chapter 8 #20
a 15 meter, 500 newton uniform ladder rests against a friction less wall making an angle of 60 degrees with the horizontal. find the horizontal and vertical forces exerted on the base of the ladder by the earth when an 800 newtons fire fighter is 4 meter from the bottom. if the ladder is just on the verge of slipping when the fire fighter is 9 meter up, what is the coefficient of the static friction between ladder and ground
2007-12-04 09:08:10 · 2 answers · asked by Don 1 in Science & Mathematics ➔ Physics
sum the torques about the bottom of the ladder
set the distance of the firefighter as d from the bottom
cos(60)*(500*7.5+800*d)-R*sin(60)*15=0
a) set d=4 and solve for R
R=267.5 N
This is a purely horizontal force since the wall is frictionless
Summing forces in the x, this is the horizontal component of the reaction force at the base of the ladder.
The y component os 800+500=1300 N
b) The normal force of the ladder base is always the sum of forces in the y, or 1300 N
When the base is on the verge of slipping, the force of friction, normal times µs, will balance R, the horizontal component as found in A.
when d=9, R=421.4657 N
since
1300*µs=421.4657
µs=0.324
j
2007-12-04 09:20:08 · answer #1 · answered by odu83 7 · 1⤊ 1⤋
Do you have an actual question to ask about this problem, or were you planning on having others do your entire homework set for you?
2007-12-04 09:13:53 · answer #2 · answered by ZikZak 6 · 0⤊ 1⤋