First establish the summation of the forces acting int the
ladder
Forces in the x direction Fx = 0 = force of friction (Ff) –
normal force in the wall(n2)
Forces in the y direction Fy =0 = normal force in floor (n1)
– (12*9.81) –( 60*9.81)
So n1 = 706.32 N
Since Ff = un1 = 0.28*706.32 = 197,77 N = n2
Torque balance along the bottom of the ladder = 0 = n2(4 m) –
(12*9.81*2.5 m) – (60*9.81 *x m)
X = 0.844 m
5/ 3 = h/ 0.844
H = 1.4 m can the 60 kg person climb berfore the ladder will
slip
a. The restoring force in the spring has magnitude
F[spring] = k (0.79 m)
which counters the weight of the mass,
F[weight] = (0.46 kg) g = 4.508 N
so that by Newton's second law,
F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m
b. Using the same equation as before, we now have
F[weight] = (0.75 kg) g = 7.35 N
so that
(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m
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