Lect17

# Now mg and t and fx will not enter into the torque

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Unformatted text preview: - Fx = 0 y: T sin θ + Fy - Mg - mg = 0 x Now use ∑ τ = 0 Just to be different, choose the rotation axis to be through the end of the rod. Now mg and T and Fx will not enter into the torque equation: L FY L − Mg = 0 2 T M m y Fy θ L/2 L/2 Fx Mg mg Physics 211 Lecture 17, Slide 16 So we have three equations and three unknowns: three T cos θ - Fx = 0 T sin θ + Fy - Mg - mg = 0 y L FY L − Mg = 0 2 x which we can solve to find: T (M / 2 + m )g T= sin(θ ) (M / 2 + m )g Fx = tan(θ ) 1 F y = mg 2 M m Fy θ L/2 L/2 Fx Mg mg Physics 211 Lecture 17, Slide 17 Physics 211 Lecture 17, Slide Same as in the pre­lecture T m M Fy θ L/2 Fx L/2 mg y Mg x (M / 2 + m )g sin(θ ) (M / 2 + m )g Fx = tan(θ ) T= Fy = 1 mg 2 Physics 211 Lecture 17, Slide 18 Physics 211 Lecture 17, Slide Preflight In case 1 one end of a horizontal massless rod of length L case is attached to a vertical wall by a hinge, and the other end holds a ball of mass M. In case 2 the massless rod holds the same ball but is twice as case long and makes an angle of 30o with the wall as shown. 30 In which case is the total torque about the hinge biggest? In A) Case 1 B) Case 2 C) Both are the same gravity 90o M 30o 2L M L Case 1 67% got this right Case 2 Physics 211 Lecture 17, Slide 19 In which case is the total torque about the hinge biggest? In A) Case 1 B) Case 2 C) Both are the same M Case 2 90o M 30o 2L Case 1 L A) the force of gravity is perpendicular to the rod, which will cause mo...
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## This note was uploaded on 03/18/2013 for the course CSCI 402 taught by Professor Tuceryan during the Fall '10 term at IUPUI.

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