Answer to Question #212132 in Mechanics | Relativity for Cadmium John

Question

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.

Answer 1

Gives

$$m_1=6,m_2=4kg,M=9kg,\theta=30°$$

Find acceleration (a)

$$(m_2+m_1)a-Ma =Mg-m_2gsin\theta -m_1gsin\theta$$

$a=\frac{Mg-(m_2+m_1)gsin\theta}{m_1+m_2-M}$

Put value

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

$a=\frac{9g-(10)gsin30°}{1}=4g$

$g=9.8m/sec^2$

Tension

$$2T=(m_1+m_2)gcos30°=10gcos30°=84N$$

T=42N

Option (d) is correct option