The figure below shows a 20 kg ladder leaning against a frictionless wall and re
ID: 1961365 • Letter: T
Question
The figure below shows a 20 kg ladder leaning against a frictionless wall and resting on a frictionless horizontal surface. To keep the ladder from slipping, the bottom of the ladder is tied to the wall with a thin wire. The tension in the wire is 29.4 N. The wire will break if the tension exceeds 200 N.
(a) If a 65.0 kg person climbs halfway up the ladder, what force will be exerted by the ladder against the wall?
kN
(b) How far up the ladder can a 65.0 kg person climb? (Measure along the length of the ladder.)
m
Can you take a moment about the force of the wall? I am confused about that part.
Explanation / Answer
Given data:
Mass of the ladder, M = 20 kg
Mass of the person, m = 65 kg
Tension, T = 29.4 N
Maximum tension, Tmax = 200 N
Vertical height, h = 5 m
Horizontal distance, d = 1.5 m ............................................................................................................. SOLUTION Let
Fw = Force exerted by laddeer against the wall
Solution:
(a)
= 0 about the bottom of the ladder:
( M + m ) ( g ) ( d/2 ) = Fw h
( 85 kg ) ( 9.81 m /s2 ) ( 1.5 / 2 ) = Fw 5
Fw = 125.07 N
= 0.125 kN
Ans:
Force exerted by ladder against the wall, Fw = 0.125 k N ................................................................................................... ................................................................................................... (b)
f = Fraction of the total length climbed
= 0 about the bottom of the ladder:
M ( g ) ( d/2 ) + m ( g ) ( d ) ( f ) = Tmax ( h )
(20 ) ( 9.8 ) ( 0.75 ) + (65 )( 9.8 ) ( 1.5 ) ( f ) = ( 200 ) ( 5 )
147 + 955.5 f = 1000
f = 0.892
Length of the ladder, L = [ (1.5)^2 + (5)^2 ]
= 5.22 m
Climbing length = ( f ) ( L )
=( 0.892 ) ( 5.22 )
= 4.66 m Ans:
Climbed length = 4.66 m
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