Question

2.Брусок массы m=12 кг зажатый между двумя колодками скользит вниз (см. рис.). Коэффициент трения между колодками и грузом f =0,3. Если каждая из прижимающих сил F=50 Н, то какая требуется дополнительная вертикальная сила Fдоп, чтобы удержать груз в покое?

Answer 1

Ответ:

90 Н

Объяснение:

Дано:

m=12 кг

μ=0,3

F=50 Н

g=10 м/c

__________

Fдоп-?

Решение

На брусок оказывают действие 6 сил: сила тяжести mg, две силы давления от поверхностей колодок F, две силы трения между бруском и поверхностями колодок, а также сила Fдоп, которая позволяет удердживать груз в состоянии покоя (рисунок).

Согласно второму закону Ньютона в проекции на вертикальную ось и учитывая, что на брусок действует сила трения с двух сторон, имеем:

Fдоп=m·g–2Fтр

Так как Fтр=μF, то Fдоп=m·g–2μF=12·10–2·0,3·50 = 120 – 30 = 90 Н

Answer 2

Problem Analysis

We have a block of mass m = 12 kg that is wedged between two blocks and is sliding downwards. The coefficient of friction between the blocks and the mass is f = 0.3. Each of the clamping forces F is 50 N. We need to determine the additional vertical force Fдоп required to keep the mass in equilibrium.

Solution

To solve this problem, we need to consider the forces acting on the block. There are three forces at play: the weight of the block (mg), the normal force (N), and the frictional force (fN).

1. Weight of the block (mg): The weight of the block is given by the formula mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Normal force (N): The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the clamping force F.

3. Frictional force (fN): The frictional force is given by the formula fN, where f is the coefficient of friction and N is the normal force.

Since the block is in equilibrium, the sum of the forces in the vertical direction must be zero. This means that the normal force N must be equal to the weight of the block (mg) plus the additional vertical force Fдоп.

Mathematically, we can express this as:

N = mg + Fдоп

To find the value of Fдоп, we can substitute the given values into the equation and solve for Fдоп.

Calculation

Given: - Mass of the block (m) = 12 kg - Coefficient of friction (f) = 0.3 - Clamping force (F) = 50 N

Using the equation N = mg + Fдоп, we can substitute the values:

50 N = (12 kg)(9.8 m/s^2) + Fдоп

Simplifying the equation:

50 N = 117.6 N + Fдоп

Subtracting 117.6 N from both sides:

Fдоп = 50 N - 117.6 N

Fдоп = -67.6 N

Answer

To keep the block in equilibrium, an additional vertical force of -67.6 N is required. The negative sign indicates that the force must be applied in the opposite direction to counteract the downward motion of the block.

Please note that the negative sign indicates the direction of the force and does not represent a negative value in this context.

Let me know if you need any further assistance!