Answer:
(c) F= 47 N
Explanation:
We apply Newton's first law for the equlibrio in x-y
ΣFx=0 Formula (1)
F-Ff=0
ΣFy=0 Formula (2)
Fn-W=0
Where:
F : The reaction force from the wall (N)
Ff : Friction force (N)
Fn : Normal force (N)
W : ladder Weight (N)
Known Data:
m= 15 Kg : mass ladder
μs: 0.32 : coefficient of static friction between the ladder and the floor
g= 9.8 m/s²
Problem development
We calculate the weight of the ladder (W):
W= m*g = 15 kg *9.8 m/s² = 147 N
look at the free body diagram of the ladder in the attached graphic
We apply formula (2) to calculate the normal force (Fn):
ΣFy=0 Formula (2)
Fn-147 N =0
Fn = 147 N
We calculate the force of static friction between the ladder and the floor (Ff):
Ff= μs*Fn = 0.32* 147 N = 47 N
We apply formula (1) to calculate The reaction force from the wall (F)
ΣFx=0
F- 47 N = 0
F= 47 N
