A dynamometer is installed in the elevator, on which a body weighing 1 kg is suspended. What will the dynamometer show if the lift goes down with an acceleration of 5 m / s2.
m = 1 kg.
g = 10 m / s2.
a = 5 m / s2.
R – ?
The dynamometer will show the force P with which the load acts on the spring, that is, the weight of the load. Since according to Newton’s 3 law, the force P with which the load pulls the spring is equal to the force N with which the spring acts on the load: P = N
Two forces act on the body: gravity Ft directed vertically downward, force N of tension of the dynamometer spring directed vertically upward.
m * a = F + N – 2 Newton’s law in vector form.
For projections onto the vertical axis 2, Newton’s law will take the form: – m * a = – Fт + N.
N = Fт – m * a.
The force of gravity Ft is determined by the formula: Ft = m * g.
N = m * g – m * a = m * (g – a).
P = m * (g – a).
P = 1 kg * (10 m / s2 – 5 m / s2) = 5 N.
Answer: the dynamometer will show the body weight equal to P = 5 N.
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