Phy 101 LAB 7-Torque & Equilibrium_ADS

pdf

School

Iowa State University *

*We aren’t endorsed by this school

Course

131X

Subject

Mechanical Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by JusticeBraveryLyrebird33

PHY 101 LAB Group Member Names Torque & Equilibrium Torque & Equilibrium: Experiment 7 Introduction Torque and equilibrium are important concepts in mechanics. In the field of structural engineering, these two concepts help explain why buildings stand up, or why bridges don’t fall down. In this experiment we will investigate the idea of torque and how it applies to the equilibrium of a rigid body. Historical Background Galileo mathematically described the lever, as shown in Figure 10.1. It should be mentioned that the lever was most probably identified first by Archimedes (287-212 B.C.). This is typical in the history of science. Ideas develop through the centuries, and even millennia. Science is an incredible collection and synthesis of ideas from all over the world and throughout human civilizations. Figure 1. The lever. The “L’s” are the distance from the fulcrum (lever arms), and the “F’s” are the forces at each end. For a relatively small force, a large force can be produced on the other side of the lever. This idea is called leverage. A small force gains leverage as its distance from the fulcrum increases. Leverage is a ‘special case’ contained in Newtonian mechanics. The quantity, l F, is called TORQUE. That is,  = l F. From Newton’s 2 nd Law, it can be shown that if an object is not rotating, then the sum of the torques on that object must be zero,   0  . L f = l F l F f fulcrum L

2 Governing Principle The expression F l = fR was generalized by Newton. Newton showed this to be a direct result of the principle of equilibrium. Newton stated that if a system was not accelerating, then the sum of the external forces must be zero. Similarly, Newton stated that if a system was not angularly accelerating, then the sum of the external torques must be zero. Figure 10.2 shows a rigid body acted upon by several forces. If the system is not rotating, then the sum of the torques caused by the forces must be zero. That is, F 1 l 1 – F 2 l 2 – F 3 l 3 = 0. A positive sign is often assumed for counterclockwise rotation and a negative sign for clockwise rotation. Figure 10.2. Rigid body under several torques. Experimental Procedure For the problems, use the following diagrams and equations. Use a ruler clamp to hang and balance a meter stick from a ring stand. F 2 F 3 l 3 l 2 l 1 F 1 0 0 2 2 1 1     gl m gl m  l 1 l 2 m 2 m 1 Setup 1 l 3 l 2 l 1 m 2 m 3 Setup 2 0 0 3 3 2 2 1 1      g m l g m l g m l  m 1 Setup 3 m 1 m cm l 1 x 0 0 1 1     gx m gl m cm 

3 Experimental Data: Experiment 7 For the given values, determine the unknown quantities. Verify the results experimentally by constructing the configurations. Show all calculations explicitly and neatly. Case 1: Setup 1 with l 1 = 5 cm, m 1 = 500 g, m 2 = 100 g. Case 2: Setup 2 with l 1 = 20 cm, l 2 = 25 cm, m 1 = 450 g, m 2 = 150 g, m 3 = 100 g. Case 3: Setup 2 with l 1 = 25 cm, l 2 = 10 cm, m 1 = 300 g, m 2 = 250 g, l 3 = 20 cm. l 2 = _________(calculated) l 2 = _________(experimental) l 3 = _________(calculated) l 3 = _________(experimental) m 3 = _________(calculated) m 3 = _________(experimental)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

4 Clamps were used to hold the weights in this experiment. Record the mass of each of the clamps used. Recalculate the lengths for Case 1 and Case 2 adding the clamp mass to m 1 and m 2 . l 2 = _________(Case 1) l 3 = _________(Case 2) m clamp1 = _________ m clamp2 = _________

5 Case 4: Setup 3 with l 1 = 10 cm, m 1 = 100gm + clamp weight . (you have to use trial and error to find x, start with x about 8 cm) (note that you have to move both clamps to keep l 1 at 10 cm) ( don’t forget to include the mass of the clamp for m 1 !) Case 5: Setup 3 with x = 10 cm, m 1 = 100gm + clamp weight . Calculate the average mass of the meter stick from Cases 4 and 5. Measure the mass of the meter stick with an electronic scale and calculate the percent error. Use the electronic scale reading as an accepted value. x = _________(experimental) m cm =________(calculated from data) 1 1 = _________(experimental) m cm = ________(calculated from data) m cm = _________(average) cm m =_________(scale) % error = _________

6 Questions: Experiment 7 Answer each question completely with explanations of all answers. Show all calculations. 1. A uniform meter stick supported at the 25 cm mark is in equilibrium when a 200-gm rock is suspended at the 0 cm end. Is the mass of the meter stick greater than, equal to, or less than the mass of the rock? Explain your reasoning. 2. A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 mark. What is the mass of the meter stick?

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

7 3. A ‘breaker’ bar is a pipe used to extend the length of a wrench. The torque required to turn a bolt is 100 N·m. If a wrench is 20 cm long and the maximum force a person can exert is 300 N, what minimum length breaker bar should be used? 4. A 50 kg child sits 1 m to the left of a seesaw’s fulcrum. A 30 kg child sits 1 m to the right of the fulcrum. Where should a 40 kg child sit to balance the seesaw?