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3. A homogeneous cylinder 3 ft in diameter and weighing 400 lb is resting on two rough inclined surfaces as shown in the figure. If the angle of friction is 15°, find the couple C applied to the cylinder that will start it rotating clockwise. Solution: FBD: w=400 lb 1:5' 1.5' NB RA 15 NA 600 RB From the force triangle: RA Sine law: RB 400 sin450 sin60° sin750 450 RA = 292.82 lb RB RB = 358.64 lb W=400 lb From the FBD of the cylinder: +U Mo = 0 C- (292.82sin 15°)(1.5))+ (358.64sin15°)(1.5) = 0 C = 252.92 lb-ft 750 600 RA

3. A homogeneous cylinder 3 ft in diameter and weighing 400 lb is resting on two rough inclined surfaces as shown in the figure. If the angle of friction is 15°, find the couple C applied to the cylinder that will start it rotating clockwise. Solution: FBD: w=400 lb 1:5' 1.5' NB RA 15 NA 600 RB From the force triangle: RA Sine law: RB 400 sin450 sin60° sin750 450 RA = 292.82 lb RB RB = 358.64 lb W=400 lb From the FBD of the cylinder: +U Mo = 0 C- (292.82sin 15°)(1.5))+ (358.64sin15°)(1.5) = 0 C = 252.92 lb-ft 750 600 RA

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please explain the moments part for the system. Thanks

GUIO CI 3. A homogeneous cylinder 3 ft in diameter and weighing 400 lb is resting on two rough inclined surfaces as shown in the figure. If the angle of friction is 15°, find the couple C applied to the cylinder that will start it rotating clockwise. Solution: FBD: 60° W=400 lb 1:5' FA 1.5' NB B. RA 75 50 450 RB NA 600 60° 459 From the force triangle: RA Sine law: RB 400 %3D sin45° sin60° sin75° 45° RA = 292.82 lb RB RB = 358.64 Ib ICE W=400 lb From the FBD of the cylinder: +U Mo = 0 C – (292.82sin15°)(1.5))+ (358.64sin15°)(1.5) = 0 C = 252.92 lb-ft 750 %3D %3D 600 RA OFICAEN
Transcribed Image Text:GUIO CI 3. A homogeneous cylinder 3 ft in diameter and weighing 400 lb is resting on two rough inclined surfaces as shown in the figure. If the angle of friction is 15°, find the couple C applied to the cylinder that will start it rotating clockwise. Solution: FBD: 60° W=400 lb 1:5' FA 1.5' NB B. RA 75 50 450 RB NA 600 60° 459 From the force triangle: RA Sine law: RB 400 %3D sin45° sin60° sin75° 45° RA = 292.82 lb RB RB = 358.64 Ib ICE W=400 lb From the FBD of the cylinder: +U Mo = 0 C – (292.82sin15°)(1.5))+ (358.64sin15°)(1.5) = 0 C = 252.92 lb-ft 750 %3D %3D 600 RA OFICAEN
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