On the floor of an elevator moving vertically upward at a constant speed

On the floor of an elevator moving vertically upward at a constant speed V during time t, there is a load of mass m. What is the modulus of weight of this load?

Two forces act on the body in the elevator: the force of gravity Ft directed vertically downward, the force N with which the floor of the elevator acts on the body directed vertically upward.

Since the lift moves uniformly in a straight line, then, according to 1 Newton’s law, the action of forces on it is compensated: Ft = N.

The force of gravity Ft is determined by the formula: Ft = m * g, where m is the mass of the body, g is the acceleration of gravity.

N = m * g.

According to Newton’s 3 laws, the force N, with which the floor acts on the body, is equal in magnitude and opposite to the directed force P, with which the body presses on the floor of the lift, that is, the weight of the body: N = – P.

P = m * g.

Answer: body weight is equal to gravity P = m * g.