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.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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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.
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.
Given
, F loor = mpig +meg
し= Gmm
M= bekg
686N
Sfyc0 Friction forcef M FNi Fico
Co. 50) C686) = 343M
FNgW
L=6m
wall
smooth
TN, Froor
mcng
Ffloor
Et=0
M= mp.g
X 2 Cose
X 4 cos.@ t Fx 6 sin@-Mx€s
mcg
X6.
X 2COS@ + 1o x9.8 x4 cos @ + 343 x6sinô- 686cose-o
1| 76 Cos Lo) t 392 (o5 6 2058 sin @ - 4116 CosO-
2548 Cos ☺.t2058Sin 6
60
こ6
cos @
cos ☺
-25 48 +2058 tan Ce) =o
+2548
> 2058 tan 8=2548
2058
2058
+ 2 548
tan e=26, =e=tan C26 ) =51.19
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 forcef M FNi Fico Co. 50) C686) = 343M FNgW L=6m wall smooth TN, Froor mcng Ffloor Et=0 M= mp.g X 2 Cose X 4 cos.@ t Fx 6 sin@-Mx€s mcg X6. X 2COS@ + 1o x9.8 x4 cos @ + 343 x6sinô- 686cose-o 1| 76 Cos Lo) t 392 (o5 6 2058 sin @ - 4116 CosO- 2548 Cos ☺.t2058Sin 6 60 こ6 cos @ cos ☺ -25 48 +2058 tan Ce) =o +2548 > 2058 tan 8=2548 2058 2058 + 2 548 tan e=26, =e=tan C26 ) =51.19 21 21 angie made byg ladder with verticol wall= 90-51.1= 38.9
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