Question

3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following correctly describes the tension in the three rope segments?
a. T1< T2 < T3
b. T1< T2 = T3
c. T1 = T2 = T3
d. T1> T2 > T3
please help.show how and which?
see attachment for more detail.​

Answer 1

Answer:

The correct option is d: T₁ > T₂ > T₃.  

Explanation:

Let's evaluate each tension.

Case T₃.

$T_{3} - W_{3} = 0$

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.          

$T_{3} = W_{3}$          

Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:                

$T_{3} = W_{3} = mg$  (1)                                                                                                     Case T₂.

$T_{2} - (T_{3} + W_{2}) = 0$    

$T_{2} = T_{3} + W_{2}$   (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

$T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg$

Case T₁.

$T_{1} - (T_{2} + W_{1}) = 0$  

$T_{1} = T_{2} + W_{1}$    (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

$T_{1} = 3mg + 3mg = 6mg$        

Therefore, the correct option is d: T₁ > T₂ > T₃.  

I hope it helps you!

Answer 2

Correct answer: D. $T_{1} > T_{2} > T_{3}$

First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:

Mass m:

$\Sigma F = T_{3}-m\cdot g = 0$ (1)

Mass 2m:

$\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0$ (2)

Mass 3m:

$\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0$ (3)

The solution of this system is: $T_{3} = m\cdot g$, $T_{2} = 3\cdot m\cdot g$ and $T_{1} = 6\cdot m\cdot g$, which means that $T_{1} > T_{2} > T_{3}$. (Correct answer: D.)