An object is in static equilibrium when the vector sum of all the forces acting on that object is equal to zero. This is

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An Object Is In Static Equilibrium When The Vector Sum Of All The Forces Acting On That Object Is Equal To Zero This Is 1 (204.31 KiB) Viewed 32 times

An Object Is In Static Equilibrium When The Vector Sum Of All The Forces Acting On That Object Is Equal To Zero This Is 2 (124.99 KiB) Viewed 32 times

An Object Is In Static Equilibrium When The Vector Sum Of All The Forces Acting On That Object Is Equal To Zero This Is 3 (107.28 KiB) Viewed 32 times

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, I or ΣΕ = 0, where i = 1, 2, 3, ..., n, F1+F2+F3+...+F„ =0_(1) or in component form, 71 ΣF=c 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 1 1. Hang 100g at both ends of the strings, as m, and m3 in Figure 1. Note that the mass hangers each have a mass of 50g. Now hang 90g at the center, m, in Figure 1. Measure the angle between string 1 and the loop attached to the center hanger, string 2. This angle is labeled 0, in Figure 1. 0₂² 117° 01, see Figure 2, is then equal to ₁ minus the 90° between the horizontal and the vertical. m3 m₁ 0₁ = 27° วา 100g 1009 Measure the angle b shown in Figure 1. Figure 2 = 4. 0₂ is then equal to 0, minus the 90° between the horizontal and the vertical. 0₂= 27° 5. T1, T2, and T3 are the tensions in each of the strings, are shown in Figure 3. In figure 3, a) Draw the x-y coordinate system; b) Label the components Tx and Ty for each T; c) Label 0₁ and 0₂. T₁ T3 T₂ Figure 3 6. Since the system is in equilibrium, the tensions are equal to the weights hanging on the 2. 3. m₂ 909

end of the strings. |I₁|=100g 11₂1= 90g |13|= 100g 7. Use the measured angles 0₁ and 0₂ to calculate the horizontal (x) and the vertical (y) components of T₁, T2, and T3. Tix = T2x = T3x = Tly = T2y = T3y = 8. Use the results in 7 to show that the equilibrium conditions of equation 2 are satisfied by your experimental values.