An 8.6 m, 267 kg uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.53 , and the ladder makes a 32.7 ◦ angle with the ground. How far up the ladder can a 1068 kg person climb before the ladder begins to slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

If we want to find how far up the ladder can a person climb before the ladder begins to slip, the total force in the y-direction and the total torque must be zero, it is an equilibrium condition.

So, let's start whit y-direction forces.

[tex]\Sigma F_{y}=0[/tex]

[tex]N_{L}-W_{L}-W_{P}=0[/tex]

NL is the normal force
WL is the weight of the ladder
WP is the weight of the person

So, we can find the normal force:

[tex]N_{L}=m_{L}g+m_{P}g=g(m_{L}+m_{P})=9.8(267+1068)=13083 N[/tex]

We know that the friction force is the product of the coefficient of static friction times the normal force. Let's find it.

[tex]F_{f}=\mu N_{L}=0.53*13083=6934 N[/tex]

Referent to the x-direction forces, we can say that the total sum must be equal to zero too.

[tex]\Sigma F_{x}=0[/tex]

[tex] -N_{wall}+F_{f}=0[/tex]

[tex] N_{wall}=F_{f}=6934 N[/tex]

Where N(wall) is the normal force due to the wall. (negative direction)

Now, as we said before, the total torque must be zero, so we have:

[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=0[/tex] (1)

We will choose the pivot point at the bottom of the ladder, then we have:

Each force acting on the ladder must be perpendicular to the ladder, so we need to use the components.

[tex]\tau_{wall}=r_{wall}*N_{wall}sin(32.7)=8.6*6934*sin(32.7)=32215.8Nm[/tex] (positive due to the counterclockwise rotation)

[tex]\tau_{L}=r_{L}*W_{L}sin(57.3)=(8.6/2)*2616.6*sin(57.3)=-9468.2 Nm[/tex] (negative due to the clockwise rotation)

[tex]\tau_{P}=r_{P}*W_{P}sin(57.3)=r_{P}*10466.4*sin(57.3)=-r_{P}8807.6 Nm[/tex] (negative due to the clockwise rotation)

We use 57.3° because is the angle between the ladder weight vector and the ladder surface, we use the same angle to the person.

Now we can use (1) to find the distance

[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=32215.8-9468.2-r_{P}8807.6=0[/tex]

Therefore [tex]r_{P}=2.58 m[/tex]

I hope it helps you!