In this lab, we will investigate the conditions for static equilibrium. Then, we will use those conditions and resulting relationships to determine an unknown mass.

• 1 low friction, rotary lab stand mount for a meter stick
• 1 aluminum meter stick with 2 hangers (10 grams each)
• Various known masses
• 1 lab stand connected to a Vernier Force sensor
• LabQuest or LabPro to read the force sensor
• A protractor
• Record data in this Google Sheets data table

At some point in an introductory physics course, after learning about kinematics and dynamics, in linear and circular motion, we get to apply this knowledge to an exciting and important application: systems that do not move¹. This may seem disappointing at first but the study of static equilibrium (and extension into dynamic equilibrium) is crucial to any field where structures need to stay intact. This includes, but is not limited to, mechanical engineering, civil engineering, orthopedics and physical therapy.

In two dimensions (so, a 3-D object but only moving or rotating in one plane) the condition for static equilibrium of an object is that net forces and torques on that object are zero, ie:

$$\Sigma F_x= 0, \Sigma F_y= 0, \Sigma \tau= 0 $$

To evaluate the torque about an axis (the point where the axis intersects the plane of rotation is the pivot or fulcrum ), we use

$$ \vec{\tau} = \vec{r} x \vec{F}, \tau=rF\sin{\theta} $$

where \(r\) is the moment arm, or distance from the point at which the force acts to the pivot. \(\theta\) is the angle between the force and a vector which points from the pivot to where the force acts (\(\vec{r}\)). The resulting vector torque \(\vec{\tau}\) points out, perpendicular to the plane created by \(\vec{r}\) and \(\vec{F}\). The cross-product relationship gives us that a torque pointing toward us seeks to create counterclockwise (ccw) rotation and a torque vector pointing away creates clockwise (cw) rotation. In cases where \(\theta = 90^\circ \), the torque is simply \(\tau=rF\); we take ccw as positive torque and cw as negative.

For Parts I and II of this experiment, one can simplify the problem even more. Masses hung to the left of the fulcrum experience a gravitational force which makes a counterclockwise torque on the system. Masses hung to the right of the fulcrum experience a \(\ F_{grav}\) which makes a clockwise torque on the system. It is the balancing of these torques which we investigate in Parts I and II.

In Part III we add a string which contributes a tension, \(F_{string}\), on the system. This string is connected to the left of the fulcrum but makes a clockwise torque because there is a component of the force \(F_{string}\sin{\theta}\) which is "up". Note that we are taking \(\theta\) to be the angle between the meter stick and the string.

Since the system is static we can choose the point about which to evaluate the torques; the sum of torques is zero about any pivot. Often we choose a complicated point, where forces which we do not know are acting and, hence, have a moment arm of zero.

For the system in Part III, we take the rotary mount of the meter stick to be the pivot and ccw torques positive, resulting in the expression:

$$\Sigma \tau = 0 = m_{extraweight}gr_{extraweight} + m_{stick}gr_{stick} - F_{string}r_{string}\sin{\theta} $$

We will be identifying positions along the meter stick as \( x \), such as \( x_{string hanger}\). In general, the relevant distance at which a torque acts, its moment arm \( r \) , is \( r = x_{where-torque-acts} - x_{pivotpoint} \) .

Part I: Balancing Torques

Begin by sliding the aluminum meter stick into its low friction, rotary mount and balancing it in the mount. This will happen when the mount indicator is close to \(50cm\); use the spirit level bubble to set this point carefully. Tighten the attachment screw then record the position.²

Slide a hanger onto the left (positions \(\lt 50cm\)) side of the ruler and add mass to the hanger as listed in your data sheet. Now add a hanger and mass to the right side (positions \(\gt 50cm\)) and adjust the position of the hanger until the system is balanced, ie. the meter stick is horizontal. Record the positions of each mass, with uncertainty.

Repeat the above steps for the other mass combinations, in all, 5 pairs. Remember that the hangers each have mass of \( 10 \) grams.

Part II: Determining the Mass of the Meter Stick

In Part I, we centered the meter stick so that it did not contribute to the counter-clockwise (ccw) or clockwise (cw) torques. In this part, we will determine the mass of the meter stick by mounting it off-center and seeing what is needed to balance it.

Remove the hanger from the left (\(\lt 50cm\)) side of the meter stick. While holding the meter stick up, slide the meter stick until the rotary mount is about the \( 70cm\) point. Now put a 100 gram mass on the right side hanger and slide until the meter stick balances. Record the balance point of the meter stick and location of the hanger plus 100 gm mass when the system is in equilibrium.

Part III Option A: Determining the Mass of the Meter Stick, again

Remove any hanger from the meter stick and take it out of the rotary mount, Weigh it on a scale and estimate the associated uncertainty.

Part III Option B: Determining the Mass of the Meter Stick, again

Click here for details of PartIII, Option B

Part I: Balancing Torques

For all 5 mass combinations, calculate the positive (counter-clockwise) and negative torques on the meterstick plus masses system. Propagate uncertainties. Do the torques balance in each case?

Part II: Determining the Mass of the Meter Stick

In our models, the weight of the meter stick is a force which acts at the center-of-mass, ideally, the \(50cm\) point. Since the system is in balance, counter-clockwise torques equal the clockwise torques, so, if \(m_{stick}\) is the mass of the meter stick and \(x_{hanger}\) is the position of 100 gram mass (plus hanger), \(m_{stick}\) can be found (in grams) by

$$m_{stick} = (x_{hanger}-x_{balance})*110/(x_{balance}-50) $$

Do the calculation for \(m_{stick}\), noting that \(x_{balance} \approx 70cm)\).

Part III, Option A: Determining the Mass of the Meter Stick, again

Compare the mass value from the scale with the value you found in Part II. Do they agree?

Part III, Option B: Determining the Mass of the Meter Stick, again

Taking Equation 3 and solving for the mass of the meter stick we get:

$$ m_{stick} = (F_{string}r_{string}\sin{\theta} - 0.210gr_{200gm} )/(gr_{stick}) $$

The above equation assumes that we use \( g = 9.80 m/s^2 \) which will give the mass of the meter stick in kg. .^{3 3}

For uncertainty in \(m_{stick}\), the proper calculation is difficult. For this lab, assume that the fractional (relative) uncertainty in \(m_{stick}\) is the same as the largest fractional uncertainty in the terms of Equation 5. Does your \(m_{stick}\) agree with the value you got from Part II?

Please see Brightspace\Assignments\Statics for (actual) possible discusssion questions

Your TA will ask you to answer some of the following questions (they will tell you which ones to answer):

Experimental questions:

1. Why did we do that?
2. Should we have done this, instead?

Theoretical questions:

1. What if the cow were spherical?
2. What if you or your lab partner were massless?
3. What would happen if gravity became repulsive during the measurements in Part III?
4. Why?

For further thought:

1. Suppose we hadn't. Could we do so, later?

Guide to Uncertainty Propagation & Error Analysis (Quick Reference)

PHY121/122 Plotting Tool

PHY133/134 Plotting Tool