Question

С наклонной плоскости, составляющей угол 60° с горизонтом, соскальзывает треугольный клин, верхняя грань которого горизонтальна. Коэффициент трения между клином и наклонной плоскостью равен 1. На клине покоится брусок массой 4 кг. Найти силу трения, действующую на брусок.

Answer 1

Ускорение клина на наклонной плоскости
a=g*(sin(alpha)-мю*cos(alpha)
система отсчета  "на клине" - ускоренная,
нужно учитывать силу инерции, нправленную противоположно ускорению (под углом alpha к горизонту)
тело на клине неподвижно
m*0=mg+N+Fтр+Fинерции (вектрная запись)
в проекции на горизонталь
0=0+0+Fтр-m*a*cos(alpha) (здесь и далее скалярная запись)
Fтр = m*a*cos(alpha)=m*(g*(sin(alpha)-мю*cos(alpha))*cos(alpha)=4*(10*(корень(3)/2-1*1/2)*(1/2) H=10*(корень(3)-1) H= 7,320508076 H ~ 7,3 Н

Answer 2

Problem Analysis

We are given a triangular wedge sliding down an inclined plane with an angle of 60° with the horizontal. The upper surface of the wedge is horizontal. The coefficient of friction between the wedge and the inclined plane is 1. A block with a mass of 4 kg is resting on the wedge. We need to find the frictional force acting on the block.

Solution

To solve this problem, we can break it down into two components: the forces acting parallel to the inclined plane and the forces acting perpendicular to the inclined plane.

Let's consider the forces acting parallel to the inclined plane first. The only force in this direction is the component of the weight of the block that is parallel to the inclined plane. This force can be calculated using the formula:

Force parallel = Weight of the block * sin(θ)

where θ is the angle of the inclined plane with the horizontal.

Next, let's consider the forces acting perpendicular to the inclined plane. The forces in this direction are the normal force and the frictional force. The normal force is equal to the component of the weight of the block that is perpendicular to the inclined plane. This force can be calculated using the formula:

Normal force = Weight of the block * cos(θ)

The frictional force can be calculated using the formula:

Frictional force = Coefficient of friction * Normal force

Substituting the value of the normal force, we get:

Frictional force = Coefficient of friction * (Weight of the block * cos(θ))

Now, let's substitute the given values into the formulas and calculate the frictional force.

Given: - Angle of the inclined plane (θ) = 60° - Coefficient of friction = 1 - Mass of the block = 4 kg

Weight of the block = mass * acceleration due to gravity Weight of the block = 4 kg * 9.8 m/s^2 = 39.2 N

Force parallel = Weight of the block * sin(θ) Force parallel = 39.2 N * sin(60°) = 33.94 N

Normal force = Weight of the block * cos(θ) Normal force = 39.2 N * cos(60°) = 19.6 N

Frictional force = Coefficient of friction * (Weight of the block * cos(θ)) Frictional force = 1 * (39.2 N * cos(60°)) = 19.6 N

Therefore, the frictional force acting on the block is 19.6 N.

Answer

The frictional force acting on the block is 19.6 N.