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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.
 Fall '10
 Tuceryan

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