Answer:
X = 5.44 m
Explanation:
First we can calculate the normal force acting from the floor to the ladder.
W₁+W₂ = N
W1 is the weigh of the ladder
W2 is the weigh of the person
So we have:
[tex] m1g+m2g=N[/tex]
[tex]N=755.37 N[/tex]
The friction force is:
[tex]F_{force}=\mu N=0.5\cdot 755.37=377.68 N[/tex]
Now let's define the conservation of torque about the foot of the ladder:
[tex]\tau_{ledder}+\tau_{person=\tau_{reaction}}[/tex]
[tex]m_{1}\cdot g\cdot X \cdot cos(53)+m_{2}\cdot g\cdot 3.75 \cdot cos(53)=F_{force}7.5sin(53)[/tex]
Solving this equation for X, we have:
[tex]X = \frac{377.68\cdot 7.5\cdot sin(53)-21\cdot 9.81\cdot 3.75 \cdot cos(53)}{56\cdot 9.81\cdot cos(53)}[/tex]
Finally, X = 5.44 m
Hope it helps!
