Answer:
Explanation:
Given
Two block are connected by rope [tex]R_1[/tex]
[tex]R_2[/tex] rope is attached to block 2
suppose [tex]F_2[/tex] is a force applied to Rope [tex]R_2[/tex]
Applied force [tex]F_2[/tex]=Tension in Rope 2
[tex]F_2=(m_1+m_2)a---1[/tex]
where a=acceleration of system
Tension in rope [tex]R_1[/tex] is denoted by [tex]F_1[/tex]
[tex]F_1=m_1a---2[/tex]
divide 1 and 2 we get
[tex]\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}[/tex]
also [tex]m_1=2.11\cdot m_2[/tex]
[tex]\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}[/tex]
[tex]\frac{F_2}{F_1}=\frac{3.11}{2.11}[/tex]
[tex]\frac{F_1}{F_2}=\frac{2.11}{3.11}[/tex]
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The ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.
Tension force:
When an object is connected with a rope such that the force exerted on the rope due to the weight of any object towards the upward direction, then such force is known as tension force.
Let R be the rope by which two blocks are connected. And R' is the rope attached to block 2.
Also, F' is the force applied to the rope R'. Then,
Applied force F' = Tension in the rope R'
[tex]F'=T\\\\ \dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}[/tex]
here, [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of ropes such that [tex]m_{1} = 2.11 \times m_{2}[/tex].
Solving as,
[tex]\dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}\\\\ \dfrac{F'}{T}= \dfrac{(2.11m_{2}+m_{2})a} {2.11 \times m_{2}a}\\\\\\ \dfrac{F'}{T}= \dfrac{3.11}{2.11} [/tex]
or
T/F' = 2.11 / 3.11
Thus, we can conclude that the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.
Learn more about the tension force here:
https://brainly.com/question/2287912
