A 12mlong ladder leans against a frictionless wall as shown

A 1.2-m-long ladder leans against a frictionless wall as shown in the figure below. The coefficient of static friction between the ladder and the floor is 0.44. What is the minimum angle the ladder can make with the floor without slipping?

_____ degrees

Center of mass 7, Tant O about 0 about this point. Weight acts at the center of mass. Static friction prevents slipping. di

Solution

Because the ladder is stationary (in equilibrium), the horizontal and vertical forces must sum to zero.

Also the ladder has no angular acceleration, the sum of the torques must also be zero.

Given: center of gravity is right in the middle of the ladder.

L = 1.2 m

Coefficient of static friction: = 0.44

Vertical forces on ladder are:

mg (down) and Fn (normal force from floor, pushing up)

Fn = mg

Horizontal forces on ladder are:

Fw (force of wall, pushing to left) and Ff (force of friction from floor, pushing to right)

Ff = uN = umg

Fw = umg

Wall torque = Fw x d = Fw*L*sinA

Weight torque = mg x d = mg*(L/2)*cosA

Torque balance

Fw*L*sinA = mg*(L/2)*cosA

umg*L*sinA = mg*(L/2)*cosA

tan A = 1/2u

A = arctan(1/2u) = arctan(1/0.88) = 48.65 deg