# Tension Problem: Finding an angle when only given the tension in two ropes [closed]

A crate is hanging from a rope which is attached to a metal ring through which a second rope runs, as shown to the right. What is the angle $\theta$ if the tension in rope 1 is $1.19$ times the tension in rope 2?

I don't understand how to find the angle without knowing the length of the rope itself.

## closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind♦, Qmechanic♦Mar 4 '15 at 1:10

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You don't need the length; just the tension.

A hint: Use Newton's 1st law in the y-direction. The force downwards must be equal to the total force upwards. And the only forces acting upwards are the y-components of the tension of each part of the rope. The expressions for these y-components will include the angle, and here you have it.

Firstly, draw a free-body force diagram.

Let the tension in rope 1 be $T_1$ , and the tension in the second rope with both the Y-components adding up be $2\times(T_2\cos\theta)$.

It is given that $T_1=1.19(T_2)$

Since the system is in equilibrium just equate the Y component forces. You should be able to get the angle by solving for $\theta$

• Are those => supposed to indicate the greater than equal to or is it supposed to be a pointing? Could you clarify that? – Kyle Kanos Mar 3 '15 at 16:40

We know: $$T_1 = 1.19 \cdot T_2 \cdots 1)$$

We can determine:

$$T_3 \text{(other rope)} = T_2 \cdots 2)$$ $$T_{1y} = T_1 \text{ (no x-component [horizontal])}\cdots 3)$$ $$T_{1x} = 0 \cdots 4)$$ $$T_{2y} + T_{3y} = -T_{1y} = -T_1 \text{(box isn't moving up or down....)} \cdots 5)$$ $$T_{2x} = -T_{3x} \text{(box isn't moving right or left)}\cdots 6)$$ $$T_2 = \sqrt{{T_{2y}}^2 + {T_{2x}}^2)}\cdots 7)$$

There are 7 equations and only 6 unknowns ($T_1$, $T_2$, $T_3$, in $x$ and $y$) So find $T_{2x}$ and $T_{2y}$ and you will have your angle.

I guess you need to assert that the angle of rope 2 with the vertical is the same as theta, due to the metal ring which allows the rope to slip through it and equalize the tensions on rope 2 and rope 3.

• I don't think this is the simplest way to approach this... – Floris Mar 3 '15 at 17:12