Question

Лестница прислонена к гладкой вертикальной стене. Коэффициент трения между ножками лестницы и полом равен 0,79. Центр тяжести лестницы находится на половине ее длины. Какой наибольший угол может образовать лестница со стеной? (ответ в градусах, округлить до целых)

Answer 1

Ответ:

57

Объяснение:

ооо

Answer 2

Problem Analysis

We are given a ladder leaning against a smooth vertical wall. The coefficient of friction between the ladder's legs and the floor is 0.79. The ladder's center of gravity is located at half its length. We need to determine the maximum angle the ladder can form with the wall.

Solution

To solve this problem, we can consider the forces acting on the ladder. There are two main forces: the weight of the ladder acting downward and the frictional force acting between the ladder's legs and the floor.

Let's denote the angle between the ladder and the wall as θ.

1. Weight of the ladder: The weight of the ladder acts vertically downward and can be represented by the force W.

2. Normal force: The normal force acts perpendicular to the floor and counteracts the weight of the ladder. It can be represented by the force N.

3. Frictional force: The frictional force acts parallel to the floor and opposes the motion of the ladder. It can be represented by the force F.

Since the ladder is in equilibrium, the sum of the forces in the horizontal and vertical directions must be zero.

In the horizontal direction: - There is no horizontal acceleration, so the sum of the horizontal forces is zero. - The only horizontal force is the frictional force F.

In the vertical direction: - There is no vertical acceleration, so the sum of the vertical forces is zero. - The vertical forces are the weight W and the normal force N.

Now, let's analyze the forces in more detail.

1. Weight of the ladder: The weight W can be split into two components: one parallel to the wall and one perpendicular to the wall. - The component parallel to the wall is Wsin(θ). - The component perpendicular to the wall is Wcos(θ).

2. Normal force: The normal force N acts perpendicular to the floor and counteracts the weight of the ladder. - The normal force is equal to the component of the weight perpendicular to the wall, which is Wcos(θ).

3. Frictional force: The frictional force F acts parallel to the floor and opposes the motion of the ladder. - The frictional force is given by F = μN, where μ is the coefficient of friction and N is the normal force. - Substituting the value of N from above, we have F = μWcos(θ).

Since the ladder is in equilibrium, the sum of the forces in the horizontal and vertical directions must be zero.

In the horizontal direction: - The only horizontal force is the frictional force F. - Therefore, F = 0.

In the vertical direction: - The vertical forces are the weight W and the normal force N. - Therefore, Wsin(θ) + N = 0.

Now, let's substitute the expressions for N and F into the equations above and solve for θ.

From the equation F = 0, we have: μWcos(θ) = 0

Since μ ≠ 0, we can divide both sides of the equation by μW: cos(θ) = 0

This equation implies that θ = 90°.

Therefore, the maximum angle the ladder can form with the wall is 90°.

Answer

The maximum angle the ladder can form with the wall is 90 degrees.

Note: The answer is rounded to the nearest whole number.

Source:

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