Note:1. Mass in units of Kgs, length in unit of metersAt equilibrium, the meter stick must be horizontal to ground.3. This experiment should be performed on a flat surface.2.Procedure: To find center of mass of the rulerExperimental setup: Part A: Prove Sum of Torques = 0CHARStep 1Step 2Step 3aStep 3Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that theruler is balanced.Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler,record the observation (indicated by arrow)The center of mass of the ruler = re =cm =CHARLYS LEStep 4Step 4: Hang m, = 50g and m;=100g mass on the ruler with a string (see fig above) followed by gentlymoving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2Step 5: Find the distance ri and rz from re, record data in observation table below.Example: In figure below, r2 = - 20 + 50 = 30 cm, r = 87 - 50 = 37 cm20305060F=mgStyrofoam cupF=migFind forceFind forceF1 of thehangingF2of thehangingmassmassFig 1: Experimental set upStep 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F;=mag andF2=m2gStep 7: Exchange weights m, and m2 and repeat from step 4Step 8: There is no step 8Distance fromTorqueT= FrSum ofF= mg (N)g= 9.8 m/s?Mass (m) in KgObs. #center of rulerTorques(1g=0.001kg)(1cm = 0.01m)T= T1+ T2F,T = F,n0.05"1 = mF2=m20.1012 =-mF10.102m20.0512 = -mIs sum of torque zero?What are the possible errors in the lab? Theory:The rotational equivalent of force is defined as Torque(7). When a force Fis applied to a rigid body at anypoint away from the center of mass, a torque is produced. Torque, 1 (Greek letter, tau), can be defined asthe tendency to cause rotation. The magnitude of the vector is:i = FrsineWhere, r is the distance from the point of rotation to the point at which the force is being applied (i.e.,lever arm), and Fr sin 0 is the component of the force perpendicular to r. The dimension of torque is [M][L] [T*] with unit of N.mIn part A, all forces will be acting normal (perpendicular) to the meter stick: 0 = 90°; therefore, sin 0 = 1.The equation for torque is simplified:i = FrEquilibrium is reached when the net force and net torque on an object are zero. The first condition is thatthe vector sum of the forces must equal zero:ΣF-ΣΕ-Σ , -Σ 0 NewtonThe second condition that must be met is that the net torques about any axis of rotation must equal zero.We will use the standard convention for summing torques. Torques that tend to cause counterclockwiserotation(Tcc) will be positive torques, while torques that tend to cause clockwise rotation(Tc), will benegative torque.Στ= Στ. - Στ - ΝmThe system under consideration for this experiment will need to not only attain equilibrium, but alsoremain in equilibrium. This will require that the object be in stable equilibrium, meaning if a slightdisplacement of the system occurs, the system will return to its original position. If the system were tomove farther from its original position when given a slight displacement, it would be in unstable and notbe in equlibrium.Apparatus:50 grams mass, 100 gram mass, Tape, 8oz Styrofoam cup, 30 cm Ruler, Thread

Aim: To determine the sum of the torque is zero. Apparatus required: 50 and 100 grams of masses,…

Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques - 0 Step 1 Step 2 Step 3 Step 3a Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler - re . cm= m CHNLES LEON je, o Step 4 Step 4: Hang m, = 50g and m,-100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance r, and re from re, record data in observation table below. Example: In figure below, r2=- 20 + 50 - 30 cm, r, - 87 - 50 - 37 cm 60 100 F=mag Styrofoam cup F:-mig Find force Find force F2of the F1 of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F-mag and F=mag Step 7: Exchange weights m, and mą and repeat from step 4 Step 8: There is no step 8 Distance from Torque center of ruleri= Fr (1cm - 0.01m) Sum of F- mg (N) g- 9.8 m/s² Mass (m) in Kg Obs # Torques (1g-0.001kg) i= 11+ 12 0.05 r = m 1 F2 m2 0.10 : =-m F 0.10 r1 = m m2 0.05 12 = - m Is sum of torque zero?

Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques - 0 Step 1 Step 2 Step 3 Step 3a Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler - re . cm= m CHNLES LEON je, o Step 4 Step 4: Hang m, = 50g and m,-100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance r, and re from re, record data in observation table below. Example: In figure below, r2=- 20 + 50 - 30 cm, r, - 87 - 50 - 37 cm 60 100 F=mag Styrofoam cup F:-mig Find force Find force F2of the F1 of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F-mag and F=mag Step 7: Exchange weights m, and mą and repeat from step 4 Step 8: There is no step 8 Distance from Torque center of ruleri= Fr (1cm - 0.01m) Sum of F- mg (N) g- 9.8 m/s² Mass (m) in Kg Obs # Torques (1g-0.001kg) i= 11+ 12 0.05 r = m 1 F2 m2 0.10 : =-m F 0.10 r1 = m m2 0.05 12 = - m Is sum of torque zero?

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