Question

Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block 2. A force is applied to R2 and the blocks accelerate forward. Find the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 respectively. Here m1 = 2.11m2.

Answer 1

Answer:

Explanation:

Given

Two block are connected by rope $R_1$

$R_2$ rope is attached to block 2

suppose $F_2$ is a force applied to Rope $R_2$

Applied force $F_2$=Tension in Rope 2

$F_2=(m_1+m_2)a---1$

where a=acceleration of system

Tension in rope $R_1$ is denoted by $F_1$

$F_1=m_1a---2$

divide 1 and 2 we get

$\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}$

also $m_1=2.11\cdot m_2$

$\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}$

$\frac{F_2}{F_1}=\frac{3.11}{2.11}$

$\frac{F_1}{F_2}=\frac{2.11}{3.11}$