Case 2: painters on a scaffold Weight of object 1: 350gWeight of object 2: 181 gWeight of stick : 100gLocation of string 1 : 4 cm Location of string 2: 88cm Center of mass of stick: 50 cmLocation of object 2: 17cm Location of string 2 : 66cmMeasured tension 1: 6N Measured tension 2 : 5.5 NPlease provide equations and calculations. **Lesson 12 Lab: Static Equilibrium****Introduction**The three conditions of static equilibrium are:\[\Sigma F_x = 0\]\[\Sigma F_y = 0\]\[\Sigma T = 0\]In words, the sum of all the forces acting in the x-direction is zero; the sum of all the forces acting in the y-direction is zero; and the sum of all the torques is zero.In this experiment, we will verify the static equilibrium conditions by re-creating three static equilibrium word problems. The choice of force magnitudes, points of application, and angles will be a matter of personal choice. The goal is to calculate all forces for each system and compare the results to measured values.**Materials**- Statics Report Sheet *(Download and print)*- A meter stick or yardstick- A 5.0 N spring scale- String- Three objects with masses between 100-500 grams- A ruler and protractor**Data Collection and Analysis****Case 1: Traffic Light**A traffic light is hanging from two wires set at known angles. What is the magnitude of the tension in each wire?Set up this system based on the figure. Hang an object of known mass from the center of a meter-long string with loops on both ends for attaching the spring scale. It may be easier to tie a second shorter string to the center of the first string and tie the object to the short string. Hold the long string in such a way that two different angles, \( \theta_1 \) and \( \theta_2 \), are formed. It may be helpful to have an assistant, or tie one end of the string to a fixed object. **Case 2: Painters on a Scaffold**Two painters with different masses are standing on a scaffold of a certain length and different distances from each end. The cables supporting the scaffold are vertical and attached to each end. What is the tension in each cable?Set up this system based on the figure using a meter stick or yard stick. Tie two strings about 50 cm long to each end of the stick. The other end of each string should have a loop in it for attaching the spring scale.### Case 2 Example 1![Image of an experimental setup for the painters on a scaffold problem.](#)*An experimental example of the painters on a scaffold word problem.*Tie or hang two objects with different masses from two points on the stick. The locations should be chosen so that the system is not symmetrical (not the same distance from each end of the stick). Hold the strings so that the stick is horizontal.Record the distances, masses, and the mass of the stick (usually around 100 grams) on the Report Sheet. Solve for the tension in each string.### Case 2 Example 2![Image of an experimental setup for the painters on a scaffold problem with a spring scale added.](#)*An experimental example of the painters on a scaffold word problem with spring scale added.*You can set the system down while you are doing this, or you can immediately measure the two tensions before doing the calculation. In either case, when measuring the tensions make sure the stick is horizontal.Carefully attach the spring scale to the loop of one string by holding the loop with the hook, making sure that the spring scale is vertical and the stick remains horizontal. Record the tension, T1, on the Report Sheet. Repeat this procedure for the other string, recording the second tension, T2, on the Report Sheet. Compare the calculated tensions with the experimental values measured by the spring scale.*Note: all angles between applied forces and the horizontal stick are 90 degrees.*

Given: Mass of object 1, m1=350 g Mass of object 2, m2=181 g Mass of stick , M=100 g Location of string 1= 4 cm Location of string 2= 88 cm Center of mass of stick= 50 cm Location of object 1= 17 cm Location of object 2= 66 cm Measured tension in sting 1,T1=6 N Measured tension in string 2, T2=5.5 NThe free body diagram of the question is drawn below Using the idea of equilibrium Force down due to gravity = Force up due to tension in the two strings. Mathematically this can be expressed as m1g+m2g+Mg=T1+T2Since the stick is not rotating , the sum of the moments about any point will be zero. The moment is defined as , Moment=Force×perpendicular distance Taking pivot at point where string 1 is attached. w1d1+w2d2=T2d Rearrange this to calculate T2. T2=w1d1+w2d2d Substitute m1g for w1 and m2g for w2. T2=m1gd1+m2gd2d d1=17 cm-4cm=13 cmd2=66 cm-4 cm=64cmd=88 cm- 4cm=84 cm Substitute the values and calculate T2. NOTE: Convert g into kg and cm into m. T2=0.35 kg×9.8 m/s2×0.13 m +0.181 kg×9.8 m/s2×0.64 m0.84 mT2=1.88 NNow substitute the value of T2 in the equation m1g+m2g+Mg=T1+T2 and calculate T1. 0.35 kg×9.8 m/s2+0.181 kg×9.8 m/s2+0.1 kg×9.8 m/s2=T1+1.88 N T1=6.18-1.88 T1=4.30 N Therefore the tension in the string 1 and string 2 are 4.30 N and 1.88 N respectively.