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» Lab 6 » 6- Timer Notes EvaluateFeedback Print ? Info Course Contents» 6-3 Torque (Loyd 10-6) Due in 23 hours, 21 minutes The question below refers to this image of the meter stick, which is a bit different from the one in Figure 10-5 of the lab manual. 10 20 30 40 50 60 70 80 90 m1 m2 The mass m0.490 kg acts at 30.0 cm. Assume the meter stick is uniform and symmetric so it balances at the 50.0 cm point where the triangular support (called a fulcrum) is shown. (6) What is the value of the mass m2 that must be placed at the position 65.0 cm, as shown, to put the system in equilibrium? Submit Answer Tries 0/15 Dashlane Word 2016 CellularApplication sargeanttlab PHY2048 Diffusion to Replace Google Exam 1 Stu. 0 Type here to search

Explanation / Answer

Solution:

By balancing the torque about 50cm point-
=>m1*(50-30)=m2*(65-50)
=>m1*20=m2*15
=>m2=0.490*4/3

=0.65kg

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