Torque worksheet (1) Jacob Roser

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Dec 6, 2023

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Page 1 CONCEPTUAL QUESTIONS TORQUE WORKSHEET PS 113L – Intro Physics I Laboratory / 50pts Name: Jacob Roser Section # : 1. _Force is the action responsible for acceleration in a straight line, while Torque is the action responsible for acceleration around an axis of rotation. 2. What conditions must be true to say an object is in equilibrium? Equilibrium is reached when the sum of forces and the sum of torques are both equal to zero net force and net force and net torque are both zero 3. What conditions must be true to say an object is in static equilibrium? Static equilibrium occurs when the equilibrium conditions are met and the object is at rest, neither translating or rotating. 4. You have two wrenches you can use to tighten a bolt: one has a handle 8” long the other a handle 12” long. Assuming whichever one you use your hand can produce the same maximum force, which wrench would allow you to produce the largest torque on the bolt? The 12 inch long would allow me to produce the largest torque on the bolt 5. Using the following image as a guide, what if the Force applied was not at a right angle to the lever arm? Would the Torque on the bolt increase, decrease, or stay the same? The torque on the bold would decrease as the value of the torque is calculated as an angle less than 90 degrees. This will decrease the value of the torque.

PART 1: METER STICK CENTER OF GRAVITY & MASS *Assume g = 9.7926 m/s 2 Please use grams for all masses and centimeters for all positions and lengths. Exact Meter Stick Mass, M = 214 ± .05 _ Exact “50g” Mass, m = 50.2 ± .05 _ Exact “100g” Mass, m = 100 ± .05 _ Exact “200g” Mass, m = 198.2 ± .05_ Exact “500g” Mass, m = 489 ± .05_ Exact Clamp#1 Mass, m = 19.5 ± .05 Exact Clamp#2 Mass, m = 18.4 ± .05 Meter Stick Center of Gravity, R CG = 50 ± .05 6. Sketch a free body diagram of the meter stick at equilibrium with a single 200g mass suspended from one hanger. Label all forces and torques. Record all moment arm lengths. 7. 50-25.4 cm 8. 25.4 cm arm length 9. 19g 10. Using your free body diagram, the values recorded, and knowledge of static equilibrium, determine a derived mass for the meter stick. Do not forget the clamp has a mass. Hint: the sum of all torques should equal zero, then solve for the “unknown” meter stick mass. F 18.5 -20.9w=0 18.5F= 20.9w = m = 18.5 (200+20)/20.9xg M = 194.73g

11. Compare the derived meter stick mass from above to the exact meter stick mass you measured earlier. What is the percent difference between the two meter stick masses? %𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = � 𝑚𝑚 1 − 𝑚𝑚 ′ � 𝑚𝑚 1 ∗ 100% 1

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M = 216g Percent diff = 216-194.73/216 x 100 = 9.84 percent

PART 2: TORQUE BALANACE WITH OPPOSING TORQUES *Assume g = 9.7926 m/s 2 Please use grams for all masses and centimeters for all positions and lengths. Exact Meter Stick Mass, M = 214 ± .05 _ Exact “100g” Mass, m = 100 ± .05 _ Exact “200g” Mass, m = 198.2 ± .05 _ Exact Clamp#1 Mass, m = 19.5 ± .05 Exact Clamp#2 Mass, m = 18.4 ± .05 _ Meter Stick Center of Gravity, R CG = 25.4 ± .02 12. Sketch a free body diagram of the meter stick at equilibrium with a 200g mass suspended from one hanger and a 100g mass suspended from another hanger opposite the pivot from the 200g mass. Label all forces and torques. Record all moment arm lengths. 41.1 cm (200) 48.9 cm (100) 41.1=cg 13. Determine the sum of all clockwise (-) torques. Determine the sum of all counterclockwise (+) torques. Use your free body diagram, the values recorded, and knowledge of static equilibrium. Do not forget the clamps have mass. Clockwise torques (200+20)x19.4 cm 4268gxcm Counterclockwise torques (100+19)x 363cm

119x36.3 cm 4319.7gx cm

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14. Compare the sum of all clockwise (-) torques to the sum of all counterclockwise (+) torques by a percent difference. 4319.7-4268x100= 1.19 percent Percent diff = 1.19 percent 15. Ideally, what should the percent difference between the two directions of torques be? Why do you believe this? Ideally the percent difference between the two torques should be more because the equilibrium equals zero.

PART 3: TORQUE BALANCE TO MEASURE AN UNKNOWN MASS *Assume g = 9.7926 m/s 2 Please use grams for all masses and centimeters for all positions and lengths. Exact Meter Stick Mass, M = 214 ± .05 _ Exact “200g” Mass, m = 198.2 ± .05 _ Exact Clamp#1 Mass, m = 19.5 ± .05 Exact Clamp#2 Mass, m = 18.4 ± .05 _ Meter Stick Center of Gravity, R CG = 41.1 ± .05 16. Sketch a free body diagram of the meter stick at equilibrium with a 200g mass suspended from one hanger and an unknown mass suspended from another hanger opposite the pivot from the 200g mass. Label all forces and torques. Record all moment arm lengths. 17. 46 (200) 18. 54 (unknown) 19.Determine a derived mass for the “unknown” mass. Use your free body diagram, the values recorded, and knowledge of static equilibrium. Do not forget the clamps have mass. 37.4 (m+19)= 29.9 (200+20) 37.4 (m+15)= 657.8 M+19= 175.88 = m 156.88g Measured Exact “Unknown” Mass, m = 153.6 ± 20.Compare the derived unknown mass from above to the exact unknown mass you just measured with the triple-beam balance. What is the percent difference between the two masses? 154-1571/154x100 = 1.94 percent

21.Explain how you would construct your own triple-beam balance. How would you figure out the correct ruling spacing for the three mass scales?

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I would have a massless plate that was moved by three weights and use a reference phase to put all the pieces together.

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