# physics lab report

User Generated

NYNW900

Science

## Description

“Cover sheet” (front page) - title of Lab - class/ section - Your Name - Professor's Name “Introduction” next page) - Summary in your own words of the Theory/Introduction of Lab “PRE Lab” “Data” (If you have the Lab manual, you may hand-write answers as long as they are legible) - chart/ table/ excel spreadsheet - calculations (if any) “Analysis” - Questions (either type the answers in complete sentences or type the Question with the answer below it) “Conclusion” -Summary

### Unformatted Attachment Preview

migdi + Magdz + m4g dy = magda Mafgelyf = magd3 - migd- Magd2 Magda - migd. - Magda dy - mig Physics 205.0072 Mo = (1109-11.5g) = 0.105 ng 34.5 ;-> callabration yake. Xg > 50-5 cm → .0.505 m g=1000g Im : 100cm Position (m) (di-/xg-xil) Table #1 Etew-migde Mass (kg) M, -0.100 %= tow-tow mg = 0.900 % Di Perence Etcw-magde X; - 0.100 X2 = 0.643 Lever arm (m) di= 0.405 d2 = 0.189 Torque (N-m) teew = 0.3969 tew - 0.36848 3% Table #2 Mass (19) Position (m) Torove % Difference taw = 0.61985 m, 0.100 X. - 0.100 Lever orm (e dio 0.405 dz=0.305 d3 = 0.455 ~2% % M2 -0.200 X2 = 0.750 tow = 0.5978 m3 = 0.050 Xz = 0.261 Table #3 Mass (kg) Position (molese % Error Xg-Xo=do 0-505m -0.239m di = 0.166 m = 0.200 X, = 0.100 Lever arm (m) d, = 0.405 do = 0.266 m (-505 - 239) Mo = 0.125 X9 - 0-505 E-K *100 Table * Mass (19) (63-105) * 100 - 105 Lever Arm (m) Equation for Torque m -0.050 Position (m) X1 -0.050 X2 = 0.300 % Error = m 2 - 0.300 13 - 0-200 X3 = 0.700 di = 0.455 da = 0.205 dz - 0.195 (de) exp = do = 0.28 Solutions Eq for du ques (da)" Xl - 0.225 mil - 0.100 M = 0.105 xg=0.505
Purchase answer to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Attached.

“Cover sheet” (front page) - title of Lab - class/ section - Your Name - Professor's Name

Introduction

Rigid body concepts are used for many engineering things like in construction of buildings,
bridge. The torque principle is there and its focus on the twist force of an object with its pivot on
certain distance. In equilibrium the object is at rest so that the net force applied on it is nearly
zero or zero. In fact this principle applied in beam. Based from rigid bodies force, it can be
rotational either sum of counter clockwise or counterclockwise forces that it’s in equilibrium that
is called rotational equilibrium. The sum of torque is zero called translational equilibrium.

PRE Lab
1. The definition of torque is the measurement of how much the force is applied to cause an
object to move on its pivot rotate along some axis in which mathematically expressed as
force multiplied by its distance(perpendiculat to its force).
𝜏 = 𝐹𝑑
2. The condition for a body in equilibrium of rigid body must the net force applied on that
body is zero (however in real there is small change force according on strength of
materials).
∑𝐹 = 0
3. The torque on 𝐹1 is clockwise direction so its negative force
Convert centimeter to meter units.
10 𝑐𝑚 = 0.1𝑚
Plug the given Force=10N an...

### Review

Anonymous
Awesome! Perfect study aid.

Studypool
4.7
Indeed
4.5
Sitejabber
4.4