Question

A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure.

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 = 4.85 m,

a weight of

W_L = 62.5 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 =	N
N2 =	Determine a good point about which to sum the torques and then see if you can apply the condition for rotational equilibrium about that point that will allow you to determine N2, the normal force exerted on the ladder by the wall. You may need to use your knowledge of trigonometry to determine the angle between the ladder and the floor. N
f1 =	N

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

N1 =	N
N2 =	N
f1 =	N

Answer 1

solution: Me mass of ladder m = mass of man. WL = weight of ladder Parta. Jaking {fx co and Efyro. we will gel mg + WL N₂ = f 1-0 and N,= fm+ Mg - taking moment about point o equat we can write. (mg twe) Nqx d = (n + M) g x ½ x coso Where was coso = J1 - ( coso = 0.64 => =) -) → N2 X 3.75 = 162.5+882.9)4.85 X0.64 N₂ = 391.26 N N2= = 391.26 N N = 945. 4 N Scanned with CamScanner

Parlb when person is 32 up the 1. ladder. so, eter Efx & Efy will be same only our moment equation will cliffer. . So, taking moment about poinro. we can write al WL. word - mg x L Coso tre x 3 L lose 2 - 3.75 Nx405 882 94.880.64+ 25x3 N27 3.75= 62:5* 485x0.647 882.9x 344-3570 => ^2 = 573,97N (N2=fit:S73-97.N] [N, = 945: 4. ) I take Please take out time like mis solulion. tout Scanned with CamScanner