+1438 What is the AMA of a machine that requires an input force of 80 N over a distance of 1.0 m to produce an output force of 150N over a distance of 0.50m?
+36916 Actual Mechanical Advantage is a more realistic number than IMA , as AMA accounts for friction etc.
80N x 1m = 80 N*m (torque) the work 'in'
Output 150 N * .5 m = 75 N* m the work 'out (so some of your input is lost in the form of friction, sqeaks,slippage etc)
AMA = output F / input F = 150/80 = 1.875 ( I have not seen this subject in a long time......pretty sure of my answers though)
+36916 Actual Mechanical Advantage is a more realistic number than IMA , as AMA accounts for friction etc.
80N x 1m = 80 N*m (torque) the work 'in'
Output 150 N * .5 m = 75 N* m the work 'out (so some of your input is lost in the form of friction, sqeaks,slippage etc)
AMA = output F / input F = 150/80 = 1.875 ( I have not seen this subject in a long time......pretty sure of my answers though)
