Question

A ladder of length L = 2.4 m and mass m = 15 kg rests on a floor with coefficient of static friction μ_s = 0.56. Assume the wall is frictionless.

1)

What is the normal force the floor exerts on the ladder? N

2)

What is the minimum angle the ladder must make with the floor to not slip?

3)

A person with mass M = 62 kg now stands at the very top of the ladder.

What is the normal force the floor exerts on the ladder? N

4)

What is the minimum angle to keep the ladder from sliding?

Answer 1

1)

the ladder is in equlibrium

along vertical

Fnet = 0

N1 - W = 0

N1 = W

normal force N1 = m*g = 15*9.8 = 147 N

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2)

along horizontal

frictional force fs = us*N1

Fnet = 0

fs - N2 = 0

N2 = fs = us*N1

In equilibrium net torque = 0

N2*L*sintheta - W*L/2*costheta = 0

0.56*147*2.4*sintheta = 15*9.8*2.4/2*costheta

tantheta = 0.893

theta = 41.7 degrees

======================

3)

along vertical

Fnet = 0

N1 - Wladder - Wperson = 0

N1 = Wladder + Wperson

normal force N1 = (15*9.8) + (62*9.8) = 754.6 N

along horizontal

frictional force fs = us*N1

Fnet = 0

fs - N2 = 0

N2 = fs = us*N1

In equilibrium net torque = 0

N2*L*sintheta - Wladder*L/2*costheta - Wperson*L*costheta = 0

0.56*754.6*2.4*sintheta = 15*9.8*2.4/2*costheta + 62*9.8*2.4*costheta

tantheta = 0.6204

theta = 31.8 degrees