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Static Equilibrium in Terms of Equilibrium Forces and Torques

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Static Equilibrium in Terms of Equilibrium Forces and Torques 6/3/14 Objectives Static • State the conditions of static equilibrium in terms of equilibrium forces and torques. • Draw a free-body diagram of a lever showing all forces. • Use the condition of equilibrium to solve two-dimensional statics problems. Assessment Physics terms 1. What two conditions must be met for an object at rest to remain at • equilibrium rest? • net force 2. Draw the free-body diagram of the lever in the illustration. • net torque 3. What support force is exerted on the lever by the triangular support? 4. What is the value of d for which the lever is in equilibrium? Equations Brainstorm Static equilibrium: If an object or structure is to remain at rest, the net force The net force is zero. on it must equal zero. But is that enough? The net torque is zero about any center of rotation. Is there another necessary condition? 1 6/3/14 Investigation Investigation Part 1: A simply supported beam What must be true if an 1. This simulation allows object or structure is to you to place masses on a remain at rest? beam. Adjust the masses by entering distances for each one. Click on this interactive 2. [Reset] clears all the simulation on masses and distances. page 241. 3. The [force] or [torque] button toggles between displaying force or torque diagrams below the bar. Investigation Investigation Questions for Part 1 a. What is the relationship between the upwards and downwards forces? Click on this second b. What is the relationship interactive between the clockwise and simulation on counterclockwise torques? page 241. c. Use the masses to create a force scale under the left support of close to 300 N. Investigation Static equilibrium means . Going further: a lever If an object is to remain at rest then 1. The simulation allows BOTH of these conditions MUST be true: you to place four masses on a lever that is free to tip. • The net force is zero. Explore this advanced model and answer the • The net torque is zero about questions on your any center of rotation. assignment sheet. 2 6/3/14 Static equilibrium means . Balancing a see-saw If an object is to remain at rest then BOTH of these conditions MUST be true: Click on this interactive calculator on • The net force is zero. page 240. • The net torque is zero about any center of rotation. You can apply these equilibrium conditions to analyze and solve problems involving objects at rest. Balancing a see-saw Balancing a see-saw A 2.0 kg mass is placed 1.0 m A 2.0 kg mass is placed 1.0 m to the left of the center of a to the left of the center of a see-saw. see-saw. Where should a 6.0 kg mass Distance 2 Where should a 6.0 kg mass Distance 2 be placed so that the see- be placed so that the see- 1.0 19.6 1.0 19.6 0.33 58.9 saw balances? saw balances? 0.33 cm to the right Load the light mass on first. 2.0 9.81 6.0 9.81 2.0 9.81 6.0 9.81 What is the torque from the 2 kg mass? Note: The see–saw is 5 meters long, so the maximum distance from fulcrum is 2.5 m. What is the torque from the 6 kg mass? Balancing a see-saw Two-dimensional problems A 2.0 kg mass is placed 1.0 m Some problems are inherently to the left of the center of a two-dimensional. see-saw. To solve 2D problems you often Where should a 6.0 kg mass Distance 2 need to consider torques. be placed so that the see- 1.0 19.6 19.6 0.33 58.9 saw balances? 0.33 cm to the right Lever and beam problems are 2.0 9.81 6.0 9.81 examples of two-dimensional What is the torque from the situations. 2 kg mass? +19.6 N m What is the torque from the 6 kg mass? -19.6 N m The net torque is zero! 3 6/3/14 Two-dimensional problems What are you asked for? A 5.0 kg mass and a 10 kg mass A 5.0 kg mass and a 10 kg mass sit on a lever. At what distance sit on a lever. At what distance should the 10 kg mass be placed should the 10 kg mass be placed so the lever is in equilibrium? so the lever is in equilibrium? What are you asked for? You are asked for the distance d. What are you given? What are you given? A 5.0 kg mass and a 10 kg mass A 5.0 kg mass and a 10 kg mass sit on a lever. At what distance sit on a lever. At what distance should the 10 kg mass be placed should the 10 kg mass be placed so the lever is in equilibrium? so the lever is in equilibrium? What are you given? You are given both masses (5 kg, 10 kg) and one of the positions (1.5 m). What relationships do you know? What relationships do you know? A 5.0 kg mass and a 10 kg mass A 5.0 kg mass and a 10 kg mass sit on a lever. At what distance sit on a lever. At what distance should the 10 kg mass be placed should the 10 kg mass be placed so the lever is in equilibrium? so the lever is in equilibrium? The equilibrium conditions: Weight: Torque: 4 6/3/14 Solution Look at forces first Always start by drawing the Always start by drawing the free-body diagram. free-body diagram. Force equilibrium tells us that: F - m1 g + m2g = 0 But we can’t use this to find d. Free-body diagram Look at torques next Look at torques next Counterclockwise (+) Counterclockwise (+) τ τ 1 = + (1.5 m) m1g 1 = + (1.5 m) m1g Clockwise (-) Clockwise (-) τ τ 2 = - d m2g 2 = - d m2g Equilibrium condition: τ τ 1 + 2 = 0 Look at torques next Reaction forces Counterclockwise (+) Reaction forces can exert torques, τ just like any other force. 1 = + (1.5 m) m1g Clockwise (-) Can you draw the free-body diagram for this massless τ 2 = - d m2g cantilever beam? Equilibrium condition: 0.75 m Hint: Two unknown reactions forces may act at the hinge, Rx τ + τ = 0 1 2 and Ry. (1.5 m) m1g - d m2g = 0 (1.5 m) (5.0 kg/10 kg) = 0.75 m d = 5 6/3/14 Reaction forces Choosing the center of rotation Reaction forces can exert torques, What can you do? Make a smart just like any other force. choice for the center of rotation! The free-body diagram shows Choose a center that eliminates the tension T, weight mg, and the torques from unknown forces. two unknown reactions forces, Rx and Ry. What center should you pick? You want to find T but you have three unknowns! What can you do? Choosing the center of rotation Choosing the center of rotation What can you do? Make a smart What can you do? Make a smart choice for the center of rotation! choice for the center of rotation! Choose a center that eliminates Choose a center that eliminates the torques from unknown forces. the torques from unknown forces. Choosing this center of rotation The weight mg creates a gets rid of the torques from the clockwise torque about this unknown reaction forces. center. What cancels this out? They will be zero. Choose this center Choose this center Choosing the center of rotation Choosing the component What can you do? Make a smart Torque is created by the component choice for the center of rotation! of a force that is perpendicular to the lever arm. Choose a center that eliminates the torques from unknown forces. • The weight mg creates an clockwise torque. The weight mg creates a • T – the perpendicular component clockwise torque about this y of the tension – creates a center. What cancels this out? counterclockwise torque. The torque from Ty Ty Set these torques equal to zero to solve for T. 6 6/3/14 Solve for the tension Investigation Part 2: Find the unknown mass Set these torques equal to zero to solve for T. 1. Attach a long, vertical track to a short, horizontal track. 2. Attach the protractor and spring scale below the horizontal track. Attach the other spring scale to a sliding pin on the horizontal track. Investigation Investigation Part 2: Find the unknown mass Questions for Part 2 3. Attach an unknown, hanging a. Draw a free-body diagram for the mass of 10–12 large washers hanging mass. to the two spring scales. b. What is the horizontal component of 4. Move the upper spring scale force exerted by the upper spring along the track until the lower scale? spring scale is horizontal. c. Use the Pythagorean theorem to 5. For each spring scale, calculate the vertical force exerted measure and record the force. on the upper spring scale. d. What is the value of the unknown mass? Going further Going further Using trigonometry to find the mass Using trigonometry to find the mass 1. Attach a spring scale to the 3. Slide the pins along the track horizontal track. Clamp a until the mass is well balanced string to a second sliding pin. and the spring scale acts at an angle of ≈ 30° from vertical.
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