Static Equilibrium

I. INTRODUCTION

Newton’s second law was introduced to you earlier this semester in conditions where a force ~F acting on an object ofmass m may lead to a linear acceleration. In the case of a static object, the sum of all forces ~Fi must equal zero. Thatis, at static equilibrium

Σ~Fi = 0 (1)

where the subscript i denotes each force applied to the object. The vector characteristics of each force are accountedfor by the component method, and in two dimensions, a force may have components along the x and/or y axis.When solving Newton’s laws we used a + or - sign to denote a direction (e.g. a vector pointing up was denoted bypositive y components, whereas one pointing down was denoted negative y components).

For an object that is free to rotate, forces may also exert a torque (τ) which may lead to a rotational motion. In astatic situation, with no rotation, we require that the sum of all torques ~τ i be zero. That is, at static equilibrium

Σ~τ i = 0 (2)

where the subscript i denotes each torque applied to the object. A torque that may result in clockwise rotationis denoted by a (-) sign, whereas a torque that may result in counterclockwise rotation is denoted by a (+) sign.This notation is commonly the one used in many texts, and is arbitrary (you often can get away with making amistake in the choice of + or -, so long as you are consistent when solving a problem). You can’t make clockwise andcounterclockwise rotations both result from negative torques. The magnitude of the torque may be determined bythe following expression:

|τ | = |~R||~F|sin(θ) (3)

Note that the above expression gives you only the magnitude of torque, |τ |. More advanced courses in engineering orphysics explore the true vector characteristics of torque, which is defined by an operation known as a cross product(we don’t cover that in algebra based physics). On the right hand side of the expression we have the magnitudeof the radial distance (|~R|) and the magnitude of the force (|~F|). The angle θ is defined as the angle between theradial distance from the point of rotation to the point where the force is applied. Occasionally some texts define themagnitude of the torque in the following manner, using a lever arm (R⊥) as follows

|τ | = |R||F|sin(θ) = R⊥F where R⊥ = |R|sin(θ) (4)

The figure below shows how these variables are defined. You should notice that the maximum torque for a givenforce and radial distance is greatest when θ is 90 degrees. Additionally, the torque is always zero when the force isapplied at the point of rotation. Imagine trying to open a door if the handle was placed exactly at the door hinge -itwould be impossible, since R=0 and the torque τ that you can generate would be zero.

FIG. 1: As an example of the geometry you may encounter with a torque calculation. The blue bar may rotate about point Q freely.The force is applied at a distance ~R from the point of rotation denoted Q. The angle θ is used for determining the magnitude ofthe torque. Note, sin(β) = sin(θ) as they are supplementary angles for any β,θ. In both cases shown, the blue bar experiences apositive torque and would rotate counterclockwise.

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One application of these concepts is in use of a torque wrench, used to tighten bolts. As an example, consider asituation where you were replacing the spark plugs or the oil pan drain plug on your car or motorcycle (or tighteningany nut or bolt anywhere). You might be inclined to tighten a bolt as much as possible, using all your strength. Thisis poor practice, as the force one would apply to a bolt would vary from person to person, and may lead to damaginga bolt by stripping the threads. A more precise method would be to quantify the exact torque that should be appliedto the bolt, which may be measured with a torque wrench (shown below). Torque wrenches come as both analog ordigital devices and have varied sensitivity ranges (and prices!). Your car or motorcycle manual will have the torquenecessary to tighten bolts (e.g. the oil pan bolt) on your vehicle. In the United States, the units used are the ft-lb,and in European vehicles the units are the Nm ( 1 Nm= 0.74 ft-lb). When you tighten a screw or a bolt, the wrenchtypically turns clockwise to tighten generating negative torque. The bolt experiences a positive torque from thethreads, and at equilibrium these two torques sum to zero.

FIG. 2: A torque wrench, used for tightening bolts or screws to a certain specification (e.g. a bolt on your car engine block)

II. MASSES ON A SEESAW

In this part of the lab we will apply equations introduced in the previous section to a situation of a seesaw. Thesimulation may be accessed at this address:

Click here

Step 1: To begin, click on the number 5 as shown below:

https://phet.colorado.edu/en/simulation/balancing-act

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Step 2: Let’s go to the lab, click on Balance Lab as shown below:

Let’s review some of the features. On the bottom right you have a few masses to choose from. These may be draggedonto the seesaw by using the mouse (right or left click, hold the mouse down and drag onto the seesaw). You canalso put masses back once you have placed them onto the seesaw.

Step 3: Let’s turn on the ruler, mass labels and level:

Step 4: Lastly, you will be able to check if your calculations worked by removing the grey supports on/off by theoption in the middle of the page.

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A. Part I: Two masses on a seesaw

Place a 10kg at a distance of 1m from the left of the pivot point. Your setup should look like this:

Computations: Where should one place a 5kg mass on the right to balance the seesaw? Show your algebraic workin how you arrived at your answer. In your answer, also indicate which of the weights cause a clockwise or acounterclockwise torque. You can (you should) test your result by removing the grey supports off (see step 4 above).Does the seesaw balance?

B. Part II: Three masses on a seesaw

Place a 10kg at a distance of 1m from the left of the pivot point. Additionally, place a 5kg mass 1.5m from the leftof the pivot point.

Computations: Determine where a 10kg mass should be placed on the right side of the pivot point so that the seesawis balanced. Show your algebraic work in how you arrived at your answer. In your answer, also indicate the weightsthat cause a clockwise and counterclockwise torque. You can (you should) test your result by removing the greysupports off (see step 4 above).

C. Part III: Five masses on a seesaw

Place a 5kg at a distance of 1.75m, and a second 5kg mass 1.5m from the left of the pivot point. Additionally, placea 10kg a distance of 1.0m from the left of the pivot point. Also place 20kg mass on the right, a distance of 0.25m,and a 15kg mass a distance of 1.25m from the right of the pivot.

Computations: Determine where a 5kg mass should be placed on the right side of the pivot point so that the seesawis balanced. Show your algebraic work in how you arrived at your answer. In your answer, also indicate the weightscause a clockwise and counterclockwise torque. You can (you should) test your result by removing the grey supportsoff (see step 4 above).

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III. A HINGED ROD IN STATIC EQUILIBRIUM

In this part of the lab you will apply equations 1 and 2 in a situation of static equilibrium of a hinged rod, wherethe mass is equally distributed. Please go to this web site:Click here

1: Setup the following parameters:Mass of bar : 40kgMass of box : 30kgLength of bar : 8mPosition of the box: 2.5mAngle of bar: 27 degrees

2: This simulation has an option to view the tension in the string. Verify the numerical value shown by explicitlysolving Newton’s laws (equations 1 and 2). Show all work for full credit.

3. Determine the magnitude of the force that the hinge exerts on the rod. Again, show all work for full credit.

https://ophysics.com/r6.html

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IV. EQUILIBRIUM OF A BAR SUPPORTED BY A CABLE

In this part of the lab you will apply equations 1 and 2 in a situation of static equilibrium of a hinged rod, wherethe mass is equally distributed. Please go to this web site:

Click here

Step 1: Set the rod weight to 15 N.

Step 2: Set the angle that the string makes with the vertical equal to 27 degrees.

Step 3: Set the distance that the string attaches to the rod from the pivot equal to 20 cm.

Your simulation should look like this:

Computations:

1. Draw a free body diagram for rod, highlighting the forces acting on the rod. In your diagram label the tensionof the string as (T) and the force the hinge makes with the rod as (F). For your lab report you can take a picture ofyour work (just draw it by hand on paper) and import it to whatever word processor you are using.

2. Determine the magnitude of the torque generated by the force that the hinge makes with the rod (F). Explainyour answer.

3. Determine the magnitude of the tension in the string. Show all work for full credit.

4. Determine the components of the force (F) that the hinge makes with the rod. Show all work for full credit.

http://physics.bu.edu/~duffy/HTML5/hinged_rod.html

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V. QUESTIONS TO BE ANSWERED IN YOUR LAB REPORT

1. For the case of a seesaw that is balanced with masses (part-one of this lab), does the number of objects on the leftof the pivot have to equal the number of objects on the right of the pivot for the seesaw to be balanced? Explainyour answer for full credit.

2. For the case of a seesaw that is balanced with masses (part-one of this lab), does the total mass on the left of thepivot have to equal the total mass on the right of the pivot for the seesaw to be balanced? Explain your answer forfull credit.

3. In part one of the lab, you able to ignore the mass of the ruler. Why? Explain your answer for full credit.

4. The figure below shows an individual trying to loosen an old rusty bolt. Explain why a long wrench would workbetter than a smaller length wrench (picture credit : https://www.thoughtco.com/torque-2699016).

5. In part one of the lab would any of the static equilibrium cases you studied change if the value of ’g’ is altered(e.g. if the experiment were done on the moon or mars). Why? Explain your answer for full credit.

IntroductionMasses on a seesawPart I: Two masses on a seesawPart II: Three masses on a seesawPart III: Five masses on a seesaw

A hinged rod in static equilibriumEquilibrium of a bar supported by a cable Questions to be answered in your lab report