STATIC EQUILIBRIUM: Concurrent Force Systems

Q₁: A 200 N force ( F ) is used to stabilize a system of pulleys within an industry’s weighing station. The system’s weighing receptacle is filled with a quantity of liquid that results in 150 N of force exerted in the vertical direction. If the pulley to the right is at a 20^o angle relative to the horizontal, what must ( θ ) be to establish static equilibrium?

A: Equilibrium is established in the system when all of its components are at rest. This requires the sum of the horizontal and vertical forces ( or force components ) to equal zero:

∑F_h = 0

∑F_v = 0

The 150 N force vector has a vertical component with no horizontal components; however, the pulleys above are each locked into a diagonal position. For this reason, trigonometry must be used to break these vectors into their horizontal and vertical components:

∑F_h = ( F_T2 )( cos θ ) – ( F_T3 )( cos θ ) = 0

∑F_v = ( F_T2 )( sin θ ) + ( F_T3 )( sin θ ) – 150 N = 0

F_T2 = 200 N

( 200 N )( cos θ ) – ( F_T3 )( cos θ ) = 0

( 200 N )( cos θ ) = ( F_T3 )( cos θ )

( F_T3 )( cos θ ) = ( 200 N )( cos 20^o )

( F_T3 )( cos θ ) = 188 N

And,

( F_T2 )( sin θ ) + ( F_T3 )( sin θ ) – 150 N = 0

( F_T3 )( sin θ ) = 150 N – ( 200 N )( sin 20^o )

( F_T3 )( sin θ ) = 150 N – 68.4 N

( F_T3 )( sin θ ) = 81.6 N

Ooops!!!

We do not have a value for F_T3, so how must we go about solving ( θ )??? This is where trigonometry can be really helpful. A ratio of the ( F_T3 )( sin θ ) and ( F_T3 )( cos θ ) expression fit very neatly into a tan ( θ ) expression:

tan ( θ ) = ( opposite / adjacent )

tan ( θ ) = ( sin θ / cos θ )

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sin θ = ( 81.6 N / F_T3 )

cos θ = ( 188 N / F_T3 )

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tan ( θ ) = [ ( 81.6 N / F_T3 ) / ( 188 N / F_T3 ) ]

tan ( θ ) = [ ( 81.6 N / 188 N)( F_T3 / F_T3 ) ]

tan ( θ ) = ( 81.6 N / 188 N)

tan ( θ ) = 0.43

θ = tan^-1 0.43

θ = 23.5^o

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Q₂: What is the value of F_T3?

( F_T3 )( sin θ ) = 81.6 N

( F_T3 )( sin 23.5^o ) = 81.6 N

F_T3 = ( 81.6 N / sin 23.5^o )

F_T3 = ( 81.6 N / 0.398 )

F_T3 = 205 N

And,

( F_T3 )( cos θ ) = 188 N

( F_T3 )( cos 23.5^o ) = 188 N

F_T3 = ( 188 N / cos 23.5^o )

F_T3 = ( 188 N / 0.917 )

F_T3 = 205 N