Answer to Question #125628 in Mechanics | Relativity for Mwansa Kunda

Question

.Three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The three masses are mA=30Kg, mB=40Kg and mC=10Kg . Determine the tension in the cord connecting B and C when the assembly was released. ii. How far does A move in the first 0.25 s (assuming it does not reach the pulley)?

Answer 1

The tension in the cord connecting B and C when the assembly was released is the weight of C. To find the weight, we must find the acceleration of the system.

Direct x-axis to the right, y-axis upward. Write Newton's second law for box A:

$$T=m_Aa.$$

We can consider the hanging boxes B and C as one body since the cord is non-inextensible. Therefore, Newton's second law for them looks like

$$T-(m_C+m_B)g=-(m_C+m_B)a,\\ m_Aa-(m_C+m_B)g=-(m_C+m_B)a,\\\space\\ a=\frac{m_B+m_C}{m_A+m_B+m_C}g=6.13\text{ m/s}^2.$$

Therefore, the tension between B and C is

$$T=W=m_C(g-a)=36.8\text{ N}.$$

To answer how far A goes in the first 0.25 s, all we need is to use this simple equation:

$$d=\frac{at^2}{2}=19.1\text{ cm}.$$