A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.30. What minimum angle can the ladder make with the floor before it slips?
A) 22 $\degree$
B) 51 $\degree$
C) 18 $\degree$
D) 59 $\degree$

Question

Quizplus · Accepted Answer

The minimum angle is when the friction force is at its maximum value, which is given by:$$f_s = \mu_s n$$where:* $f_s$ is the friction force* $\mu_s$ is the coefficient of static friction* $n$ is the normal forceThe normal force is equal to the weight of the ladder, which is:$$n = mg$$where:* $m$ is the mass of the ladder* $g$ is the acceleration due to gravitySubstituting these expressions into the equation for the friction force, we get:$$f_s = \mu_s mg$$The ladder will slip when the friction force is less than the force of gravity pulling the ladder down, which is:$$f_s < mg \sin heta$$where:* $ heta$ is the angle the ladder makes with the floorSubstituting the expression for the friction force into this inequality, we get:$$\mu_s mg < mg \sin heta$$Dividing both sides by $mg$, we get:$$\mu_s < \sin heta$$Taking the inverse sine of both sides, we get:$$ heta > \sin^{-1} \mu_s$$Substituting the given value of $\mu_s = 0.30$, we get:$$ heta > \sin^{-1} 0.30 = 17.5^\circ$$Therefore, the minimum angle the ladder can make with the floor before it slips is 17.5 degrees.

A 100-N Uniform Ladder, 8 °\degree° B) 51 °\degree° C) 18 °\degree° D) 59

A 100-N Uniform Ladder, 8 $\degree$ B) 51 $\degree$ C) 18 $\degree$ D) 59