A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0-kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. What is the magnitude of the friction force that the floor exerts on the ladder?
A) 538 N
B) 706 N
C) 1290 N
D) 833 N
E) 601 N

Question

A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0-kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. What is the magnitude of the friction force that the floor exerts on the ladder? 
A) 538 N 
B) 706 N 
C) 1290 N 
D) 833 N 
E) 601 N

Quizplus · Accepted Answer

To solve this problem, we need to draw a free body diagram of the ladder and bucket system as shown below:

![ladder diagram](https://i.imgur.com/quSYJ4L.png)

First, we need to calculate the force exerted by the bucket on the ladder. Since the bucket is not moving vertically, the tension in the rope must be equal to the weight of the bucket:

Tension = Weight of bucket = mg = (80.0 kg)(9.81 m/s^2) = 784.8 N

Next, we can use the torque equation to find the force of friction between the floor and the ladder. Since the ladder is in equilibrium, the torque about any point must be zero. We can choose the point where the ladder contacts the floor as our pivot point.

The forces acting on the ladder are shown in the diagram above. The force of friction Ff and the normal force N both act at the point where the ladder contacts the floor. The weight of the ladder acts downward at its center of mass, which is 1.25 m from the pivot point. The weight of the bucket acts downward at the top rung of the ladder, which is 2.10 m from the pivot point. We can write the torque equation as:

(Ff)(2.50 m) - (10.0 kg)(9.81 m/s^2)(1.25 m) - (80.0 kg)(9.81 m/s^2)(2.10 m) + (784.8 N)(2.10 m) = 0

Solving for Ff, we get:

Ff = 537.8 N ≈ 538 N

Therefore, the magnitude of the friction force that the floor exerts on the ladder is 538 N, which is closest to choice A (538 N).

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