The load is lifted up on the cord with an acceleration of 6 m / s2, then lowered with the same
The load is lifted up on the cord with an acceleration of 6 m / s2, then lowered with the same acceleration, determine the ratio of the cord elongations
a = 6 m / s2.
g = 9.8 m / s2.
x1 / x2 -?
Let’s write Newton’s 2 law for a load: m * a = N + m * g, where m is the mass of the load, a is the acceleration of the load, N is the elastic force of the cord, m * g is the force of gravity.
According to Hooke’s law: N = k * x, where k is the stiffness of the cord, x is the elongation of the cord.
Let’s write 2 Newton’s law for projections on the vertical axis OU directed upwards.
1) For movement with acceleration a directed upwards.
m * a = N – m * g.
m * a = k * x1 – m * g.
k * x1 = m * (a + g).
x1 = m * (a + g) / k.
2) For movement with acceleration a directed downward.
– m * a = N – m * g.
– m * a = k * x2 – m * g.
k * x2 = m * (g – a).
x2 = m * (g – a) / k.
x1 / x2 = m * (a + g) * k / k * m * (g – a) = (a + g) / (g – a).
x1 / x2 = (9.8 m / s2 + 6 m / s2) / ((9.8 m / s2 – 6 m / s2) = 4.2.
Answer: x1 / x2 = 4.2.