Question

A ladder of mass 30 kg and length 5m leans against a frictionless wall. The angle between the ladder and the wall is 37 degree. (a) what is the force between the ladder and the wall? (b) a 60 kg man now climbs up to the midpoint of the ladder, what is the force at the base of the ladder (magnitude and direction).? (c) what must be the coefficient of friction so that the mass can climb all the way to the top of the ladder without slipping?

Answer 1

a) At the base, where ladder is touching the floor, there are two force acting one in horizontal direction N3 (towards right)and one in vetical direction N1 (upward). This is because of the friction at the base.

While at the top, where ladder is touching the wall, on;y normal force (normal to both the ladder and the wall) is acting. This acts in the horizontal direction. This is denoted by N2 (acting towards left).

At the centre of the ladder gravitational force is acting downward.

since it is in equilibrium, net vertical force must be equal to zero. Therefore m*g=N1, g=10 m/s^2

and m=30 kg. Therefore N1=30*10=300 N

Net horizontal force is zero. therefore, N3=N2----------------------(1)

Net moment about the base or bottom point is = 0. Therefore m*g*(L/2)cos37=N2*L*sin37--------------(2), where is the lenght of ladder. By this N2=30*10*0.5*cot37=199.06 N (toeards left)

b) now at centre of the ladder total gravitational force = (30+60)*10=900 N (downward)

N1= 900N, (upward)

N3=N2 and............... by equation 1

by equation (2)......N2=(30+60)*10*0.5*cot37=597.17 N (towards left)

Therefore N3 =597.17 N (towards right), here N1 and N3 are the components of forces acting at the base of the ladder

c) friction coefficient = N3/N2 radian = 597.17/900 = 0.664 radian=38.04 dgreees