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Any time you have to climb a ladder, you want the ladder to remain in static equilibrium. At what angle should a 60-kg painter place his ladder against the wall in order to climb two-thirds of the way up the ladder and have the ladder remain in static equilibrium? The ladder's mass is 10 kg and its length is 6.0 m. The exterior wall of the house is very smooth, meaning that it exerts a negligible friction force on the ladder. The coefficient of static friction between the floor and the ladder is 0.50.

Any time you have to climb a ladder, you want the ladder to remain in static equilibrium. At what angle should a 60-kg painter place his ladder against the wall in order to climb two-thirds of the way up the ladder and have the ladder remain in static equilibrium? The ladder's mass is 10 kg and its length is 6.0 m. The exterior wall of the house is very smooth, meaning that it exerts a negligible friction force on the ladder. The coefficient of static friction between the floor and the ladder is 0.50.

Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter14: Static Equilibrium, Elasticity, And Fracture
Section: Chapter Questions
Problem 82PQ
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Physics Problem about Torque and forces. Show the location of the axis of rotation and r for each force that produces torque in the diagram. Also, check calculations, I am off by 1 degree.
Given , F loor = mpig +meg し= Gmm M= bekg 686N Sfyc0 Friction force M FNi Fioo Co. 50) C686) = 343M FNgW L=6m wall smooth TN, Froor Et=0 X 2 Cose + X 4 cos.@ + Fx 6 sin@-Mx€2 mcg X6. X 2C0S@ + 1o x9.8 x4 cos O + 343 x6sinô- 686 cose-o 1| 76 Cos Ce) t 392 (oS 6 + 2058 sin @ - 41! 6 Cose- 2548 Cos ☺.t2058Sin 6 60 こ6 cos e cos e -25 48 +2058 tan Ce) =o +2548 > 2058 tan 8=2548 2058 2058 + 2 548 tan e=26, =e=tan C26 ) =51.1 21 21 angie made byg ladder with verticol wall= 90-51.1= 38.9
Transcribed Image Text:Given , F loor = mpig +meg し= Gmm M= bekg 686N Sfyc0 Friction force M FNi Fioo Co. 50) C686) = 343M FNgW L=6m wall smooth TN, Froor Et=0 X 2 Cose + X 4 cos.@ + Fx 6 sin@-Mx€2 mcg X6. X 2C0S@ + 1o x9.8 x4 cos O + 343 x6sinô- 686 cose-o 1| 76 Cos Ce) t 392 (oS 6 + 2058 sin @ - 41! 6 Cose- 2548 Cos ☺.t2058Sin 6 60 こ6 cos e cos e -25 48 +2058 tan Ce) =o +2548 > 2058 tan 8=2548 2058 2058 + 2 548 tan e=26, =e=tan C26 ) =51.1 21 21 angie made byg ladder with verticol wall= 90-51.1= 38.9
Any time you have to climb a ladder, you want the ladder to remain in static equilibrium. At what angle should a 60-kg painter place his ladder against the wall in order to climb two-thirds of the way up the ladder and have the ladder remain in static equilibrium? The ladder's mass is 10 kg and its length is 6.0 m. The exterior wall of the house is very smooth, meaning that it exerts a negligible friction force on the ladder. The coefficient of static friction between the floor and the ladder is 0.50.
Transcribed Image Text:Any time you have to climb a ladder, you want the ladder to remain in static equilibrium. At what angle should a 60-kg painter place his ladder against the wall in order to climb two-thirds of the way up the ladder and have the ladder remain in static equilibrium? The ladder's mass is 10 kg and its length is 6.0 m. The exterior wall of the house is very smooth, meaning that it exerts a negligible friction force on the ladder. The coefficient of static friction between the floor and the ladder is 0.50.
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