Understanding Torque and Equilibrium in Static Physics

State of an extended object in which net force and net torque are equal to zero 1 Static equilibrium The twisting force that tends to cause rotation of an object 2 Torque The average position of all parts of an object or system weighted according to mass 3 Center of mass Distance between the pivot point and point of force application 4 Radial distance Axis along which a force is applied 5 Line of action Axis of an object around which the object rotates 6 Pivot point Physics 111L Torque and Equilibrium Final Report Student Name Kyra Pell Student ID 291291 Lesson Torque and Equilibrium Institution University of Southern Mississippi Session Fall 2023 Course Physics 111L Instructor Sidney Gautrau Test Your Knowledge Match each term to the best description.

Identify each statement as true or false. True False 1 Torque is a vector quantity. 2 The rotational velocity must be zero Torque is positive when it induces during static equilibrium. rotation in the clockwise direction. A force applied parallel to the moment arm produces the largest possible torque. Exploration A force applied at a 90° angle to the moment arm produces the largest possible torque. True False Torque can be found using the equation . τ = rF ⊥ τ = r ⊥ F τ = rF sin( θ ) All of the above The is the average position of all parts of the object or system weighted according to mass. center of gravity center of mass pivot point None of the above

The moment of inertia depends on how the of an object is distributed. gravity inertia mass All of the above is the rotational equivalent of force and causes rotational motion. Center of mass Moment of inertia Torque All of the above Static equilibrium occurs when the is zero. net force net torque velocity All of the above Exercise 1

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Four conditions must be met in order for equilibrium, static equilibrium, to be achieved. The net force on the object must be zero. This condition requires that the linear acceleration of the object is zero, following Fnet = ma. The linear velocity of the object must be zero. The net torque on the object must be zero. This condition requires that the angular acceleration of the object is zero, following τ net = I α . The rotational velocity must be zero. When the pivot point of a balance is not at the center of mass, the net torque on the balance is calculated by considering the torques from all the forces acting on it. Torque itself is found with the equation τ=rF=rmg, with r being the radial distance, F being the positive or negative force applied, m being the mass, and g being the acceleration due to gravity. Do note that if the balance is in a state of equilibrium, the total net torque will equal zero. If a force is applied directly to the pivot point, the torque due to that force is zero. This is because the distance between the pivot point and the line of action of the force is zero, resulting in the term "Distance" becoming zero in the torque formula. Also, the sine of zero degrees is also zero, making the torque zero overall. Describe the conditions necessary for equilibrium on a balance. When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated? When a force is applied directly to the pivot point of a balance, what is the torque due to that force?

As shown in data table 3 in order for the torque to be maintained as the mass on the balance increased, the position of the mass had to be moved closer to the pivot point, in this case the center of gravity. This is because torque relies on the radial distance, and the force applied. As the mass increased the force of gravity also increases. So, to counter this and maintain the torque, the radial distance was lessened. Torque is defined as the force applied to an extended object multiplied by the perpendicular distance from the pivot point to the line of action of the force. To simplify this definition, think of torque as a measurement of the force that can cause an object to rotate about an axis. Anyone who has ever opened a door has an intuitive understanding of torque. When a person opens a door, they push on the handle, on the side of the door farthest from the hinges. Pushing on the side closest to the hinges requires considerably more force. Although the work done is the same in both cases (the larger force would be applied over a smaller distance) people generally prefer to apply less force, hence the usual location of the door handle. During this experiment my present errors seemed to all correlated with torque in some way. Therefore, I am under the impression that main experimental factors that contributed to the errors found in my experiment are the factors of torque. These factors are the radial distance and the forced applied, or the mass of an object times the acceleration of gravity. I could have mismeasured the masses or I could have misjudged the radial distance needed to maintain equilibrium, leading to margins of error. To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass change? Use your measurements recorded in Data Table 3 to support your answer. Why is the handle on a door far from the hinge? Use the definition of torque in your answer. What are the main experimental factors that contributed to the error you found in each part of your experiment?

Data Table 1: Mass of Washers and Ruler Mass of 5 washers (g) Average mass of 1 washer (g) Mass of ruler (g) 26.25 5.25 12.87 Data Table 2: Initial Data Center of mass position (cm) Mass m 1 (g) Ruler Position x 1 (cm) Radial Distance r 1 (cm) Torque τ 1 (Nm) 15.20 10.50 10.00 5.20 0.00535626 Data Table 3: Measurements and Calculations for Torque About the Center of Mass Number of washers for m 2 1 2 3 Mass m 2 (g) 5.25 10.50 15.75 Ruler position x 2 (cm) 28.20 20.80 18.90 Radial distance r 2,exp (cm) 13.00 5.60 3.70 Torque τ 2,exp (Nm) -0.006695325 -0.005768280 -0.005716778 Torque τ 2,theory (Nm) -0.00535626 -0.00535626 -0.00535626 Radial Distance r 2,theory (cm) 10.40 5.20 3.47 % Error 25.0% 7.7% 6.6% Data Table 4: Center of Mass Equilibrium with Hanging Mass

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

x 0 = 10.00 cm Measured mass m (g) Ruler Position x (cm) Radial Distance r (cm) Torque τ (Nm) Ruler center of mass 12.87 15.20 5.20 -0.00679833 4 washers 21 6.70 3.30 0.00679833 x 0 = 10.00 cm Calculated mass (g) Percent error (%) Ruler center of mass 13.33 3.6 4 washers Data Table 5: Center of Mass Equilibrium with Vertical Spring Scale x 0 = 10.00 cm Ruler position x (cm) Radial distance r (cm) Spring scale reading or mass (g) Torque τ ,exp (Nm) Spring scale 25.00 15.00 5.0 0.0073575 Ruler center of mass 15.20 5.20 12.87 -0.00679833 x 0 = 10.00 cm Torque τ ,theory (Nm) Percent error (%) Spring scale 0.00679833 8.2 Ruler center of mass

As shown in Data Table 7, two objects with different masses can be in balanced equilibrium on a beam with a pivot at the beam center of mass, if the total net forces acting on each side of the pivot are equal to one another. To achieve this balance, the heavier object needs to be positioned closer to the pivot point compared to the lighter object. This way, the torque produced by the heavier object is counterbalanced by the torque produced by the lighter object. As shown in this experiment torque and equilibrium can be used to determine the mass of an unknown object. To do this, we first need to use the equilibrium equation Tnet = 0= tA + tbrick and solve for the torque of whichever mass we are trying to find. Then, using this torque and the equation t = rF=rmg, we can solve for the mass by dividing the torque by the radial distance time the acceleration force. Data Table 6: Center of Mass Equilibrium with Spring Scale at an Angle Location Center of mass Spring scale Pivot point Spring scale reading or mass (g) 12.87 5.5 Angle θ (°) 90 60 70 Perpendicular Force F □ (N) (-)0.126 (+)0.0467 (+)0.0793 Parallel Force F ‖ (N) 0 (+)0.0270 (-)0.0270 Net Force F net (N) 0.126 0.0539 0.0838 Torque τ (Nm) (-)0.00655 0.00701 0 Exercise 2 How can two objects with different masses be in balanced equilibrium on a beam with a pivot at the beam center of mass? Refer to your measurements recorded in Data Table 7 to support your answer. Describe how torque and equilibrium determine the mass of an unknown object.

When I compare the two exercises, I can see very little difference in how I found the mass measurements of my ruler in Exercise 1 vs the mass measurements of the objects in Exercise 2. In both of these exercises I had to use the same equilibrium equation to solve for the torque and, then had to use the formula of torque to solve for the masses. The only difference here was that I had a scale measurement of my ruler mass as well in Exercise 1. Compare the mass measurements you made using the simulation in Exercise 2 to the mass measurement of the ruler you made in Exercise 1. How were the measurements similar, and how were they different? Data Table 7: Determining the Mass of a Mystery Object Mystery object Object distance (m) Brick mass (kg) Brick distance (m) Brick torque (Nm) Object mass (kg) A 1.00 20 1.00 (-)196.20 20 B 1.00 20 0.25 (-)49.05 5 C 1.00 20 0.75 (-)147.15 15 D 1.00 20 0.50 (-)98.10 10 Mystery object F object (N) F bricks (N) F pivot point (N) A 196.20 196.20 (-)392.40 B 49.05 196.20 (-)245.25 C 147.15 196.20 (-)343.35 D 98.10 196.20 (-)294.30

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Competency Review is calculated by finding the product of the applied force and the radial distance. Center of mass Moment of inertia Torque All of the above The spans the distance from the to the location of the applied force. moment arm; pivot point moment of inertia; center of mass torque; center of mass None of the above A net force of zero is the only requirement for static equilibrium. True False Torque can be calculated if the and angular acceleration are known. mass moment arm moment of inertia None of the above Torque is the rotational equivalent of force and causes translational motion. True

False A hanging system can only be balanced if the pivot point is placed at the center of mass. True False A door is pushed with a force of 5 N at a distance of 30 cm from the hinge. The torque is . 6 N/m 150 N m 1.50 N m None of the above A 40 cm ruler with non-uniform mass is hung from its center of mass, at 21 cm. A force is applied at the 40 cm mark of the ruler. The length of the moment arm is . 40 19 21 None of the above A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is , if the ruler is assumed to have uniform mass. 0.48% 0.52% 4% None of the above

In this instance torque can be found using the formula T=rFsin(theta), were r is the radial distance, F is the magnitude of the force, and sin(theta) is the sine angle between them. In diagram one we are given an angle of 90 degrees, a radial distance of 17.5 (cm) and the magnitude force of 800 (N). Using the previous values and the equation above we get the follow torque value of 140 (Nm). In diagram two we are given an angle of 135 degrees, a radial distance of 17.5 (cm) and the magnitude force of 800 (N). Using the previous values and the equation above we get the follow torque value of 98.99 (Nm). As you can see the 90-degree position gives us a lager torque value. The mass of a ruler can be determined by hanging the ruler from its center of mass. True False A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is kg. 20 30 40 None of the above Extension Questions In cycling, the torque generated about the crank axis is determined by the magnitude of the force from a foot pushing the pedal as well as the angle between the crank arm and the force vector. The diagram shows the 17.5 cm crank pedal at two positions: 90 degrees and 135 degrees. What is the torque at each of the points if the foot applies a force of 800 N? Which position creates the larger torque? Note: Use the angle between the force application and crank arm in your calculations.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Physics111L_TorqueandEquilibrium_291291-converted

Related Documents