Question

A 25.5 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.22 . The angle between the horizontal and
the ladder is θ . The person wants to climb
up the ladder a distance of 0.98 m along the
ladder from the ladder’s foot.What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 0.98 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s
2.Answer in units of ◦. Don't round answer.

Answer 1

$$W=(25.5)(9.8)=249.9\ N$$

$$F_f=(0.22)(249.9)=54.978\ N$$

Counter clockwise torque:

$$249.9(0.98)\sin\theta$$

Clockwise torque:

$$54.978(0.98)\cos\theta$$

They need to be equal.

$$\tan{\theta}=\frac{54.978}{249.9}$$

$$\theta=12.4\degree$$