The force acting on the gear tooth is F=45 KN. Resolve this force into twocomponents acting along the lines aa and bb *F.80°50°F_a=65.46 KN, F_b=D35 KNnone of aboveO F_a=32.46 KN, F_b=D40 KNF_a=45.17 KN, F_b=45 KNO F_a=57.85 KN, F_b=45 KN

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The force acting on the gear tooth is F=45 KN. Resolve this force into two components acting along the lines aa and bb * F 80° 50° b' O F_a=65.46 KN, F_63D35 KN none of above O Fa=32.46 KN, F_63D40 KN O F_a=45.17 KN, F_b=D45 KN O F_a=57.85 KN, F_b=45 KN

The force acting on the gear tooth is F=45 KN. Resolve this force into two components acting along the lines aa and bb * F 80° 50° b' O F_a=65.46 KN, F_63D35 KN none of above O Fa=32.46 KN, F_63D40 KN O F_a=45.17 KN, F_b=D45 KN O F_a=57.85 KN, F_b=45 KN

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter7: Dry Friction

Section: Chapter Questions

Problem 7.75P: The single-plate clutch transmits the torque C from the input shaft on the left to the output shaft...

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Power Transmission Elements

It is defined as the energy transfer from one location to another location. The energy will start moving from the energy generation source to the energy applied position to obtain the work output. From this method, a large amount of electric energy moves from the power station to various electrical substations.

Question

The force acting on the gear tooth is F=45 KN. Resolve this force into two
components acting along the lines aa and bb *
F.
80°
50°
F_a=65.46 KN, F_b=D35 KN
none of above
O F_a=32.46 KN, F_b=D40 KN
F_a=45.17 KN, F_b=45 KN
O F_a=57.85 KN, F_b=45 KN

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