Question

Тело массой m1 =2 кг расположено на наклонной плоскости с углом наклона 30 градусов к горизонту. к телу прикреплена нить, перекинутая через блок, закрепленный на вершине наклонной плоскости. ко второму концу нити прикреплено тело массой m2=8 кг. если коэффициент трения между первым телом и плоскостью √3, то модуль силы натяжения нити составляет ...Н

Answer 1

В приложении дана аналогичная задача.
Надо только свои цифры подставить.

Answer 2

Calculation of the Tension Force in the String

To calculate the tension force in the string, we need to consider the forces acting on the two masses, m1 and m2, and the angle of inclination of the inclined plane.

Let's denote the tension force in the string as T. We can break down the forces acting on m1 and m2 as follows:

For m1: - The weight of m1 acts vertically downward and can be calculated as m1 * g, where g is the acceleration due to gravity. - The normal force exerted by the inclined plane on m1 acts perpendicular to the plane and can be calculated as N1 = m1 * g * cos(θ), where θ is the angle of inclination.

For m2: - The weight of m2 acts vertically downward and can be calculated as m2 * g. - The normal force exerted by the inclined plane on m2 acts perpendicular to the plane and can be calculated as N2 = m2 * g * cos(θ).

Since the system is in equilibrium, the sum of the forces in the x-direction and y-direction must be zero.

In the x-direction: - The component of the weight of m1 along the x-axis is m1 * g * sin(θ), and it is balanced by the tension force T. - The component of the weight of m2 along the x-axis is m2 * g * sin(θ), and it is also balanced by the tension force T.

Therefore, we can write the equation for the x-direction as: m1 * g * sin(θ) + m2 * g * sin(θ) = 2T

In the y-direction: - The normal force N1 cancels out the component of the weight of m1 along the y-axis, which is m1 * g * cos(θ). - The normal force N2 cancels out the component of the weight of m2 along the y-axis, which is m2 * g * cos(θ).

Therefore, we can write the equation for the y-direction as: N1 + N2 = m1 * g * cos(θ) + m2 * g * cos(θ) = (m1 + m2) * g * cos(θ)

Since the coefficient of friction between m1 and the inclined plane is given as √3, we can calculate the maximum frictional force as: f_max = μ * N1 = √3 * m1 * g * cos(θ)

In equilibrium, the tension force T must be equal to the maximum frictional force f_max to prevent m1 from sliding down the inclined plane. Therefore, we have: T = f_max = √3 * m1 * g * cos(θ)

Now, let's substitute the given values into the equation to calculate the tension force T.

Given: m1 = 2 kg m2 = 8 kg θ = 30 degrees μ = √3

Using the acceleration due to gravity g = 9.8 m/s^2, we can calculate the tension force T as follows:

T = √3 * m1 * g * cos(θ) T = √3 * 2 kg * 9.8 m/s^2 * cos(30 degrees) T = √3 * 2 * 9.8 * 0.866 T ≈ 30.1 N

Therefore, the magnitude of the tension force in the string is approximately 30.1 N.