Question

Location of the center of mass x0 = 50.0 cm

Two masses are to be suspended from the meter stick. Given the position of the first mass, calculate the position of the second mass such that the system is in equilibrium.

Case 1:

m1 = 50.0 g 

x1 = 25.0 cm 

d1 = |x1-x0| = 

 τccw = 

m2 = 35.0 g 

d2 = 

x2 = 

τcw = 

Case 2:

m1 = 50.0 g 

x1 = 15.0 cm

d1 = |x1-x0| = 

τccw = 

m2 = 125.0 g

d2 = 

x2 = 

τcw = 

Case 3:

m1 = 100.0 g 

x1 = 45.0 cm

d1 = |x1-x0| = 

τccw = 

m2 = 

x2 = 70.0 cm

d2 = |x2-x0| =

τcw =

Case 4: 

m1 = 120.0 g 

x1 = 40.0 cm

d1 = |x1-x0| = 

m2 = 50.0 g 

x2 = 90.0 cm 

d2 = |x2-x0| = 

m3 = 50.0 g 

d3 = 

x3 = 

τccw = 

τcw = 

Answer 1

1)

$$d_1=25\ cm\\\tau_{ccw}=0.25(9.8)(0.05)=0.1225\ Nm\\\\\tau_{cw}=0.1225\ Nm\\d_2=\frac{0.1225}{9.8\cdot0.035}=0.357\ m=35.7\ cm\\x_2=85.7\ cm$$

2)

$$d_1=35\ cm\\\tau_{ccw}=0.35(9.8)(0.05)=0.1715\ Nm\\\\\tau_{cw}=0.1715\ Nm\\d_2=\frac{0.1715}{9.8\cdot0.125}=0.14\ m=14\ cm\\x_2=64\ cm$$

3)

$$d_1=5\ cm\\\tau_{ccw}=0.05(9.8)(0.1)=0.049\ Nm\\\\\tau_{cw}=0.049\ Nm\\d_2=\frac{0.049}{9.8\cdot0.07}=0.071\ m=7.1\ cm\\x_2=57.1\ cm$$

4)

$$d_1=10\ cm,d_2=40\ cm,\\\tau_{ccw}=0.1(9.8)(0.12)=0.1176\ Nm\\\\\tau_{cw}=0.4(9.8)(0.05)=0.196\ Nm\\d_3=\frac{0.196-0.1176}{9.8\cdot0.05}=0.16\ m=16\ cm\\x_3=34\ cm$$