Question

If the coefficient of static friction is 0.40, and the same ladder makes a 51° angle with respect to the horizontal, how far along the length of the ladder can a 57.0-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

To solve this problem, we apply the concepts related to the sum of forces and balance in a diagram that will be attached, in order to identify the behavior, direction and sense of the forces. The objective is to find an expression that is in terms of the mass, the angle, the coefficient of friction and the length that allows us to identify when the ladder begins to slip. For equilibrium of the ladder we have,

$\sum F_x = 0$

$\sum F_y = 0$

$\sum M_o = 0$

Now we have that

$f_1 = N_2$

$N_1 = mg$

And for equilibrium of the two forces we have finally

$mgdcos\theta = N_2lsin\theta$

Rearranging to find the distance,

$d = \frac{N_2}{mg}ltan\theta$

$d = \frac{f_1}{mg}ltan\theta$

So if we have that the frictional force is equivalent to

$f_1 = \mu N_1$

$f_1 = \mu mg$

$f_1 = (0.4)(57*9.8)$

$f_1 = 223.44N$

With this value we have that

$d = \frac{(0.4)(57)(9.8)}{57*9.8}(7.5) tan(60\°)$

$d = 5.19m$

Therefore can go around to 5.19m before the ladder begins to slip.