Transcript 101KnightChp7_22.pptx

Textbook - Knight
Chapter 7: 22,23,26
(pages 220-221)
Chap 7:22 (pg 220)
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Distance from the pivot to F1  L
Distance from the pivot to F2  L/2
Torque about pivot due to F1  = F x d
= F1 ⋅ L ⋅ sin 45° = 20 ⋅ L ⋅ .707 = 14.14 L
Torque about pivot due to F2  = F x d
= − F2 ⋅ L/2
Total Torque = 0 = 14.14 L − F2 ⋅ L/2
Solve the Equation
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Total Torque = 0 = 14.14 L − F2 ⋅ L/2
F2 ⋅ L/2 = 14.14 L
Cancel L’s  F2 / 2 = 14.14
F2 = 28.28 newtons
7:23
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Torque
Tom: Fxd = 50 x 3 x sin60°
Jerry: Fxd = − 35 x 3 x sin80°
Total: 50 x 3 x sin60°
− 35 x3 x sin80°
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= 150 x .866 − 105 x .985
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=
129.9 − 103.4
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=
26.5 N-m
7:26
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At right:
F x d = 10N ⋅75cm ⋅ sin 150° = 375 N-cm
At left:
F x d = − 8N ⋅25cm ⋅ sin 140° = −128.6 N-cm
Total = 246.4 N-cm = 2.464 N-m