Question

Тело массой 2кг скользит с ускорением 1,5 м/с2 по горизонтальной поверхности стола под действием груза массой 0,5 кг, прикрепленного к концу шнура, привязанного к телу и перекинутого через неподвижный блок, который укрепоен на краю стола. Определите силу трения между телом и поверхностью стола.

Answer 1

Тело:

m1a=T-Fтр

0=N-m1g

N=m1g

Fтр=µm1g

m1a=T-µm1g

Груз:

m2a=m2g-T

T=m2g-m2a

 

m1a=m2g-m2a-µm1g

µ=(m2g-m2a-m1a)/m1g=(0.5*10-0.5*1.5-2*1.5)/2*10=0.0625

Answer 2

1-ое тело: R=m1a1
                 m1g +N+Fтр+T1=m1a1 (всё это с векторами сверху)
                 OX: T1-Fтр=m1a1 (Fтр можно расписать как мю * N )
                 OY: N=m1g
                  (подставляем N в OX)
                  T1-мю*m1g=m1a1
2-ое тело: R=m2a2
                 T2+m2g=m2a2 (всё это с векторами)

                  OY: m2g - T2=m2a2
                   m2g - T2=m2a2
т.к нить невесома и нерастяжима, а тела связаны то - a1=a2=a и T1=T2=T
пишешь систему уравнений из двух этих уравнений(просто здесь нормально фигурную скобку не изобразить) - T -мю*m1g=m1a и -T + m2g=m2a и складываешь их!
у тебя получается m2g -мю*m1g=a*(m1+m2)
и выражаешь отсюда мю : -мю=(a*(m1+m2)-m2g)/m1g= -0.0625 (получается с минусом, так как и выражал с минусом, то есть минус уходит)
и находишь N=m1g=20
и Fтр =мю*N= 0.0625 * 20 = 1.25
Fтр=1.25 

Answer 3

Calculation of Frictional Force

To determine the frictional force between the body and the surface of the table, we need to consider the forces acting on the body.

The given information is as follows: - Mass of the body (m1) = 2 kg - Acceleration of the body (a) = 1.5 m/s^2 - Mass of the hanging load (m2) = 0.5 kg

Let's analyze the forces acting on the body:

1. Force due to gravity (Weight): The weight of the body is given by the formula: Weight = mass x acceleration due to gravity. Since the body is on a horizontal surface, the vertical component of the weight is balanced by the normal force from the table. Therefore, we only consider the horizontal component of the weight, which is equal to m1 x g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Force due to the hanging load: The hanging load exerts a force on the body in the horizontal direction. This force is equal to m2 x a, where a is the acceleration of the body.

3. Frictional force: The frictional force between the body and the surface of the table opposes the motion of the body. It can be calculated using the formula: Frictional force = coefficient of friction x normal force.

To find the normal force, we need to consider the forces acting in the vertical direction:

4. Normal force: The normal force is equal to the vertical component of the weight of the body, which is m1 x g.

Now, let's calculate the frictional force:

- Weight of the body = m1 x g = 2 kg x 9.8 m/s^2 = 19.6 N- Force due to the hanging load = m2 x a = 0.5 kg x 1.5 m/s^2 = 0.75 N- Normal force = m1 x g = 19.6 N- Frictional force = coefficient of friction x normal force

Unfortunately, the coefficient of friction is not provided in the given information. Without this value, we cannot calculate the exact frictional force. The coefficient of friction depends on the nature of the surfaces in contact (e.g., roughness, material), and it is necessary to know this value to calculate the frictional force accurately.

Please provide the coefficient of friction between the body and the surface of the table so that we can calculate the frictional force.