Question

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
NO LINKS. ​

Answer 1

correct is C

Step-by-step 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