Question

Мальчик тянет санки с грузом общей массой 40 кг по горизонтальной поверхности, прилагая силу, направленную под углом 30° к горизонту. Определить величину этой силы, если коэффициент трения 0,3​

Answer 1

Ответ:

Вот решение

Объяснение:

Данные: α (угол между вектором перемещения санок и вектором силы) = 30º; m (масса санок) = 40 кг; равномерное движение; μ (коэффициент трения санок о снег) = 0,1.

Постоянные: g (ускорение свободного падения) ≈ 10 м/с2.

Для определения модуля действующей силы воспользуемся равенством: m * a = 0 = F * cosα - Fтр = F * cosα - μ * (m * g - F * sinα).

F * cosα = μ * (m * g - F * sinα).

F * cos30º = 0,1 * (40 * 10 - F * sin60º).

F * 0,866 = 40 - 0,1F * 0,5.

0,916F = 40.

F = 40 / 0,916 = 43,67 Н.

Answer 2

I found some information that might help you with your question. Here is a detailed answer:

To solve this problem, we need to draw a free-body diagram of the forces acting on the sled and the load. The forces are:

- The weight of the sled and the load, which is equal to mg, where m is the mass and g is the gravitational acceleration. This force acts vertically downward from the center of mass. - The normal force of the ground on the sled and the load, which is equal to N. This force acts vertically upward from the point of contact with the ground. - The friction force of the ground on the sled and the load, which is equal to μN, where μ is the coefficient of friction. This force acts horizontally opposite to the direction of motion. - The pulling force of the boy on the sled and the load, which is equal to F. This force acts at an angle of 30° above the horizontal.

We can apply Newton's second law of motion to find the net force and acceleration of the system. Since the system is moving horizontally, we only need to consider the horizontal component of the forces. We can use trigonometry to find that:

- The horizontal component of F is F cos 30°. - The vertical component of F is F sin 30°.

The net horizontal force is:

F cos 30° - μN = ma

where a is the horizontal acceleration.

The net vertical force is:

N - mg - F sin 30° = 0

since there is no vertical acceleration.

We can solve for N from the second equation and substitute it into the first equation to get:

(F cos 30° - μ(mg + F sin 30°)) / m = a

We can rearrange this equation to solve for F:

F = (ma + μmg) / (cos 30° - μ sin 30°)

Now we can plug in the given values to find F:

- m = 40 kg - g = 9.8 m/s^2 - μ = 0.3 - a = 0

We get:

F = (0 + 0.3 × 40 × 9.8) / (cos 30° - 0.3 × sin 30°)

F ≈ 147 N

Therefore, the boy needs to apply a force of about 147 N at an angle of 30° above the horizontal to pull the sled and the load at a constant speed.

I hope this answer helps you understand how to solve this problem. You can also check out this website for a similar example with a diagram and explanation.