Question

Груз массой m=300 кг начинает подниматься канатом равнмерно ускоренно и проходит 5,4 м за 3 с. Канат проходит через блок С к лебедке. Определить реакции опоры рамы ABC. Начальная скорость груза ровнялась нулю. от книге першино

Answer 1

Вес груза P= T.
Ускорение найдено: a = 2·s/t² = 1 м/с² 
m·a = m·g − T 
P = T = m·(g − a) = 120·(9,8 − 1) = 120·8,8 = 1,06·10³ Н = 1,06 кН

Answer 2

Problem Statement

A load with a mass of m = 300 kg is being lifted by a cable with uniform acceleration. The load travels a distance of 5.4 m in 3 s. The cable passes through a block C attached to a winch. We need to determine the reactions at the support points of the frame ABC. The initial velocity of the load is zero.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. We'll start by finding the acceleration of the load.

The equation for displacement in uniformly accelerated motion is given by:

s = ut + (1/2)at^2

where: - s is the displacement - u is the initial velocity - a is the acceleration - t is the time

In this case, the displacement (s) is 5.4 m, the initial velocity (u) is 0 m/s, and the time (t) is 3 s. We can rearrange the equation to solve for acceleration (a):

a = (2s) / t^2

Substituting the given values, we find:

a = (2 * 5.4) / (3^2) = 3.6 / 9 = 0.4 m/s^2

Now that we have the acceleration, we can find the tension in the cable using Newton's second law:

T - mg = ma

where: - T is the tension in the cable - m is the mass of the load - g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the values, we have:

T - (300 * 9.8) = 300 * 0.4

Simplifying the equation, we find:

T - 2940 = 120

T = 3060 N

The tension in the cable is 3060 N.

To determine the reactions at the support points of the frame ABC, we need to consider the forces acting on the load. There are two support points, so we'll consider the vertical forces at each point separately.

At point A, the vertical force is the tension in the cable (T) acting upwards. Therefore, the reaction at point A is equal to the tension:

Reaction at A = T = 3060 N

At point C, the vertical force is the weight of the load (mg) acting downwards. Therefore, the reaction at point C is equal to the weight of the load:

Reaction at C = mg = 300 * 9.8 = 2940 N

The reactions at the support points of the frame ABC are 3060 N at point A and 2940 N at point C.

Please let me know if you need any further assistance!