A 3.53 m long uniform ladder weighing 195 N is leaning against a smooth wall at

Question

A 3.53 m long uniform ladder weighing 195 N is leaning against a smooth wall at an angle theta = 62.7 degree. A 77.1 kg person climbs a distance x = 2.26 m up the ladder. (The distance x is measured along the ladder as shown in the drawing.) What are the components of the force acting on the bottom of the ladder, at point A, if the ladder is in equilibrium? (Use "+" if the force component is to the right or up, "-"if it is to the left or down, and g 9.80 m/s^2.) A_X A_Y = What is the minimum coefficient of friction between the ladder and the ground to just keep the ladder from slipping when the 77.1 kg person is 2.26 m up the ladder? Mu =

Explanation / Answer

Weight of person = 77.1 * 9.8 = 755.58 N
Total weight = 195 + 755.58 = 950.58 N
Since there is no friction between the top of the ladder and the wall, this is the force that causes the friction.

Ff = * 950.58

The other horizontal force is the force that the wall exerts on the top of the ladder. The friction force is equal to this force.

This is a torque problem. Let the pivot point be at the bottom of the ladder. The weights of the ladder and person will produce clockwise torque. The horizontal force that the wall exerts on the top of the ladder will produce counter clockwise torque. The weight of the ladder is at its center. This is 1.765 meters from the bottom of the ladder.

Clockwise torque = 195 * 1.765 * cos 62.7 + 755.58 * 2.26 * cos 62.7 = 941.05 Nm

Vertical distance from point A to point B = 3.53 * sin 62.7
This is approximately 3.14 meters.

Counter clockwise torque = F * 3.53 * sin 62.7

F * 3.53 * sin 62.7 = 941.05
F = 300N This is the friction force.

300 = * 950.58

= 0.3156

Ax = 300N

Ay = 950.58N

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