Question

A uniform ladder of length l rests against a smooth, vertical wall. If the coefficient of static friction is 0.50, and the ladder makes a 53⁰ angle with respect to the horizontal, how far along the length of the ladder can a 56.0-kg person climb before the ladder begins to slip? The ladder is 7.5 m in length and has a mass of 21 kg.

Answer 1

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:

$m1g+m2g=N$  

$N=755.37 N$

The friction force is:

$F_{force}=\mu N=0.5\cdot 755.37=377.68 N$

Now let's define the conservation of torque about the foot of the ladder:

$\tau_{ledder}+\tau_{person=\tau_{reaction}}$

$m_{1}\cdot g\cdot X \cdot cos(53)+m_{2}\cdot g\cdot 3.75 \cdot cos(53)=F_{force}7.5sin(53)$

Solving this equation for X, we have:

$X = \frac{377.68\cdot 7.5\cdot sin(53)-21\cdot 9.81\cdot 3.75 \cdot cos(53)}{56\cdot 9.81\cdot cos(53)}$

Finally, X = 5.44 m

Hope it helps!