

Spring Force

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Spring Force


  • When springs extend or compress, they "want" to return to their equilibrium position. The force points back to the original position of the spring, and is called a restoring force. (It wants to restore the initial position.)

  • The magnitude of the force is proportional to the distance away from the equilibrium position, Δx=x−xo\Delta x =x-x_o where equilibrium is a point at which the net force on spring is zero.
  • The constant of proportionality is called the spring constant, k.


F⃗s=−kΔx⃗\boxed{ \hspace{3cm} \vec{F}_s = -k\Delta \vec{x} \hspace{3cm}}


  • The negative sign indicates that the force points to the initial position of the spring.
  • This is called Hooke's Law.
  • We can see from this equation that the spring constant must have units of N/m.







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Example: Hooke's Law (Finding Spring Constant)


A stretchy rope suspended from the ceiling is 59 cm long when a 70 N weight hangs from it, but is 80 cm long when a 205 N weight hangs from it. Determine the spring constant “k” of the rope.

Solution:


X1=0.59 m
F1=70N=mg
X2=0.8 m
F2=205N
k=?


F⃗=−kΔX⃗\vec{F}=-k\Delta\vec{X}
FH⃗ Hooke′s law=−kΔx⃗ \vec{F_H}\ Hooke's\ law=-k\Delta\vec{x}\
F=mg=FH → Because of equillibrium .F=mg=F_H\ \rightarrow\ Because\ of\ equillibrium\ .

magnitude
F1=kΔx1=k(0.59−xo)F_1=k\Delta x_1=k\left(0.59-x_o\right) k(0.59−xo)=70k\left(0.59-x_o\right)=70
⟹ \Longrightarrow\
F2=kΔx2=k(0.8−xo)F_2=k\Delta x_2=k\left(0.8-x_o\right) k(0.8−xo)=205k\left(0.8-x_o\right)=205

2 equations , 2 unknowns  ↗k\ \nearrow k
↘xo\searrow x_o
0.59k−kxo=700.59k-kx_o=70
0.8k−kxo=205           ⟹   k=643 Nm0.8k-kx_o=205\ \ \ \ \ \ \ \ \ \ \ \Longrightarrow\ \ \ k=643\ \frac{N}{m}
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Example: Coffee on Springs

Clara is measuring out some coffee beans. Her scale is composed of 4 springs on the corners of a square platform. She measures out 24.0 g of coffee beans.

If each spring compresses evenly, and has spring constants of k=1.50N/mk = 1.50 N/m , how much lower is the square platform with the beans on it?





Let's draw the free-body diagram. The four springs apply a force upward, while the weight of the beans push downward.



Since the system is in equilibrium, we know that there is no acceleration, and we have ΣF⃗=0\Sigma \vec F = 0 . The change in platform height is given by Δy\Delta y .

ΣF⃗=0 4kΔy−mg=0 Δy=mg4k=(24.0×10−3kg)(9.81m/s2)4×1.5N/m=0.0392m\begin{array}{c} \Sigma \vec{F} = 0 \\~\\ 4 k \Delta y -mg = 0 \\~\\ \Delta y = \frac{mg}{4k} = \frac{ (24.0 \times 10^{-3} kg ) (9.81m/s^2 )}{4 \times 1.5 N/m } = 0.0392 m \end{array}

So the springs are compressed about 4cm after the coffee beans are sitting on them.


Which of the following curves best describes the dependence of the magnitude of the spring force as a function of displacement from equilibrium?


Practice: Hooke's Law

A spring with a spring constant (k) of 300 N/m is compressed 15 cm from its equilibrium position.

What is the magnitude of the restorative force acting on the spring?
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A stretchy rope suspended from the ceiling is 59 cm long when a 70 N weight hangs from it, but is 80 cm long when a 205 N weight hangs from it. Determine the spring constant “k” of the rope.

Solution:


X1=0.59 m
F1=70N=mg
X2=0.8 m
F2=205N
k=?


F⃗=−kΔX⃗\vec{F}=-k\Delta\vec{X}
FH⃗ Hooke′s law=−kΔx⃗ \vec{F_H}\ Hooke's\ law=-k\Delta\vec{x}\
F=mg=FH → Because of equillibrium .F=mg=F_H\ \rightarrow\ Because\ of\ equillibrium\ .

magnitude
F1=kΔx1=k(0.59−xo)F_1=k\Delta x_1=k\left(0.59-x_o\right) k(0.59−xo)=70k\left(0.59-x_o\right)=70
⟹ \Longrightarrow\
F2=kΔx2=k(0.8−xo)F_2=k\Delta x_2=k\left(0.8-x_o\right) k(0.8−xo)=205k\left(0.8-x_o\right)=205

2 equations , 2 unknowns  ↗k\ \nearrow k
↘xo\searrow x_o
0.59k−kxo=700.59k-kx_o=70
0.8k−kxo=205           ⟹   k=643 Nm0.8k-kx_o=205\ \ \ \ \ \ \ \ \ \ \ \Longrightarrow\ \ \ k=643\ \frac{N}{m}