FORCE AND ACCELERATION: Net Force Exerted on a Ring

Q: How may we determine the net force ( F ) exerted on the ring below?

A: We must first reduce the F₁ and F₂ vectors into their x/y-components:

The F_1y and F_2y components of F₁ and F₂ oppose the motion of F₃. The net force in the y-direction is as follows:

F_nety = F_2y + F_1y – F₃

F_nety = F₂ cos 45⁰ + F₁ cos 30⁰ – 966N

F_nety = ( 500N )( cos 45⁰ ) + ( 707N )( cos 30⁰ ) – 966N

F_nety = 354N + 612N – 966N

F_nety = 966N – 966N = 0

There is no acceleration in the y-direction. The x-axis components are derived with the ( sin θ ) function:

F_netx = F_2x – F_1x

F_netx = F₂ sin 45⁰ – F₁ sin 30⁰

F_netx = ( 500N )( sin 45⁰ ) – ( 707N )( sin 30⁰ )

F_netx = 354N – 354N = 0

There is no acceleration in the x-direction, and the ring will remain at rest until an unbalanced force is exerted on the system. Since the sum of forces is zero, tip-to-tail addition will give us a closed triangle:

The angles do indeed add up to 180⁰.