# A Spring Having An Unstretched Length Of 2 Ft

• A it has a speed of 10 ft s. If the spring to which it is attached has an unstretched length of 3 ft and a stiffness of, determine the normal force on the collar and the acceleration of the collar at this instant. K = 10 lbft SOLUTION Ans. A = 2(180.2)2 + (8.9443)2 a n = v2 r = (10)2 11.18 = 8.9443 fts2 a t = 180.2 fts2 +R©F t = ma.
• The collar has a mass of 30 kg and is supported on the rod having a coefficient of kinetic friction of 0.4. The attached spring has an unstretched length of 0.2 m and a stiffness of k =50 N/m. Determine the speed of the collar after the applied force F= 200N causes it to be displaced x=1.5 m from point A. When x=0 the collar is held at rest.

Determine the stretch in each of the two springs required to hold the 20-kg crate in the equilibrium position shown. Each spring has an unstretched length of 2 m and a stiffness of k = 300 N/m.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.This question involves expressing forces in Cartesian vector form. If you are unsure on how to do this, read the detailed guide on expressing forces in Cartesian notation.

The position vector for OC:

1 Answer to Each spring has an unstretched length of 2 m and a stiffness of k = 330 N/m.(Figure 1) Part A Determine the stretch in OA spring required to hold the 18- kg crate in the equilibrium position shown. Express your answer to two significant figures and include the appropriate units. Part B Determine.

r_{OC},=,left{6i+4j+12kright} m

A position vector, denoted mathbf{r} is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was (x_A,y_A,z_A) and the coordinates of point B was(x_B,y_B,z_B), then r_{AB},=,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

Now, we calculate the magnitude:

magnitude of r_{OC},=,sqrt{(6)^2+(4)^2+(12)^2},=,14 m

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r,=,ai+bj+ck, then the magnitude would be, r_{magnitude},=,sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

Now, we write the unit vector for the position vector.

u_{AB},=,left(dfrac{6}{14}i+dfrac{4}{14}j+dfrac{12}{14}kright)

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was

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

r,=,ai+bj+ck, then unit vector, u,=,dfrac{a}{sqrt{(a^2)+(b^2)+(c^2)}}+dfrac{b}{sqrt{(a^2)+(b^2)+(c^2)}}+dfrac{c}{sqrt{(a^2)+(b^2)+(c^2)}}

We can now express the force in the cable OC in Cartesian form:

F_{OC},=,left{0.428F_{OC}i+0.286F_{OC}j+0.857F_{OC}kright}

Now the other forces:

F_{OA},=,left{0i-F_{OA}j+okright}
F_{OB},=,left{-F_{OB}i+0j+okright}
F,=,left{0i+0j-(20)(9.81)kright},=,left{0i+0j-196.2kright}

Now that we have written all forces in Cartesian vector form, we can write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium. sum text{F},=,0
F_{OC}+F_{OA}+F_{OB}+F,=,0

As all the forces added together equals zero, then each component (x,y, z-components) added individually must also equal zero.

x-components: 0.429F_{OC}-F_{OB},=,0
y-components:

0.286F_{OC}-F_{OA},=,0
z-components:

0.857F_{OC}-196.2,=,0

Solving the three equations gives us:

F_{OC},=,229 N

F_{OB},=,98 N

F_{OA},=,65.5 N

We can now find the stretch of the springs using the following equation (Hooke’s Law):

F=ks

Starting with the force in spring OA:

F=ks
65.5,=,300s

s,=,0.218 m

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

Now, the force in spring OB:F=ks

## A Spring Having An Unstretched Length Of 2 Ft Extension Cord

98,=,300s

s,=,0.327 m

Stretch of spring OA = 0.218 m

Stretch of spring OB = 0.327 m