Question #212132

A system comprising blocks, a light frictionless pulley, a frictionless incline at 30°, and a rope connecting 6.0-kg and 4.0-kg blocks on the incline with the 6.0-kg block being lower than the 4.0-kg block. The rope extends to a 9.0-kg block. The 9.0-kg block hangs over the pulley on the side of the incline and accelerates downward when the system is released from rest. The tension in the rope connecting the 6.0-kg block and the 4.0-kg block is closest to


  • A. 30 N.

 

  •  B. 39 N.

 

  •  C.  36 N.


  •  D. 42 N.
1
Expert's answer
2021-07-01T04:24:01-0400

Gives


m1=6,m2=4kg,M=9kg,θ=30°m_1=6,m_2=4kg,M=9kg,\theta=30°


Find acceleration (a)


(m2+m1)aMa=Mgm2gsinθm1gsinθ(m_2+m_1)a-Ma =Mg-m_2gsin\theta -m_1gsin\theta

a=Mg(m2+m1)gsinθm1+m2Ma=\frac{Mg-(m_2+m_1)gsin\theta}{m_1+m_2-M}

Put value

a=9g(6+4)gsinθ6+49a=\frac{9g-(6+4)gsin\theta}{6+4-9}

a=9g(10)gsin30°1=4ga=\frac{9g-(10)gsin30°}{1}=4g

g=9.8m/sec2g=9.8m/sec^2

Tension


2T=(m1+m2)gcos30°=10gcos30°=84N2T=(m_1+m_2)gcos30°=10gcos30°=84N

T=42N


Option (d) is correct option


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