Static Equilibrium. Theory: The conditions for the mechanical equilibrium of a rigid body are (a) (b)

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LPC Physics A 00 Las Positas College, Physics Department Staff Purpose: To etermine that, for a boy in equilibrium, the following are true: The sum of the torques about any point is zero The sum of

1 LPC Physics A 00 Las Positas College, Physics Department Staff Purpose: To etermine that, for a boy in equilibrium, the following are true: The sum of the torques about any point is zero The sum of forces is zero Equipment: Non-Concurrent Forces Apparatus Hooke Mass Set Long Ros () Pulleys () Right-Angle Clamps () Table Clamps Balance String Carpenter s Level Theory: The conitions for the mechanical equilibrium of a rigi boy are F = 0 = 0 (a) (b) Eq. a an b That is, the (vector) sums of the forces F an the torques acting on the boy are zero. The first conition, F = 0, is concerne with translational equilibrium an ensures that the object is at a particular location (not moving linearly) or that it is moving with a uniform velocity (Newton s first law of motion). In this experiment, the rigi boy (the non-concurrent forces apparatus) is restricte from linear motion an F is automatically satisfie. To be in static equilibrium, a rigi boy must also be in rotational static equilibrium. Although the sum of the forces on the boy may be zero an it is not moving linearly, it is possible that it may be rotating about some fixe axis of rotation. of 9

2 LPC Physics A 00 Las Positas College, Physics Department Staff However, if the sum of the torques is zero, = 0, the object is in rotational equilibrium, an either it ies not rotate (static case) or it rotates with a uniform angular velocity. (Forces prouce linear motion an torques prouce rotational motion.) A torque, or moment of force, results from the application of a force acting at a istance from an axis of rotation. The magnitue of the torque is equal to the prouct of the force s magnitue an the perpenicular istance from the axis of rotation to the force s line of action, or = F (Figure ). The istance is calle the lever arm or the moment arm of the force. Axis of rotation (perpenicular to plane of paper) Force Line of action Figure The magnitue of the torque is equal to the prouct of F an the perpenicular from the axis of rotation to the force s line of action = F. Relative to an axis of rotation, a rigi boy can rotate in only two irections: clockwise an counterclockwise. It is therefore customary to refer to clockwise torques an counterclockwise torques, that is, torques that may prouce clockwise rotations an torques that may prouce counterclockwise rotations. For example, in Figure, F an F prouce counterclockwise torques an F an F prouce clockwise torques, but no rotation woul take place if the system were in rotational equilibrium. F 50 cm m m m m F = m g F = m g F = m g F = m g Figure An example of torque in ifferent irections. F an F give rise to counterclockwise torques an F an F to clockwise torques. The conition for rotational equilibrium is = cc = 0 Eq. where cc an are counterclockwise an clockwise torques, respectively. Designating the irections arbitrarily by plus an minus signs, Eq. can be written = 0 cc or of 9

3 LPC Physics A 00 Las Positas College, Physics Department Staff cc = Eq. sum of counterclockwise torques = sum of clockwise torques For example, for the ro in Figure, we have Counterclockwise Clockwise = or F F = F F The forces are ue to weights suspene from the ro. Then, with F = mg, m g m m m g = m = m g m m g Eq. Center of Gravity an Center of Mass The gravitational torques ue to iniviual mass particles of a rigi boy efine what is known as the boy s center of gravity. The center of gravity is the point of the boy about which the sum of the gravitational torques aroun an axis through this point is equal to zero. For example, consier the ro shown in Figure. If the ro is visualize as being mae up of iniviual mass particles an the point of support is selecte such that = 0, then or cc i cc ( m g) i = = an ( m g) (m m m L ) cc = ( m m m L) With the ro in rotational equilibrium, it may be supporte by a force equal to its weight, where the support force is irecte through the center of gravity. Hence, it is as though all of the object s weight (Mg) is concentrate at the center of gravity. That is, if you were blinfole an supporte an object at its center of gravity on your finger, weightwise you woul not be able to tell if it were perhaps a ro or a block of equal mass. If an object s weight were concentrate at its center of gravity, so woul be its mass, an we often refer to an object s center of mass instea of center of gravity. These points are the same as long as the acceleration ue to gravity, g, is constant (uniform gravitational fiel). Notice how g can be factore an ivie out of the previous weight equations, leaving mass equations. Also, it shoul be evient that for a symmetric object with a uniform mass istribution, the center of gravity is locate at the center of symmetry. For example, if a ro has a uniform mass istribution, its center of gravity is locate at the center of the ro s length. Why? i i of 9

4 LPC Physics A 00 Las Positas College, Physics Department Staff Center of gravity m g m g Mg Figure A ro can be consiere as being mae up of iniviual masses in rotational equilibrium when the vertical support force is irecte through the center of gravity m g m g Experiment: x = 0 x xcg θ θ mg c.g aj. m m m m x x Figure. Determine the mass of the non-concurrent force apparatus.. Suspen the apparatus horizontally by strings attache to weights, as shown in the sketch. Weights shoul not be attache symmetrically.. Ajust weights an the center of gravity ajustment to obtain equilibrium.. Measure the istances an angles necessary to etermine the torques. Recor uncertainties for all measurements. The theory section of this lab was shamelessly borrowe from: Jerry D. Wilson. Physics Laboratory Experiments, n Eition. Lexington MA: D.C. Heath an Company, 986. of 9

5 LPC Physics A 00 Las Positas College, Physics Department Staff 5. Remove m an m an ajust m an m until the apparatus is again horizontal. 6. DO NOT change the center of gravity ajustment. Measure θ an calculate the center of gravity. Analysis:. Carefully make a large sketch of the apparatus inicating the forces acting on the apparatus an their corresponing angles.. Assuming the pulleys are frictionless, write Newton's n Law for torques an forces as applie to the equilibrium of the apparatus.. Which of the torques are positive? Calculate each positive torque an its uncertainty. A the positive torques to fin the total positive torque.. Calculate the error in the total positive torque. 5. Repeat Step in the analysis for each negative torque. Calculate the total negative torque an its uncertainty. 6. Does the absolute value of the total negative torque equal the total positive torque within the uncertainty? Why or why not? Note: How woul we calculate the uncertainty in sinθ when the uncertainty in θ, δθ, is known? One way is to calculate: sin(θ δθ) an sin(θ - δθ) i.e., δ(sinθ) = Sin(θ δθ) - Sin(θ - δθ) 7. Calculate % ifference between positive an negative torques as well as the % uncertainty in your measurements. Results: Write at least one paragraph escribing the following: what you expecte to learn about the lab (i.e. what was the reason for conucting the experiment?) your results, an what you learne from them Think of at least one other experiment might you perform to verify these results Think of at least one new question or problem that coul be answere with the physics you have learne in this laboratory, or be extrapolate from the ieas in this laboratory. 5 of 9

6 LPC Physics A 00 Las Positas College, Physics Department Staff Clean-Up: Before you can leave the classroom, you must clean up your equipment, an have your instructor sign below. If you o not turn in this page with your instructor s signature with your lab report, you will receive a 5% point reuction on your lab grae. How you ivie clean-up uties between lab members is up to you. Clean-up involves: Completely ismantling the experimental setup Removing tape from anything you put tape on Drying-off any wet equipment Putting away equipment in proper boxes (if applicable) Returning equipment to proper cabinets, or to the cart at the front of the room Throwing away pieces of string, paper, an other etritus (i.e. your water bottles) Shutting own the computer Anything else that nees to be one to return the room to its pristine, pre lab form. I certify that the equipment use by has been cleane up. (stuent s name),. (instructor s name) (ate) 6 of 9

7 LPC Physics A 00 Las Positas College, Physics Department Staff DATA TABLES AND ANALYSIS mass of apparatus, m app : x : δx : x : δx : θ : δθ : x : δx : x : δx : θ : δθ : measuring center of gravity: calculations: θ : position of center of gravity, x cg : Newton s Secon Law for torques an forces: Positive Torques an their uncertainties: Total Positive Torque an its uncertainty: Negative Torques an their uncertainties: Total Negative Torque an its uncertainty: Positive Torque = Negative Torque?: Percent Difference: 7 of 9

8 LPC Physics A 00 Las Positas College, Physics Department Staff Experimental Equilibrium Data m m m m θ θ θ θ x x x x xcg Net Torque Positive Torque ( ) Negative Torque ( ) % Difference ( ) Error Analysis δθ δθ δθ δθ δx δx δx δx δx cg δ δ δ % Uncertainty δ 00 8 of 9

9 LPC Physics A 00 Las Positas College, Physics Department Staff Sketch of Apparatus 9 of 9

Torque OBJECTIVE INTRODUCTION APPARATUS THEORY

Torque OBJECTIVE INTRODUCTION APPARATUS THEORY Torque OBJECTIVE To verify the rotational an translational conitions for equilibrium. To etermine the center of ravity of a rii boy (meter stick). To apply the torque concept to the etermination of an