How do I find the mass of the meter stick? I set up a meter stick at equilibrium on a triple beam balance at 49.5 cm. Wi
Posted: Mon May 09, 2022 6:17 pm
How do I find the mass of the meter stick? I set up a meter
stick at equilibrium on a triple beam balance at 49.5 cm. With 5g
added, it balanced at 65cm. 10g added was at 76 cm, and at 20g it
was 11cm.
The exact mass of the meter stick = 78.5g
Calculations
Center of gravity of the meter stick is dms = SRms -
position of fulcrum
For trial 1 - dms = 65cm - 50cm = 15cm
For trial 2 - dms = 76cm - 50cm = 26cm
D1 = 50cm - 10cm (position of fulcrum - position of m1
on the meter stick) = 40cm ???
M1=5g Tms=T1
Mms * g * dms = m1 g d1 / g dms
Mms = m1d1/dma
Mms = 5g*15cm/40cm = 1.875 g This value cannot be
correct……
stick at equilibrium on a triple beam balance at 49.5 cm. With 5g
added, it balanced at 65cm. 10g added was at 76 cm, and at 20g it
was 11cm.
The exact mass of the meter stick = 78.5g
Calculations
Center of gravity of the meter stick is dms = SRms -
position of fulcrum
For trial 1 - dms = 65cm - 50cm = 15cm
For trial 2 - dms = 76cm - 50cm = 26cm
D1 = 50cm - 10cm (position of fulcrum - position of m1
on the meter stick) = 40cm ???
M1=5g Tms=T1
Mms * g * dms = m1 g d1 / g dms
Mms = m1d1/dma
Mms = 5g*15cm/40cm = 1.875 g This value cannot be
correct……