Answer :
Answer:
[tex]T=mg/2cos \theta[/tex]
Step-by-step explanation:
#Given that each rope is tied to the cylinder top and makes the same [tex]\angle \theta[/tex] with the vertical,:
Let tension on the rope be [tex]T[/tex]
#The vertical component of each T is, [tex]Tcos \theta[/tex]
Total vertical component is [tex]2Tcos \theta[/tex]
We know that
[tex]F=ma\\F=2Tcos \theta\\ma=mg\\\\\therefore T=mg/2cos \theta[/tex] #where [tex]g[/tex] is gravitational acceleration.
Hence the tension of each rope is [tex]T=mg/2cos \theta[/tex]
Answer:
[tex]\boxed {\displaystyle T=\frac{m.g}{2cos\theta}}[/tex]
Step-by-step explanation:
Net Force
The net force exerted on an object is the sum of all vector forces applied to it. If no acceleration is achieved, the net force is zero.
We have three forces applied to the cylindrical weight: Its own weight and the two components of the tension of each rope. Knowing they form an angle [tex]\theta[/tex] with the vertical, then each component is given by
[tex]T_y=Tcos\theta[/tex]
The sum of all forces in the vertical direction is
[tex]F_y=2T_y-m.g[/tex]
The object is not accelerated, thus
[tex]2T_y-m.g=0[/tex]
[tex]2Tcos\theta=m.g[/tex]
Solving for T
[tex]\boxed {\displaystyle T=\frac{m.g}{2cos\theta}}[/tex]