How do you solve for h in #P=mgh#?

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Answer 1 · 2016-04-06T12:34:53

# color(blue)(h = P / (mg)#

#P = mg color(blue)(h#

#P = mg * color(blue)(h#

#P / (mg) = color(blue)(h#

# color(blue)(h = P / (mg)#

Answer 2 · 2016-04-06T12:53:10

DM used the shortcut method. I am using first principles.

#h=P/(mg)#

There are two principles used to solve this question.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("The first and most important principle.")#

#color(green)("What you do to one side of an equation you do to the other side")#

Suppose I had #6=6 color(red)(" Is True")#
Suppose I subtracted 2 from the left#->4=6" "color(red)("is False")#

But #6-2=6-2color(red)(" is now True")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Moving something to the other side of the equals")#

For multiply or divide change it into 1. It then shows up on the other side.

For add or subtract change it into 0. It then shows up on the other side.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given #P=mgh#

#color(blue)("Objective: ")# To have only one #h# and for it to be on its own on one side of = and everything else on the other side.

#color(brown)("Change "m" and "g" into 1 so that "h " is on its own")#

Divide both sides by #mg# giving:

#P/(mg) =(mgh)/(mg)#

#P/(mg) =m/m xx g/g xxh#

But #m/m" and "g/g# both = 1

#P/(mg) =1 xx 1 xxh#

#h=P/(mg)#