Question

A uniform meter stick pivoted at its center has a 155-g mass suspended at the 18.5-cm position and a25.0-g mass suspended at the 38.4-cm position. At what position should a 275-g mass be suspendedto put the system into equilibrium?

Answer 1

We have the next diagram

In order to have the system in equilibrium, we will analyze the torque on the pivot

$\sum ^{}_{}\tau=F_1d_1-F_2d_2+F_3d_3=0$

In this case

F1=0.155(9.8)

F2=0.025(9.8)

F3=0.275(9.8)

The distances

d1=50-18.5=31.5cm=0.315 m

d2=50-38.4=11.6cm=0.116 m

We substitute

$(0.155)(9.8)(0.315)-(0.025)(9.8)(0.116)+(0.275)(9.8)d_3=0$

then we isolate the d3

$d_3=\frac{0.155(0.315)+(0.025)(0.116)}{0.275}=0.188m_{}$

d3=0.188m=18.8cm

From the beginning of the uniform meter stick will be 50+18.8=68.8cm

ANSWER

d3=18.8cm

From the beginning of the uniform meter is 68.8cm