Question #125628
.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)?
1
Expert's answer
2020-07-10T10:31:15-0400


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=mAa.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(mC+mB)g=(mC+mB)a,mAa(mC+mB)g=(mC+mB)a, a=mB+mCmA+mB+mCg=6.13 m/s2.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=mC(ga)=36.8 N.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=at22=19.1 cm.d=\frac{at^2}{2}=19.1\text{ cm}.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS