A meter stick is balanced at its centre (50cm). when 2 coins, each of mass 5g are put one on the top of other at 12 cm mark it is found to be balanced at 45cm what is mass of stick?

1 Answer
Neefgi
Mar 21, 2018

#"m"_"stick"=66"g"#

Explanation:

When using center of gravity to solve for an unknown variable, the general form used is:

#(weight_"1")*(displacement_"1")=(weight_"2")*(displacement_"2")#

It's very important to note that the displacements, or distances, used are relating to the distance the weight is from the fulcrum (the point the object is balanced at). That being said, since the axis of rotation is at #45"cm":#

#45"cm"-12"cm"=33"cm"# #color(blue)(" Fulcrum" - "distance" = "displacement"#
#5"g"*2=10"g"# #color(blue)(" 2 coins of 5g each = 10g")#

It's important to remember that we can't neglect the original center of gravity of #50"cm"#, meaning that since there was a #5"cm"# shift:

#(50"cm"-45"cm") = 5"cm"# #color(blue)("Displacement due to coins")#

So, to follow our original equation of

#(weight_"1")*(displacement_"1")=(weight_"2")*(displacement_"2")#

We substitute with:

#(10"g")*(33"cm") = (weight_"2")*(5"cm")#
#(330g*cm)=(5"cm")(weight_"2")# #color(blue)("Solve for unknown weight")#
#(weight_"2")=66"g"# #color(blue) (( 330"g"* cancel("cm"))/ (5cancel("cm")))#

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