Answer Happy

INTRODUCTION An object is in static equilibrium when the vector sum of all the forces acting on that object is equal to zero. This is stated mathematically as, n Σ F = 0, where i=1, 2, 3, ..., n, or F₁+F2 +F3 +...+F₂=0 (1) i=1 or in component form, ΣF=0 and ΣF=0 (2) where the F; are the individual forces and the x and y components separately sum to zero. If an object suspended by strings is stationary then there is no net force acting on it. The forces in this case are the tensions in the strings, which suspend the object (the knot). In this exercise mass will be hung from mass hangers and the tension in the strings will be generated by the mutual gravitational attraction between the masses and the earth. The force of gravity on the masses at the surface of the earth is given by, F(N) = m(kg) x 9.80(m/s²) (3) and is called the weight of the masses. Notice that Equation 3 gives the appropriate units to be used in the SI system of units.

Part 3 1. Each group will be given a different set of values for this part of the experiment. Ask your instructor for your group number. Highlight the my, m3, and 0; values for your group in the table shown below. Group #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 mi 55g 60g 65g 70g 75g 80g 85g 90g 95g 100g m3 65g 70g 75g 80g 85g 90g 95g 100g 105g 110g 0₁ 25° 30° 35° 40° 45° 50° 55° 60° 65° 70° 2. For the given m₁, m3, and 0₁, use static equilibrium conditions to predict the m₂ and the 0₂ values required for the system to reach static equilibrium. Show your work! 25° my 659 ? Figure 2 65g Your predicted values: m₂ = and 0₂= 40° 3. Now hang my and m3, Adjust m₂ until the above given 8, is obtained. Measure and record m2 and 0. Calculate the angle 02. 0₂ = 0₂= m₂ = 4. Your predicted values in step 2 and your experiment results in step 3 should be close to each other. If not, identify the mistake(s) and make correction. m₁ 559 0₁ m₂ 0₂