When a system is in static equilibrium, the sum of the forces acting upon the system must equal zero. In the diagram below, forces F1 and F2 exert torques upon the system:
Recall that a torque ( π = Fr sin ΞΈ ) has the ability to make a system rotate, and it is the product of a force being applied to a lever ( r ) with a length equal to ( x ). The magnitude of torque is maximized when it is applied to a lever at a 900 angle. Since sin 900 = 1, the torques in this system are expressed as π = Fr = ( mg )( x ). The center of massβ location is chosen as the point of reference for convenience. If we designate the clockwise direction as negative, and we designate the counterclockwise direction as being positive, the systemβs torques may be expressed as follows:
π1 + ( – π2 ) = 0
π1 = π2
( m1g )( x1 ) = ( m2g )( x2 )
Since m1 = m2, and x1 = x2, the system remains at rest. If we doubled the mass of m2 and allowed m1 to remain unchanged, this would necessitate doubling the length of the x1 lever to maintain balance:
( mg )( 2x ) = ( 2mg )( x )
( m )( 2x ) = ( 2m )( x )
2mx = 2mx
2x = 2x
In the former example, it was necessary to eliminate the ( mg ) term via division to solve for ( x ). There are other circumstances, however, where an appropriate substitute for ( mg ) must be derived. The system below is one such example:
If any of the forces acting on this system were unbalanced, they would cause the system to rotate in either a clockwise or counterclockwise manner; thus, if T1 > T2, the system would rotate clockwise ( and vice versa for T2 ). Likewise, if T1 + T2 > mg, the system would move upward in the vertical direction ( and vice versa ). Since the net forces acting upon the system must equal zero, the following conclusion can be made:
T1 + T2 – mg = 0
T1 + T2 = mg
Additionally, notice that the center of mass is not located at an equidistant location between T1 and T2. As a consequence, a more detailed process must be used to determine the value of T1 and T2:
π1 = π2
π1 = ( T1 )( β L )
π2 = ( T2 )( β L )
( T1 )( β L ) = ( T2 )( β L )
We may now substitute for T1 and T2 in order to expand and subsequently simplify the equation:
T1 + T2 – mg = 0
T1 + T2 = mg
T1 = mg – T2
T2 = mg – T1
( mg – T2 )( β L ) = ( mg – T1 )( β L )
( mg )( β L ) – ( T2 )( β L ) = ( mg )( β L ) – ( T1 )( β L )
( mg )( β L ) – ( mg )( β L ) = ( T2 )( β L ) – ( T1 )( β L )
The ( L ) term is on both sides of the equation, but the T1 and T2 terms are not on the left-hand side of the equation. Since we are solving for ( T ), it will be useful to substitute for ( mg ) on the left-hand side of the equation:
( T1 + T2 )( β L ) – ( T1 + T2 )( β L ) = ( T2 )( β L ) – ( T1 )( β L )
( β T1L ) + ( β T2L ) – [ ( β T1L ) + ( β T2L ) ] = ( T2 )( β L ) – ( T1 )( β L )
( β T1L ) + ( β T2L ) – ( β T1L ) – ( β T2L ) = ( β T2L ) – ( β T1L )
( β T1L ) – ( β T2L ) = 0
( β T1L ) = ( β T2L )
β T1 = β T2
2T1 = T2
We may now substitute T1 + T2 = mg for either T1 or T2:
T1 + 2T1 = mg
3T1 = mg
T1 = β mg
T2 = β mg
π1 = π2
( β mg )( β L ) = ( β mg )( β L )
2/9 mgL = 2/9 mgL
1 = 1
RENAISSANCE PHYSICS 