2066 Words9 Pages

RYERSON UNIVERSITY
DEPARTMENT OF PHYSICS
LAB REPORT
FOR
PCS 211 SECTION **
EXPERIMENT: STATIC EQUILIBRIUM - FORCES AND TORQUES
EXPERIMENTERS: ***** ********* ***** *********
AUTHORS OF THIS REPORT *** ***
EXPERIMENT PERFORMED ON: ***
REPORT SUBMITTED ON: ***
INSTRUCTOR: ***
PRE-LAB QUESTIONS:
1) What is meant by static equilibrium?
The meaning of static equilibrium can be explored by first examining the definition of equilibrium. Equilibrium means that an object is at rest or that the objects center of mass moves at constant velocity relative to the observer.
Static equilibrium is the case where an object is at rest. An object is at rest or at static equilibrium when the net force*…show more content…*

5. Data was recorded in a table. PART 3: EQUILIBRIUM WITH PARALLEL FORCES 1. Started with a similar mass of M=500g at a distance l, along the ruler from the pivot. 2. The end of the ruler was supported with a second newton scale P. 3. F and P were adjusted to achieve static equilibrium. 4. Figure 3. Part III: Equilibrium with parallel forces F, Mg and P were recorded in a table. 5. Steps 3 and 4 were repeated at other values of M and l. 6. A full table was filled with all the performed products. OBSERVATIONS & CALCULATIONS: Mass of ruler: m = 0.2150±5.0e-6kg CALCULATIONS PART I: PART I: DETERMINATION OF THE CENTER OF GRAVITY: Trials | d (m)±0.005m | F (N)±0.05N | Tg (Nm)Nm | a | 0.60 | 1.9 | 1.14 | b | 0.70 | 1.7 | 1.19 | c | 0.80 | 1.5 | 1.20 | xcm=position of center of gravity on ruler 0-mgxcm+Fd=0 xcm=Fdmg a)xcm=(1.9N)(0.6m)0.2150kg9.8ms2=0.54m b)xcm=(1.7N)(0.7m)0.2150kg9.8ms2=0.56m c)xcm=(1.5N)(0.8m)0.2150kg9.8ms2=0.57m Average xcm=0.54m+0.56m+0.57m3=0.56±0.005m PART II: EQUILIBRIUM WITH PARALLEL FORCES OBSERVATION TABLE PART II Table 2. Equilibrium with Parallel Forces Tg=mgxcm a) 0.2150kg9.8ms20.54m=1.14Nm b) 0.2150kg9.8ms20.56m=1.18Nm c) 0.2150kg9.8ms20.57m=1.20m Mgl a) 0.500kg9.81ms20.20m=1.0Nm b) 0.400kg9.81ms20.30m=1.2Nm c) 0.300kg9.81ms20.25m=0.7Nm

5. Data was recorded in a table. PART 3: EQUILIBRIUM WITH PARALLEL FORCES 1. Started with a similar mass of M=500g at a distance l, along the ruler from the pivot. 2. The end of the ruler was supported with a second newton scale P. 3. F and P were adjusted to achieve static equilibrium. 4. Figure 3. Part III: Equilibrium with parallel forces F, Mg and P were recorded in a table. 5. Steps 3 and 4 were repeated at other values of M and l. 6. A full table was filled with all the performed products. OBSERVATIONS & CALCULATIONS: Mass of ruler: m = 0.2150±5.0e-6kg CALCULATIONS PART I: PART I: DETERMINATION OF THE CENTER OF GRAVITY: Trials | d (m)±0.005m | F (N)±0.05N | Tg (Nm)Nm | a | 0.60 | 1.9 | 1.14 | b | 0.70 | 1.7 | 1.19 | c | 0.80 | 1.5 | 1.20 | xcm=position of center of gravity on ruler 0-mgxcm+Fd=0 xcm=Fdmg a)xcm=(1.9N)(0.6m)0.2150kg9.8ms2=0.54m b)xcm=(1.7N)(0.7m)0.2150kg9.8ms2=0.56m c)xcm=(1.5N)(0.8m)0.2150kg9.8ms2=0.57m Average xcm=0.54m+0.56m+0.57m3=0.56±0.005m PART II: EQUILIBRIUM WITH PARALLEL FORCES OBSERVATION TABLE PART II Table 2. Equilibrium with Parallel Forces Tg=mgxcm a) 0.2150kg9.8ms20.54m=1.14Nm b) 0.2150kg9.8ms20.56m=1.18Nm c) 0.2150kg9.8ms20.57m=1.20m Mgl a) 0.500kg9.81ms20.20m=1.0Nm b) 0.400kg9.81ms20.30m=1.2Nm c) 0.300kg9.81ms20.25m=0.7Nm

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