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
By Newton's second law, the net force on the object acting parallel to the surface is
∑ F = F[applied] - F[friction] = (8k g) (3 m/s²)
If F[applied] = 30 N, then
30 N - F[friction] = 24 N ⇒ F[friction] = 6 N
so the answer is B.
may i ask your answer soloutions... and i believe that the correct answer is mass
