Question

Since the floor is rough, it exerts both a normal force N₁ and a frictional force f₁ on the ladder. However, since the wall is frictionless, it exerts only a normal force N₂ on the ladder. The ladder has a length of L = 4.1 m, a weight of W_L = 58.5 N, and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the following.

(a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.)

N1=

N2=

F1=

(b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.)

N1=

N2=

F1=

Answer 1

a)
Angle made by the ladder with horizontal, theta = sin^-1(d/L)

= sin^-1(3.75/4.1)

= 66.1 degrees

As the ladder is in equilibrium, net force and net torque acting on the ladder msut be zero.

Apply, Fnety = 0

N1 - WL - m*g = 0

N1 = WL + m*g

= 58.5 + 90*9.8

= 940.5 N <<<<<<<<-------------Answer

Apply, net torque about bottom = 0

N2*L*sin(66.1) - m*g*(L/2)*sin(90-66.1) - WL*(L/2)*sin(90-66.1) = 0

N2*sin(66.1) - 0.5*m*g*sin(23.9) - 0.5*WL*sin(23.9) = 0

N2 = (0.5*m*g*sin(23.9) + 0.5*WL*sin(23.9))/sin(66.1)

= (0.5*90*9.8*sin(23.9) + 0.5*58.5*sin(23.9))/(sin(66.1))

= 208 N <<<<<<<<-------------Answer

Apply, Fnetx = 0

F1 - N2 = 0

F1 = N2

= 208 N <<<<<<<<-------------Answer

b) N1 = 940.5 N <<<<<<<<-------------Answer

Apply, net torque about bottom = 0

N2*L*sin(66.1) - m*g*(3*L/4)*sin(90-66.1) - WL*(L/2)*sin(90-66.1) = 0

N2*sin(66.1) - 0.75*m*g*sin(23.9) - 0.5*WL*sin(23.9) = 0

N2 = (0.75*m*g*sin(23.9) + 0.5*WL*sin(23.9))/sin(66.1)

= (0.75*90*9.8*sin(23.9) + 0.5*58.5*sin(23.9))/(sin(66.1))

= 306 N <<<<<<<<-------------Answer

Apply, Fnetx = 0

F1 - N2 = 0

F1 = N2

= 306 N <<<<<<<<-------------Answer