1. Equilibrium & Newton’s Laws of Motion Tensional Forces

2. Equilibrium Newton’s 1 st Law of Motion  When the forces on an object are balance, it is said to be in equilibrium.  When an object is in equilibrium, it is not accelerating.  An object that is not accelerating is stationary or moving at constant speed in a straight line. You balanced forces during the force table lab.

3. Equilibrium in 2-Dimensions When objects are in equilibrium:  a x and a y = 0 It then follows that:  ΣF x and ΣF y = 0 Because F net = ma and a = 0.

4. Ex. 1: Equilibrium A 100 N sign is hung by two wires as seen below. What is the tension in the wires? Physics is Fun F g = 100 N FAFA FBFB    = 15 °

5. Diagram the Problem Physics is Fun F g = 100 N FAFA FBFB  y x System   = 15 ° y x F Bx F By F Ay F Ax F g = 100 N FAFA FBFB

6. State the Known & Unknown What is known?  F g = 100N  θ = 15 ° What is not known? FAFA FBFB

7. Perform Calculations Isolate the x and y components separately. Since the sign is not moving, F net = ma = 0 in both the x and y directions. x – direction:  -F Ax + F Bx = 0  -F A cosθ + F B cosθ = 0  F A cosθ = F B cosθ   = 15 ° y x F Bx F By F Ay F Ax F g = 100 N FAFA FBFB

8. Perform Calculations y – direction:  F Ay + F By – F g = 0  F A sinθ + F B sinθ – F g = 0  2F A sinθ = F g (due to symmetry, F Ay = F By )  F A = (100N)/((sin15 ° )(2))  F A = 193 N   = 15 ° y x F Bx F By F Ay F Ax F g = 100 N FAFA FBFB

9. Ex. 2: Equilibrium An engine has a weight of 3150 N. The engine is positioned above the engine compartment using rope, a pulley and a ring as shown. Find the tension in each of the sections of rope

10. State the Known & Unknown What is known?  W = 3150 N  θ 1 = 10 °  θ 2 = 80 ° What is not known? T1T1 T2T2

11. Perform Calculations Isolate the x and y components separately. Since the engine is not moving, F net = ma = 0 in both the x and y directions. x – direction:  -T 1 sin θ 1 + T 2 sin θ 2 = 0  -T 1 sin 10 ° + T 2 sin 80 ° = 0(1) y – direction:  T 1 cos θ 1 - T 2 cos θ 2 – W = 0  T 1 cos 10 ° - T 2 cos 80 ° - W = 0(2) Solve (1) for T 1 and substitute into (2).

12. Perform Calculations (cont.)