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A Fire Fighter and a Ladder?
A 15.0 m, 471 N uniform ladder rests against a frictionless wall, making an angle of 60.0° with the horizontal.
a) Find the horizontal and vertical forces exerted on the base of the ladder by the Earth when an 811 N fire fighter is 4.00 m from the bottom.
b) If the ladder is just on the verge of slipping when the fire fighter is 9.00 m up, what is the coefficient of static friction between ladder and ground?
is this static problem , i am not guaranteed i can do it right but i will try
a)
i let Torque is T
vertical force at the base is Ax
horizontal force at base is Ay
normal force at the top of ladder and wall is N, N is pointing to the left
apply sum of T =0
Torque = force (moment arm)
.............= 471 x 7.5 x cos 60 ( clockwise)
.............= 811x 4 x cos60 (clockwise)
.............=N x 15 x sin60 (counterclockwise)
let clockwise is -, so counter clockwise is +
sum of torque at A is
-471 x 7.5 x cos 60 -811x 4 x cos60+N x 15 x sin60 = 0
-3388.25+ 12.99N =0
N= 3388.25/12.99 = 260.8 N
Now apply the sum of force on X direction and Y direction
sum Force on x direction
Ay - N =0
therefore Ay = N = 260.8 N ( horizontal at base)
sum Force on Y direction
Ax - 471-811 = 0
Ax = 1282 N (vertical)
part b)
let u is coefficient static friction
sum of torque again
T =0
............= 471 x 7.5 x cos 60 ( clockwise)
.............= 811x 9 x cos60 (clockwise)
.............=N x 15 x sin60 (counterclockwise)
-471 x 7.5 x cos 60 -811x 9 x cos60+N x 15 x sin60 =0
-5415.75+12.99N =0
N = 5415.75/12.99 = 416.92 N
Ay = N = 416.92 N
Ay = u x Ax
u = Ay/Ax = 416.92 / 1282 = .3252
