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

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Introduction An Object Is In Static Equilibrium When The Vector Sum Of All The Forces Acting On That Object Is Equal To 1 (224.26 KiB) Viewed 37 times

Introduction An Object Is In Static Equilibrium When The Vector Sum Of All The Forces Acting On That Object Is Equal To 2 (170.11 KiB) Viewed 37 times

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, (1) Σ F = 0, where i = 1, 2, 3, ..., n, ΣΕ or F₁+F₂+F3+...+F₁ = 0 or in component form, {Fix ΣF = 0 and (2) = 0 #1 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 2 1. The masses used in Part 1 should still be hanging from the strings. Change my to 120g, and m3 to 150g (see Figure 1). Obtain measurements for 0₁ and 02 using the same procedure described in Part 1. 90° m₁ = 120g m3 = 150g Ob 0₂= 0₁ = Ba 00=126° 0₂=_36° m₂=90g 2. Use equation 3 to calculate the new magnitudes of T₁, T2, and T3. |T₁| | T₂| = |T3| = = 3. Use the measured angles 0₁ and 0₂ to calculate the components of T1, T2, and T3. Tix = T2x = T3x = Tly = T2y = T3y = 4. Use the results in 3 to show that the equilibrium conditions of equation 2 are satisfied by your experimental values.