lab report 10 - PHY2053L

docx

School

Broward College *

*We aren’t endorsed by this school

Course

2053L

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

docx

Pages

4

Uploaded by inuyasha078

Report
Cristian Acuna Vasquez 03/06/2020 PHY2053L Title : Balancing Act Purpose: The purpose of this simulation is to investigate and learn about the concepts of balancing torques with regards to rotational equilibrium. Background Information: A force’s tendency to produce the rotation of a rigid body is called the torque. It is equal to the product of the force and momentum arm, the perpendicular distance from the axis of rotation to the force’s line of action. An important basic concept when dealing with torques is the idea of equilibrium or balance. In general, an object can be acted on by several torques at the same time. A torque is also a vector quantity which means that it has both a magnitude (size) and a direction associated with it. If the size and direction of the torques acting on an object are exactly balanced, then there is no net torque acting on the object and the object is said to be in equilibrium. Because there is no net torque acting on an object in equilibrium, an object at rest will stay at rest, and an object in constant angular motion will stay in angular motion. Hence, at totational equilibrium , it should be true that T net = Σ T =I T left I – I T right I =0 Materials: computer PhET software Procedure: Using the PhET software. We need to place a variety of object at various positions on a seesaw and try to balance by making a torque summation. The objects have different masses and shapes.
Results/Calculations/ Analysis Part I. Balancing two masses. Mass 1 (kg) Position 1 (m) T right = T left Mass 2 (kg) Position 2 (m) 20 0.5 (20)(9.8)(.5)=(5)(9.8)X X=2.0 5 2.0 20 1.0 (20)(9.8)(1.0)=(5)(9.8)X X=2.0 10 2.0 80 0.25 (80)(9.8)(0.25)=(20)(9.8)X X=1.0 20 1.0 30 0.5 (30)(9.8)(.5)=(10)(9.8)X X=1.5 10 1.5 10 1.0 (10)(9.8)(1.0)=(20)(9.8)X X=0.5 20 0.5 5 2.0 (5)(9.8)(2.0)=(20)(9.8)X X=0.5 20 0.5 15 1.0 (15)(9.8)(1.0)=(60)(9.8)X X=0.25 60 0.25 Part II Balancing multiple masses Mass 1 (kg) Position 1 (m) Mass 2 (kg) Position 2 (m) T right = T left Mass 3 (kg) Positio n 3 (m) 20 1.5 m left 30 0.50 m left (20)(9.8)(1.5)+(30)(9.8)(0.5)= (60)(9.8)X X=0.75 60 0.75 m right 20 1.0 m left 60 2.0 m left (20)(9.8)(1.0)+(60)(9.8)(2.0)= (80)(9.8)X X=1.75 80 1.75 m right 80 1.5 m left 30 0.5 m right (80)(9.8)(1.5)=(30)(9.8)(0.5)+ (60)(9.8)X X=1.75 60 1.75 m right 60 1.0 m left 80 0.25 m right (60)(9.8)(1.0)=(80)(9.8)(0.25)+ (20)(9.8)X X=2.00 20 2.00 m right
Part III Mystery packages Know n Mass 1 (kg) Positio n 1 (m) left T right = T left Mystery Package Positio n (m) Right Mass of Packag e (kg) 60 0.5 (60)(9.8)(0.5)= X (9.8)(1.5) X=20 A 1.5 20 30 1.0 (30)(9.8)(1.0)= X (9.8)(2.0) X=15 C 2.0 15 30 0.25 (30)(9.8)(0.25)= X (9.8)(0.75) X=10 D 0.75 10 60 0.25 (60)(9.8)(0.25)= X (9.8)(2.0) X=7.5 H 2.0 7.5 Know n Mass 1 (kg) Positio n 1 (m) Know n Mass 2 (kg) Positio n 2 (m) T right = T left Mystery Package Positio n (m) Mass of Packag e (kg) 20 1.75 m left 80 0.5 m left (20)(9.8)(1.75) + (80)(9.8)(0.5) = X(9.8)(1.5) X= 50 F 1.5m right 50 60 0.5 m left 15 1.75 m right (60)(9.8)(0.5) = (15) (9.8)(1.75) + X(9.8) (1.25) X= 3 E 1.25 m right 3 20 0.25 m left 5 1.0 m left (20)(9.8)(0.25) + (5) (9.8)(1.0) = X(9.8) (2.0) X= 50 B 2.0 m right 5 Part IV Balancing Act – Game Level 1; Best score = 12 Level 2; Best score = 12 Level 3; Best score = 12
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Level 4; Best score = 12 Results/Calculations/ Analysis Error Analysis: In this lab, there were no errors since it is a simulation. Conclusion: In conclusion., I was able to identify the relationship between moment arms, torques, mass, and how to find unknowns for moments arms and mass. Torque, also called moment or moment force, is the tendency of a force to rotate an object about an axis. A system is in equilibrium if it has no acceleration gained or lost, meaning rotational or translational.