067-dt-content-rid-7778734 1/courses/PH YS.2610.40666.2017 Fall/Physics 2610 HW9
ID: 1781301 • Letter: 0
Question
067-dt-content-rid-7778734 1/courses/PH YS.2610.40666.2017 Fall/Physics 2610 HW9F 11072017 Balancing Torques 10-2.0 3.0 4.0 5 3.2 N 1.5 N F 2.5 N As in #7, a balance is set up with the fulcrum (pivot point) at the 3.0 meter inark. Fonts vertically downward are applied to the 0.0 meter mark (3.2 N), the 4.0 meter mark (1.5 N), and the 6.0 meter mark (2.5 N). What foree (in Newtons) must be applied to the 5.0 moter mark to balance the system (i.e. make the net torque about the pivot point zero O A. 0.3 N O B. 0.6N O C. 1.2 N O D. 18 N #9 Torque in greater generality A foro, F is applied to a system where F = (2.3 N, 3.4 N, 5.2 N). The position vector R from the axis of rotation to the point of application of the foree is (1.5 m,22 m, 1.8 m). Using the relation? = 1 x F, calculate the torqae developed in the system. Ruo all that to calculate the cross product tix i, one may use. O A. (532 N·m,3.66 N·m, 0.04 Nm) O B. (-1.52 N m,0.97 N-m, 1.21 N-m) O C. (072 N·m, 1.29 N-m,-2.45 N-m) O D. (2.84 N-m, 2.13 N-m,-..08 N-m)Explanation / Answer
8: net torque = (3.2 x 3) - (1.5 x 1) - (2 F) - (2.5 x 3) = 0
F = 0.3 N
Ans(A)
9. torque = r x F = (1.5, 2.2 , 1.8) x (2.3, 3.4, 5.2)
= i(2.2 x 5.2 - 1.8 x 3.4) + j (1.8 x2.3 - 1.5 x 2.3) + k (1.5 x 3.4 - 2.2 x 2.3)
= 5.32i + 0.69j + 0.04k
Ans: (5.32, 0.69, 0.04)
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