MCHT 111 Final Review General comments: closed book and notes but optional one page crib sheet allowed. STUDY: old exams, homework and power point lectures! Key: make sure you can solve your homework problems and exam problems. There will be approx. 6 – 8 problems and possibly some multiple choices. Good luck! 1. Chapter 2: Finding resultants of vectors (i.e. addition of vectors), resolving vectors into rectangular components (i.e. Cartesian vector notation). a. Graphical approaches (2D, recommend for 2 forces): triangle method, polygon method, parallelogram method. b. Multiple forces – find resultant by resolving forces into their rectangular components c. You should be able to express resultants by magnitude and direction OR by rectangular components, I, j and k (Cartesian vector notation). Example 1 (example 2‐1) : Find resultant force using triangle method. Specify resultant by magnitude and direction. Also, specify resultant in Cartesian vector notation (i.e. i, j rectangular components): Answer: Example 2: Find the resultant of the 3 force system by breaking forces into rectangular components and adding. Express resultant in Cartesian vector notation and by magnitude and direction (angle it makes with the x axis: 1 MCHT 111 Final Review Step 1 is to resolve each force into its components. Step 2 is to add all the x‐components together, followed by adding all the y components together. These two totals are the x and y components of the resultant vector.
FR = F1 + F2 + F3 = F1x i + F1y j  F2x i + F2y j + F3x i  F3y j
= (F1x  F2x + F3x) i + (F1y + F2y  F3y) j
= (FRx) i + (FRy) j Step 3 is to specify resultant by magnitude and direction: Or Cartesian vector notation: F = (F ) i + (F ) j R
Rx
Ry
Example 3: What if 3D?? Same approach as example 2 except need to break a 3D force up into components. How do you break a 3D force up into components??? Option 1 – if given coordinate (directional) angles: F = Fcos(θ ) F = Fcos(θ ) F = Fcos(θ )
x
x
y
y
z
z
2 MCHT 111 Final Review Option 2 – If know line of action: F = F(d /d) F = F(d /d) F = F(d /d)
x
x
y
y
z
z
Where d = total displacement = d  d x2  d y2  d z2
Option 2a – Same as 2 except use position vectors (books method): a) Find the position vector, r
, along two points on that line.
AB
b) Find the unit vector describing the line’s direction, u
= (r /r ).
AB
AB AB
c) Multiply the unit vector by the magnitude of the force, F = F u
.
AB
Then add up I, j and k components and express resultant vector: FR = (FRx) i + (FRy) j + (FRz) k what if want to specify resultant as magnitude and direction? Example 3: Find the resultant of the two 3D forces shown below: 3 MCHT 111 Final Review 2. Chapter 3: springs, coplanar force systems, 2D static particle equilibrium, intro to FBDs and reaction forces. Example 1: Find tension in each cable CA and CB for an angle theta = 40 degrees and a 10 kg cylinder. Note, can also give you tension forces and ask you to calculate mass or give you tension forces, mass and ask you to calculate angle, etc… Step 1: Draw FBD joint C Step 2: break each force up into components as detailed above. Step 3: Apply scalar equations of static equilibrium and solve for unknowns: ΣFx = 0
ΣFy = 0
ΣFz = 0 (if 3D)
Other 2D static equilibrium problems we’ve solved in class: Example 2: For the 3D system shown, determine cable forces necessary to support the 40 lb crate. Solve this problem on board!! End of Exam 1 Material!!
4 MCHT 111 Final Review 3. Chapter 4: principle of moments, moment of a force, force couples, equivalent force systems (recall a general system of forces and moments acting on a rigid body can be reduced to a single equivalent force and single equivalent moment) Example: 4. (simplification of a force and couple system). Given: A 2-D force system with geometry as
shown.
Find: The equivalent resultant force and
couple moment acting at A and then the
equivalent single force location measured
from A.
4. Chapter 5: 2D rigid body equilibrium, reaction forces, stability, FBD’s, 2‐Force members: Example: Find reactions at A, C and D. Draw FBD’s for beam A‐C‐B and 2 Force member D‐C. Other 2D Rigid Body Equilibrium examples we did in class ‐ find reactions for the following: 5 MCHT 111 Final Review More 2D Equilibrium: Recall, this problem was a stability problem. What counterweight is necessary to prevent tipping? Key: at point of tipping, Nc = 0. Chapter 5, Part 2 – 3D Rigid Body Equilibruim (study powerpoint, know 3D reactions, how to draw FBDs and simple 3D rigid body equilibrium problems): Recall for 3D there are six scalar equations of equilibrium (EofE). These are:
 F =  F =  F = 0 X
Y
Z
 M =  M =  M = 0 X
Y
Z
End of Exam 2 Material!!
6 MCHT 111 Final Review Chapter 6: Structural analysis for trusses, zero force members, assumptions for truss analysis, method of joints. Example: Determine the force in each member of the truss and state if the members are in tension or compression: Structural analysis for frames and machines, pulleys, mechanical advantage, etc. Examples we did in class: 7 MCHT 111 Final Review End of Exam 3!! 5. Chapter 8 Friction: Just sections 8.1, 8.2, 8.3. Dry friction, inclined planes, sliding vs. static equilibrium (i.e. friction high enough to prevent sliding), tipping, wedges, etc. Determine the minimum horizontal force, P, required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the create and plane is 0.25) 8 MCHT 111 Final Review What angle can the ladder be before it slips if the coef of friction at A between the friction pad and ground is 0.4. The man weighs 180 lb. Neglect friction at B.
6. Chapter 9: centroids and moment of inertia. Know how to calculate centroids for composite shapes. Know how to calculate moment of inertia for composite shapes using parallel axis theorem. The moment of inertia calculated w.r.t. centroid (note for the inertia problem I may give you the centroid location. Example: Find centroid location, then find moment of inertia, Ix thru centroid: Note, due to symmetry, centroid is located at x = 0 so you only need to find y bar. 9