8 Rotational Equilibrium




Boom Problem

rotation1.jpgPart 1:

Tnet = FTd – Perpendicular component of weight of boom times distance – perpendicular component of the weight of the load times distance.

Part 2

Rx base – FT Cos angle of the cable = 0


Part 3

Add up the vertical components of all the forces on the boom as seen on the diagram to the right.

Picture frame Problem


Fx = FT1cos ϴ - F(lower Right) = 0
Therefore: FT1cos ϴ = F(lower Right)
Fy = FT1 sin ϴ + FT2 – Fg
Pivot point at lower left.
Τtotal = All the above forces times the distance from their corresponding axis.

Look at the diagram: place the pivot point at FT,2. This will make the distance for the torque at that point zero. Since the torque due to gravity is down and the Torque FT,1x will cause rotation downward they should be negative. The FT,1y is upward it should be positive. The sum of these 3 torques is equal to zero.
Now part 2, the sum of the vertical forces (not torque) = zero.
Now part 3, the sum of the horizontal forces = zero.


I will assign each person one of these labs.

Lab 1

Ladybug Revolution
Click to Run

Lab 2

Do the lab on page 313 Machine and Efficiency. Do the Inclined Plane. Do the Analysis and Interpretation on page 315. Do a paragraph about what you learned.

Lab 3

Do the lab on page 312 do Alternative Assessment #3 (Build it). Write a purpose, procedure, analysis, calculations, and conclusion paragraph about what you learned.
Lab 4

Lab 5