<Dd> M A = F B F A = V A V B = n . (\ displaystyle MA = (\ frac (F_ (B)) (F_ (A))) = (\ frac (V_ (A)) (V_ (B))) = n. \!) </Dd> <P> This shows that the force exerted by an ideal block and tackle is n times the input force, where n is the number of sections of rope that support the moving block . </P> <P> Mechanical advantage that is computed using the assumption that no power is lost through deflection, friction and wear of a machine is the maximum performance that can be achieved . For this reason, it is often called the ideal mechanical advantage (IMA). In operation, deflection, friction and wear will reduce the mechanical advantage . The amount of this reduction from the ideal to the actual mechanical advantage (AMA) is defined by a factor called efficiency, a quantity which is determined by experimentation . </P> <P> As an ideal example, using a block and tackle with six ropes and a 600 pound load, the operator would be required to pull the rope six feet and exert 100 pounds of force to lift the load one foot . Both the ratios F / F and V / V from below show that the IMA is six . For the first ratio, 100 pounds of force in results in 600 pounds of force out; in the real world, the force out would be less than 600 pounds . The second ratio also yields a MA of 6 in the ideal case but fails in real world calculations; it does not properly account for energy losses . Subtracting those losses from the IMA or using the first ratio yields the AMA . The ratio of AMA to IMA is the mechanical efficiency of the system . </P>

How does increased friction affect the actual mechanical advantage of a machine