+0

# I got some decimal answer but I don't think it's right.

+1
152
1

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?

Feb 28, 2019

#1
+2

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)

Feb 28, 2019

#1
+2

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)

ElectricPavlov Feb 28, 2019