If the skater coasts only around 65 degrees of the circle, find the magnitude of his displacement vector.
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Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m
AC = distance of person on the ladder from the bottom end = x
W = weight of the ladder = 240 N
= weight of the person = 710 N
F = force by the wall on the ladder
N = normal force by ground on the ladder = ?
Using equilibrium of force along the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f =static frictional force on the ladder
Static frictional force is given as
f = μ N
f = (0.55) (950)
f = 522.5 N
Force equation along the horizontal direction is given as
F = f
F = 522.5 N
using equilibrium of torque about point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
