A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod? a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass. e.42 cm
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correct  is C

Explanation:

In this exercise we will use the rotational equilibrium relationships. Let's set the datum at the pivot point and counterclockwise rotations as positive

as the bar is uniform its center of mass coincides with its geometric center

          x_{cm} = 0.25 cm

the distance from the fulcrum is

          x₁ = 0.40 - xcm

          x₁ = 0.40 - 0.25

          x₁ = 0.15 m

         Στ = 0

         w_bar 0.15 - W x = 0

         x = 0.15 w_bar / W

         x = 0.15 m_{bar} / M

let's calculate

         x = 0.15 0.2 /0.6

         x = 0.05 m

this distance is measured from the pivot point of the positive side, if we measure from the left end of the bar

          X = 0.4 + 0.05

          X = 0.45 m = 45 cm

correct  is C

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14