Question

55 kg student stands on a bathroom in 3000 N elevator cabin that is supported by a cable as the cabin starts moving the scale read the mass of student as 45 kg. What is the acceleration vektor of cabin? What is tension in cable ?

Answer 1

55 kg student stands on a bathroom scale in 3000 N elevator cabin that is supported by a cable as the cabin starts moving the scale read the mass of student as 45 kg.

a) What is the acceleration vector of cabin?

b) What is tension in cable?

Solution.

$$\begin{array}{l} m _ {0} = 5 5 k g, G = 3 0 0 0 N, m = 4 5 k g; \\ a -? F _ {t} -? \\ \end{array}$$

a)

Newton's second law for the student in vector form:

$$m _ {0} \vec {a} = m _ {0} \vec {g} + \vec {N};$$

$m_0$ - the mass of the student.

Projection on Y:

$$m _ {0} a = m _ {0} g - N;$$

The scale read the mass of student $m$ .

$$N = m g;$$ $$m _ {0} a = m _ {0} g - m g;$$ $$a = g - \frac {m}{m _ {0}} g;$$ $$a = \left(1 - \frac {m}{m _ {0}}\right) g.$$ $$a = \left(1 - \frac {4 5}{5 5}\right) 9. 8 = 1. 7 8 \left(\frac {m}{s ^ {2}}\right).$$

b)

The weight of the elevator cabin at rest:

$$G = M _ {0} g.$$

The mass of the cabin:

$$M _ {0} = \frac {G}{g}.$$

Newton's second law for the student and the elevator cabin in vector form:

$$(M _ {0} + m _ {0}) \vec {a} = (M _ {0} + m _ {0}) \vec {g} + \overrightarrow {F _ {t}};$$

Projection on Y:

$$(M_0 + m_0)a = (M_0 + m_0)g - F_t;$$ $$F_t = (M_0 + m_0)(g - a);$$ $$F_t = \left(\frac{G}{g} + m_0\right)(g - a);$$ $$F_t = \left(\frac{3000}{9.8} + 55\right)(9.8 - 1.78) = 2896(N).$$

Answer: $a = 1.78\frac{m}{s^2}; F_t = 2896N.$