Springs

Hooke’s LawAny spring has a natural length at which it exerts

no net force on the mass m . This length, or position, is called the equilibrium

position and is labeled x0.

If the mass is moved away from this position, in either the left or right direction, the spring will exert a force on the mass that will act in the direction of returning it to the equilibrium position. This force is called the restoring force.

The magnitude of the restoring force is directly proportional to the distance that the spring has been stretched or compressed away from x0. Thus the equation for the restoring force can be written as:

m

m

Spring ExampleA 5kg object is hung from a spring and stretches the spring 15cm from it’s equilibrium position. What is the spring constant (stiffness of the spring)?

How far would a 7kg object stretch the spring?

€

F = −k(Δx)

k = −F

Δx= −

mg

Δx

k = −49

(−.15)= 327 N

m

Dx

€

F = −k(Δx)

Δx =F

−k

Δx =7 *9.8

−327= −.21m

Spring Example

m

x1

An block of mass m hangs from a spring with spring constant k. The spring is stretched to a length x1 as shown in the figure. What would be the new length (call it x3) of the spring if we added two more identical blocks?

Spring Example

m

x3

An block of mass m hangs from a spring with spring constant k. The spring is stretched to a length x1 as shown in the figure. What would be the new length (call it x3) of the spring if we added two more identical blocks?

m

m

€

rF s = −k

v x

x =Fs

−k=

Fg

k

x =3mg

k= 3

mg

k

⎛

⎝ ⎜

⎞

⎠ ⎟

x3 = x0 + x = x0 + 3mg

k

⎛

⎝ ⎜

⎞

⎠ ⎟

Spring Example

m

x1

An block of mass m hangs from a spring with spring constant k. The spring is stretched to a length x1 as shown in the figure. What would be the new length (call it x3) of the spring if we added two more identical blocks?

€

rF s = −k

v x

x1 − x0 =Fs

k

x0 = x1 −Fs

k= x1 −

mg

k

x0

x

Spring Example

m

x3

An block of mass m hangs from a spring with spring constant k. The spring is stretched to a length x1 as shown in the figure. What would be the new length (call it x3) of the spring if we added two more identical blocks?

m

m

€

Δx3,1 =(2m)g

k

€

x =(3m)g

kx3 = x0 + x

x3 = x1 −mg

k

⎛

⎝ ⎜

⎞

⎠ ⎟+ 3

mg

k

⎛

⎝ ⎜

⎞

⎠ ⎟

x3 = x1 + 2mg

k

⎛

⎝ ⎜

⎞

⎠ ⎟

StretchingRigid objects like steel rods behave a lot like springs- Stretch them a just a little and after you release them, they’ll return to their original shape-If they’re very stiff, so you don’t notice much of a stretch-If you stretch them a lot, part or all of the stretch can become permanent, with potentially catastrophic results

View of MatterMatter can be viewed as a series of atoms connected by springs.The stiffer the springs the more tightly the atoms are held together in the material.

The material can be stretched

…as long as we don’t break the springs

The material can be compressed

…as long as we don’t smash the springs

In both cases the material returns to its original shape

Young’s Modulus

The “stress” applied to a material is the Force/Area.

How much the material responds to that stress is called the “strain” DL/L.

The two are related by Young’s Modulus Y.

Y

[Y] = N/m2 = Pascal

A

ExampleA car is suspended from the ceiling by a steel rod 1 cm in diameter and 2 m long. If the car has a mass of 1200kg, by how much does the steel rod stretch?

€

F = mg =1200(9.8) =11760N

2522 1085.7)005.0(

rod theof Area

mxRA €

Ysteel = 200x109 N /m2

€

F

A= Y

ΔL

L

ΔL =FL

AY=

11760(2)

7.85x10−5(200x109)

ΔL = 0.00150m or (1.5mm)

X0

m

Where does Y come from?

X0 ΔX

€

rF 1

€

rF 2

€

rF 3

€

rF G

€

rF net =

r F 1 +

r F 2 +

r F 3

= −kΔr x − kΔ

r x − kΔ

r x

= − 3k( )r Δ x

= −keff

r Δ x

m

€

rF N

Where does Y come from?

Where does Y come from?

W

L

€

F

A= αk0( )

ΔL

L

= YΔL

L

Where does Y come from?

W

L

€

rF = −keff Δ

r L

=W

L

⎛

⎝ ⎜

⎞

⎠ ⎟k0

⎡

⎣ ⎢

⎤

⎦ ⎥Δ

r L

F

W= k0( )

ΔL

L

€

F

A= k0( )

ΔL

L

= YΔL

L

ΔL

Block on the wallAn 8 N horizontal force F pushes a block weighing 12.0 N against a vertical wall . The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially.

€

rF G = m

v g =12N

∴ m =12

9.8=1.22kg

NF 8N

€

rf s ≤ μ sN

Nf kk Horizontally the forces add up to be

zero (no motion in this direction)

Therefore, N and F have the same magnitude but opposite directions

NNf

NNf

ss

kk

52.5)8(69.

92.3)8(49.

Since the weight is larger than the static friction force then the block is sliding down the wall (kinetic).

Block on the wallAn 8 N horizontal force F pushes a block weighing 12.0 N against a vertical wall . The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially.

€

rF G = m

v g =12N

∴ m =12

9.8=1.22kg

NF 8N

€

rf s ≤ μ sN

NNf kk 92.3 Vertically the forces don’t add up to be

zero (acceleration)

€

rf k +

r F G =

r F net

μ kN − mg = ma

a =μ kN − mg

m

=.49(8) −1.22(9.8)

1.22 = −6.59 m

s2

Block on the wall (revisited)How much force must the person exert against the same block to hold it in place on the wall? The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially.

€

rF G = m

v g =12N

∴ m =12

9.8=1.22kg

?FN

Vertically the forces have to add up to be zero.

€

r N +

r F = 0

−17.39 +r F = 0

r F =17.39

€

r f s ≤ μ sN

fsμ s

≤ N

120.69 ≤ N

17.39 ≤ N

€

rf s =12N

Reminders

• Reading: Chapter 5.1 – 5.4• HW 03 Due WEDNESDAY at 10 PM– Additional office hours Tuesday at 1:30

• Midterm 1 is Friday during recitation time!– Held in Osmond 119 (next door)– Covers everything so far