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**

**-------------------------**

**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**

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