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Unformatted text preview: L/4 L/2
1
Mg
3 2
T 2 = Mg
3 Mg y
x
Physics 211 Lecture 17, Slide 11 What if you choose a different axis?
r
∑τ = 0 Choose the rotation axis to be out of the
page at the left end of the beam:
page The torque due to the string
on the right about this axis is: 3L
τ 2 = T2
4 T1 The torque due to the string on
the left is zero τ1 = 0
Torque due to gravity: L
τ g = − Mg
2 T2
M x cm
L/4 L/2 Mg y
x
Physics 211 Lecture 17, Slide 12 You end up with the same answer!
The sum of all torques must be 0: τ1 + τ 2 + τ g = 0
T2 = T2 3L
L
− Mg = 0
4
2 2
Mg
3 T1 T2 We already found that
T 1 + T2 = Mg L/4 L/2
T1 = M x cm 1
Mg
3 2
T 2 = Mg
3 Mg y
x
Physics 211 Lecture 17, Slide 13 Approach to Statics: Summary
Approach to Statics: Summary
In general, we can use the two equations r
∑F = 0 r
∑τ = 0 to solve any statics problem.
When choosing axes about which to calculate torque, we
When
can sometimes be clever and make the problem easier....
can Physics 211 Lecture 17, Slide 14 Hanging Lamp Again
Hanging Lamp Again
A lamp of mass m hangs from the end of plank of mass M and
length L. One end of the plank is held to a wall by a hinge, and
the other end is supported by a massless string that makes an
angle θ with the plank.
a) What is the tension in the string?
b) What are the forces supplied by the hinge on the plank? M θ
hinge
L m
Physics 211 Lecture 17, Slide 15
Physics 211 Lecture 17, Slide First use the fact that ∑ F = 0 in both x and y directions: x: T cos θ...
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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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