How Do You Calculate Torque in a Physics Lab Setting?

[john560]

Homework Statement

B. Torque Due to Perpendicular
Component of F
4. The tension in string 1 (on the left) is not
known, so we will calculate the torques
about this point (10.0 cm). (That way,
there is no torque due to the unknown
string 1.) Calculate the torque due to the
weight of the meter stick about 10.0 cm.

5. Calculate the perpendicular component
of the tension in string two (T2sinθ) for
the five different values of θ. Is the
perpendicular component of the tension
the same for each trial? Calculate the
torques due to string 2 about x = 10.0
cm. Is the net torque acting on the
meterstick about 10.0 cm equal to zero
for each trial?

Mass of meter stick= 149g
Clamps(these are the things that connect the string to the meter stick) = 17g
right side weight = 200g

left side clamp at 10.0cm
Center of mass of ruler = 50cm
right side clamp at 90 cm

Homework Equations

t= fl = rfsin($\theta$) = F_tanr

The Attempt at a Solution

Calculate the torque due to the
weight of the meter stick about 10.0 cm.

4. The center of mass of the meter stick was 50 cm
So the only thing that is torquing it is the gravity of the meter stick, from the 10cm point
The weight of the meter stick(including 2 clamps = 1.79N
t= rfsin(90) $\ t= (.40m) (1.792 N) (sin(90))$
$\ t= .717NM$
Negative 717NM since the torque is clockwise
Can someone confirm that I am doing this correctly?5. I am just going to do the calculations for trail 1 to see if i am doing it correct Calculate perpendicular component of tension in string 2
$\ t2sin = opposite$

$\ (1.98N) sin 31 = 1.02 N$Calculate the torques due to string 2 about x = 10.0 cm.

t= F_tanr
$\ t= (1.02 N) (.80 m)$
$\ t= .816 NM$
Positive since torque is counterclockwise.

Calculate the torque due to the
weight of the meter stick about 10.0 cm.

For this step i just added the string 2 torque from the torque measured in 4.

net torque = .-717NM + .816NM = .099

Can someone check if i did this correctly?

 

[tms]

4. The center of mass of the meter stick was 50 cm
So the only thing that is torquing it is the gravity of the meter stick, from the 10cm point

What about the bit of the meter stick to the left of the string? Doesn't it have weight, too?