A weightless inextensible cord is thrown through the block, suspended from the dynamometer, at the ends of which weights of mass m1 = 2 kg and m2 = 8 kg are fixed. What is the dynamometer reading when the goods are moving? Disregard the mass of the block and the friction forces.
Acceleration of the system, we write as -a, then we need a formula for the solution:
(m1 + m2) * a = m2 * g – m1 * g;
Let’s transform the formula:
a = (m2 – m1) * g / (m1 + m2);
Let the tension of the thread be T, then the movement of the load is m2.
Let’s write the formula:
m2 * a = m2 * g – T;
T = m2 * (g – a) = m2 * g * (1 – (m2 – m1) / (m1 + m2)) = 2 * m1 * m2 * g / (m1 + m2)).
Find the dynamometer reading F:
F = 2 * T = 4 * m1 * m2 * g / (m1 + m2);
F = 4 * 2 * 8 * 10 / (2 + 8) H = 64 H /.
Answer: F = 64 H
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