Determine the acceleration with which a body of mass m = 50 kg, suspended from the cable, rises, if the tension of the cable is T = 0.6kN, and the force of resistance to movement is Fc = 40N.
m = 50 kg.
g = 10 m / s2.
T = 0.6 kN = 600 N.
Fc = 40 N.
m * a = Fс + T + m * g – 2 Newton’s law in vector form ..
Where T is the pulling force of the cable, Fc is the resistance force, m * g is the force of gravity.
Since the force of gravity m * g and the force of resistance Fc are directed vertically downward, then m * a = – Fc + T – m * g – 2 Newton’s law for projections.
The acceleration of the body will be determined by the formula: a = T / m – Fc / m – g.
a = 600 N / 50 kg – 40 N / 50 kg – 10 m / s2 = 1.2 m / s2.
Answer: the body is lifted with an acceleration of a = 1.2 m / s2.
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