0
$\begingroup$

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.

A visual representation of the problem

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, ACuriousMind, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\theta$

$\endgroup$
  • $\begingroup$ Are those => supposed to indicate the greater than equal to or is it supposed to be a pointing? Could you clarify that? $\endgroup$ – Kyle Kanos Mar 3 '15 at 16:40
0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I don't think this is the simplest way to approach this... $\endgroup$ – Floris Mar 3 '15 at 17:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.