The Seesaw(Torque and Rotation)Activity No. 12

I Objectives To determine the condition that must be satisfied for a body to be in rotational equilibrium. To find out the factors affecting torque.

II AbstractEquilibrium is a goal of the builders of many different structures. In physics an object is said to be in mechanical equilibrium if it is in a state of transitional and rotational equilibrium. For a system to be in transitional equilibrium the sum of the forces acting on the system, in all planes, must be equal to zero. This is expressed mathematically in the following equations.Fx=0, Fy=0, Fz=0In this lab rotational equilibrium will be examined. For rotational equilibrium the summation of all of the torques must equal zero. This is expressed mathematically in the following equation.=0Torques are a measure of the rotational force that an object has about a pivot point. They are calculated by multiplying the force by the distance from the force to the point of rotation. The equation for calculating torques is the following equation. =Frsin This means that a larger force will cause greater rotation than a smaller force that is applied to an object that is free to pivot about a point. Also, a force applied far from the pivot point will cause greater rotation than a force applied closer to the pivot point.In this lab a meter stick will be balanced on a support stand. Weights will then be added to the meter stick. These weights will create forces and torques on opposite sides of the meter stick. Also, forces will be applied that cause the meter stick to become unbalanced. In these situations the meter stick will be moved so that the pivot point changes to keep the ruler balanced. It is hypothesized that in all of the situations tested rotational equilibrium will be maintained. This is because the meter stick will remain balanced so the net torques will have to equal zero.This lab will teach the concepts of rotational equilibrium and torques. It will increase understanding of the principals of rotational equilibrium because various rotational equilibrium scenarios will be tested. It will also provide practice for calculating the net torques acting on a system, and calculating and identifying clockwise and counterclockwise torques. Rotational equilibrium is when the net torque acting on a system is equal to zero. In this lab rotational equilibrium will be established in multiple experimental scenarios. This was done using a meter stick with masses suspended from it at various positions. The positions of the masses and the pivot point were altered in order to maintain rotational equilibrium. The clockwise and counterclockwise torques were calculated for each of these scenarios. By calculating the percent difference between these values it was determined that the systems were in rotational equilibrium. Possible sources of error were non-uniformly distributed mass in the meter stick, and differences from the average mass in the hangers. The lab taught the concepts of torques, and rotational equilibrium.

III Schematic DiagramFor the construction of the see saw

Figure I WoodFigure II Meter Stick

Figure III NailsFigure IV Instant Glue

For the Experiment

Figure V Digital Weighing ScaleFigure VI Weight Hangers

Figure VII Set of WeightsFigure VIII The See Saw

Figure IX Set up

IV Data Tale

A. Center of Gravity of the meter stick system: __00_cm mark

Table I: Rotational Equilibrium of Balanced Forces

TrialNumberMass(kg)Force(N)MomentArm (m)ClockwiseCounter ClockwisePercentage Difference

10.17551.71990.150.2600%

0.17551.71990.1500.26

20.050.490.200.09800%

0.050.490.2000.098

30.225752.2120.200.4400%

0.225752.2120.2000.44

Table II: Rotational Equilibrium of Unbalanced ForcesTrialNumberMass(kg)Force(N)MomentArm (m)ClockwiseCounter ClockwisePercentage Difference

10.17551.71990.200.3402.86%

0.225752.2120.15700.34

20.17551.71990.070.12039300.55%

0.450.443450.2700.120393

30.17551.71990.140.24078600.55%

0.905o.88690.2700.240786

V Sample Computation

Table I: Trial 1 Force TorquePercentage DifferenceF = maT = Fr= x 100 = 0.1755 kg (9.8m/s2) = 1.7199 N (0.15m)= x 100 = 1.7199 N = 0.26 N.m= 0

VI ConclusionIt was found that in all scenarios there was rotational equilibrium. This is because the value of the percent difference between the conditions is low. After the distribution of the mass is considered the percent difference between the torques is reduced. The forces are conserved in the first three cases because the force of the masses is acting on the support piece. That piece applies a normal force back on the meter stick. This force is in the opposite direction of the force of gravity caused by the masses. Therefore when the forces are added the net force acting on the system is zero. The data collected from case three is indicative of how a see saw works, it works by moving set masses along the end of it, away from the pivot point. On the other side of the pivot point is the substance being weighed. By moving the mass the torques are changes because the length of the leaver arm is changed. Once the see saw is balanced rotational equilibrium has been reached. At this point the torques acting on each side of the pivot point are equal, and the mass of the substance can be determined by reading the values from the see saw.This lab taught the concepts of rotational equilibrium and torques. It increased understanding of the principals of rotational equilibrium through the calculation of rotational equilibrium in numerous experimental scenarios. It also provided practice for calculating the net torques acting on a system, and calculating and identifying clockwise and counterclockwise torques. This was accomplished through calculating and identifying these values in multiple scenarios tested in the lab.Some possible sources of error in this experiment include a non-uniformly distributed mass of the meter stick, and the possibility for anomaly in the masses of the hangers. Often the pivot point of a meter stick is not at the exact position. This is because the meter stick is made of natural wood, so the mass is not evenly distributed across the meter stick. A possible method of reconciling the non-centered pivot point is to find the mass per centimeter, and add the torque due to the mass of the meter stick on each side of the pivot point to the net torques. This does not fix the problem all the way though, because it still fails to compensate for the areas of higher and lower density in the wood. The other possible source of error is that the mass of the hangers used in calculations was the average mass of the hangers. This assumes that all of the hangers are of equal mass, which is unlikely. If the mass of each was found separately, and used separately in the calculations, some error could be reduced.VII Question and Analysis

1. Three identical uniform rods are each acted on by two or more forces, all perpendicular to the rods and all equal in magnitude. Which of the rods could be in static equilibrium if an additional force is applied at the center of mass of the rod? Explain.

Ans: 0nly 3, because 1 and 2 will never be in static equilibrium since the forces at the edges create non-zero net torque.

2. Discuss what happens to the moment arm when the force is varied (increased or decreased) so as to maintain a constant torque.The force about the moment arm increases as the distance of the force from the moment increases3. To tighten a bolt, a force of 100N is applied at the end of a wrench handle that is 0.25m from the axis of the bolt. (a) How much torque is exerted? (b) To achieve the same amount of torque, how much force is needed if its applied 0.10m from the bolt? (c) Do the answers depend on the direction of the force relative to the direction of the wrench handle? Explain.

4. A seesaw supports two children (A and B) on both sides and is in equilibrium. If a child A is twice as heavy as child B, what can be said about the distance of child A from the pivot as compared to the distance of Child B from the pivot?Child A is the distance from the pivot than Child B. This can be shown by the formula = F * r. When the children are in static equilibrium, the torque produced by each child is equal. Since FA = 2 FB the distance must be ; rA = rB.

5. The meter stick shown below rotates about an axis through the point marked , 20cm from one end. Five forces act on the stick: one at each end, one at the pivot point, and two 40cm from one end, as shown. The magnitudes of the forces are all the same. Rank the forces according to the magnitudes of the torques they produce about the pivot point, least to greatest. [Hint: the forces could have the same magnitude]

Ans:F2 and F5 tie, then F4, the F3 and F1 tie

VIII RecommendationsIt is recommended to use your data and your diagram to recalculate the net torque on the system using the end of the meter stick as the point of reference for calculating torques. That is, take the end of the stick as the assumed axis of rotation (x0 = 0) and calculate the lever arms from this point. This time you must include the torque created by the upward force of the support. (The supports location has not changed only our reference point has changed. Why did we not consider this force before?)Show algebraically that if the system is in equilibrium, the net torque will sum to zero about any point. (After all, if the system is in equilibrium it is not rotating and there is no axis of rotation. We have been using the point of support because it is convenient and because that is the point about which the system rotates when it is not in equilibrium.)

IX References1. Online Referenceshttp://faculty.wwu.edu/vawter/PhysicsNet/Topics/RotationalDynamics/RotEquilibrium.htmlhttp://spiff.rit.edu/classes/phys211/lectures/torq/torq_all.htmlhttps://docs.google.com/document/d/1V5eSjqY_87ZmtQUfZqxAsn6glS2WVZNf-k1Ug1DFfOE/edit?pli=1#