What is the difference between particles and rigid bodies?

 • Particles

→ dimensions are not critical
→ does not respond to torque

• Rigid body

→ made up of many particles
→ responds to torque (thus, rotates)

Diff. Eq. - Solve the following equation using Laplace transforms

Exmpl. #1) Solve the following equation using Laplace transforms. Where y(0) = 0, y'(0) = 0 and x(t) is a step change of magnitude 2.

Standard procedure of these problems, taking the Laplace transform of both sides!

But let's start with the left side first:

(*Laplace transforms of derivatives must be known before calculating... click here!)

After plugging in f(0) and f'(0), and then simplifying, we get:

Now let's do the Laplace transform of the right side of the equation. Remember that it was given to us that x(t) is a step change of magnitude 2. Laplace transform of a step function is always 1/s, and we will multiply it by 2 to represent the change of magnitude.

Time to factor out Y, which represents the Laplace transform of a function y(t).

---> (*Apply a/b/c/1 fraction rule!)

Now, we will use partial fraction decomposition to solve for Y.

Let's set this up.

Now distribute and simplify.

Group and equate the terms by power (right and left side of the equal sign)

You will find that,

A = 1/3

B = -1

C =
2/3

Now we take the inverse laplace transform of the equation to get our answer!

A standard laplace transform table, should help you easily calculate these values

*Inv. Laplace of variable is the same variable, but in the time domain. So, Inverse Laplace of Y is y(t)

and here is our answer:

Diff. Eq. - The Laplace Transform - Exmpl. 6

 Example 6 , The Laplace transform of the function f(t) = e^(at)

Diff. Eq. - The Laplace Transform

Chapter 8 - Laplace transforms

Laplace transforms are a tool for solving initial value problems where the differential equation is linear and has constant coefficients. 

The idea is to transform the differential equation into an algebraic equation whose solution is easier to obtain and then apply the inverse transform to solve the differential equation.

f(t)  →  F(s)

*f(t) being the original function

*F(s) representing a Laplace transform

There are two ways to represent this equation:

... or,

*ℒ - Laplace transform

The Laplace transform ℒ is a linear in the sense that ℒ[C1f1 + C2f2] = C1ℒ[f1] + C1ℒ[f2], where C1, C2 are constants.

Some more theory & example problems:

> Laplace Proofing #1

> Example 8.6

> Example 8.2

> Example 8.3

Diff. Eq. - finding particular solution using the Method of Undetermined Coefficients

 Given the equation:

y'' - 4y' + 3y = e^(3x) * (6+8x)

(a) Find a particular solution of the equation by using the Method of Undetermined Coefficients.

(b) Find the general solution of the equation.

(Start by finding the auxiliary equation of the given diff. equation)

y'' - 4y' + 3y = 0  →  r^2 - 4r + 3y = 0...

► r = 3, 1

► y = e^x & e^3x

Based off your general solution answers, find a particular solution that matches the f(x) to the original function: (e^(3x) * (6+8x))

* Remember that e^(3x) was one of the roots, and we should avoid creating two roots that are exactly the same.

Thus, multiply by 'x'.

yp = xe^3x * (A+Bx) or, e^3x * (Ax+Bx^2)

(Let's start differentiating to get y'p and y''p)

* differentiate by using product rule.

y'p = [e^3x * (Ax+Bx^2)'] + [e^3x' * (Ax+Bx^2)]

►  = e^3x * (A+3Ax+2Bx+3Bx^2)

y''p = [e^3x * (A+3Ax+2Bx+3Bx^2)'] + [e^3x' * (A+3Ax+2Bx+3Bx^2)]

►  = e^3x * (6A+2B+9Ax+12Bx+9Bx^2)

Now, plug in your values into the given differential equation you were first handed.

y'' - 4y' + 3y = e^(3x) * (6+8x)

► (e^3x * (6A+2B+9Ax+12Bx+9Bx^2)) - 4[e^3x * (A+3Ax+2Bx+3Bx^2)] + 3[e^3x * (A+3Ax+2Bx+3Bx^2)] = e^(3x) * (6+8x)

Cancel / divide term e^(3x)

► ((6A+2B+9Ax+12Bx+9Bx^2)) - 4[(A+3Ax+2Bx+3Bx^2)] + 3[(A+3Ax+2Bx+3Bx^2)] = (6+8x)

Now distribute and simplify (this is a very tedious process!)

2A + 2B + 4Bx = 6+8x

►  (2A + 2B) = 6

►  4B = 8

→ B = 2, A = 1

Blue in A & B values for the yp and make your general solution as well.

Answers are....

(a) yp = xe^(3x) * (1+2x)

(b) y = C1e^x + C2e^3x + e^3x * (x+2x^2)

Statics - Fundamental Problem 7.6

   
From Chapter 7, 

Assume A is pinned and B is a roller in (Figure 1). Take that w = 4.5 kN/m . Determine the normal force, shear force, and moment at point C. Follow the sign convention.

▼ Three Part Question:

Determine the normal force, shear force, and moment at point C. Express your answer to three significant figures and include the appropriate units.

Step 1 - always make your free body diagram! In the instructions given, point A is a pinned support and point B is a roller. The triangular and distributed load need to be expressed as a concetrated point load as well:

Make your free body diagram! In the instructions given, point A is a pinned support and point B is a roller. The triangular and distributed load need to be expressed as a concetrated point load as well.

Substituting 4.5 for w, We get that By = 

Statics - Problem 6.15

  

From Chapter 6, Structural Analysis

In (Figure 1), a = 10 ft.$\mathrm{<span\; style="color:\; \#333333;\; font-family:\; arial;"> Members\; AB\; and\; BC </span>}$can each support a maximum compressive force of 950 lb, and members AD, DC, and BD can support a maximum tensile force of 

▼ PART A

Determine the greatest load  P the truss can support.

Express your answer to three significant figures and include the appropriate units.

 STEP 1  

Start by taking calculating the reaction forces about both support points.

Take moment abt. point A (which is a pin support.)

Solve for Cy;

Now, equate all vertical forces.

We have already solved for Cy, plug in its value;

 STEP 2 

Now, Create your FBD for the truss surrounding joint A.

Force AB is the force in member AB

& Force AD is the force in member AD.

Calculate the angles subtended by BAx & DAx.

BAx calculation:

*SOH CAH TOA

Use tangent for this angle calculation since we know the height (a) and the base (a).

DAx calculation:

*SOH CAH TOA

Use tangent for this angle calculation since we know the height (a/4) and the base (a).

 STEP 3 

Now, equate all horizontal forces.

Equate the forces along the vertical direction.

= 0.647P (tension)

 STEP 4 

Calculate the force in member AB:

= 0.943P (compression)

Due to symmetrical shape of ABD and BDC, the tensions in both members are equivalent.

= 0.647P

= 0.943P