# A Spring Having An Unstretched Length Of 2 Ft

The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

The 2-ft cord AB and spring AC. If the spring has an unstretched length of 2 ft, determine the angle for equilibrium.Take k = 15 lbft. U P = 10 lb 2 ft k 2 ft A B C P u. The spring has an unstretched length l o = 2 ft and a stiffness k = 50 lb/ft. Draw the free body diagram for the system, clearly labeling all forces and moments. Determine the angle θ for equilibrium. This will require solving a trig equation; you may use your calculator (if possible), or you may.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Let us first determine the stretched length of spring AB. We can use Pythagorean theorem to figure this out.

Length of spring =

### A Spring Having An Unstretched Length Of 2 Ft X 8

sqrt{4^2+3^2}

Length of spring = 5 m

We know from the question that the unstretched length of the spring is 3 m. That would mean the spring stretched by 2 m (5 m – 3 m = 2m) after the block was hung at ring A.

We can now figure out the force of spring AB using Hook’s Law.

Hook’s Law states:F,=,ks

F,=,(30)(2)

F,=,60 N

Let us now draw our free body diagram.

The angles were calculated using the inverse of tan (arctan).

The brown angle was found by:
text{tan}^{-1}left(dfrac{3}{3}right),=,45^0

The orange angle was found by:

### A Spring Having An Unstretched Length Of 2 Ft Lauderdale

text{tan}^{-1}left(dfrac{3}{4}right),=,36.87^0

The next step is to write our equations of equilibrium.

rightarrow ^+sum text{F}_text{x},=,0
60text{cos},(36.87^0),-,F_{AC}text{cos},(45^0),=,0

F_{AC},=,67.88 N

+uparrow sum text{F}_text{y},=,0
60text{sin},(36.87^0),+,F_{AC}text{sin},(45^0),-,W,=,0

W,=,84 N

To figure out the mass, remember that W=mg.  m,=,dfrac{W}{g}
m,=,dfrac{84}{9.81}

m,=,8.56 kg

mass = 8.56 kg

###### This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-19.
[SOLVED] Electric Force II

## Homework Statement

You have a lightweight spring whose unstretched length is 4.0 cm. You're curious to see if you can use this spring to measure charge. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm.
What is the magnitude of the charge (in nC) on each bead?

## Homework Equations

F = -k1 * x
F = k2 * q1q2 / r^2
g = a = -9.81
k1 = Spring constant. Unknown.
k2 = Electrostatic constant = 9*10^9

## The Attempt at a Solution

In the first experiment F = ma = -k1*x
0.001*-9.81 = -k1*(0.05 - 0.04)
Therefore the spring constant k1 = 0.981
Now the second experiment:
F = -k1*x
F = -(0.981)(0.045 - 0.04)
F = -0.004905
This is the force applied on a spring of k1 = 0.981, to stretch it 0.5cm
Similarily, this should also be the force between the two beads:
F = k2*q1q2 / r^2, however q1 = q2 according to the question.
F = k2*(2q) / r^2, solving for q.
Fr^2/(2*k2) = q (-0.004905)(0.045)^2/(2*9*10^9) = q
q = -0.0000000000000005518125 C

### A Spring Having An Unstretched Length Of 2 Ft =

q = -0.0000005518125 nC