A dynamometer is installed in the elevator, on which a body weighing 1 kg is suspended. What will the dynamometer

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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