Question

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

Answer 1

Answer:

at x=0

$T_{l}$=W*L/L

and

$T_{l}$=W-$T_{r}$

Explanation:

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

moments is the product of force and the perpendicular distance in line of the action of the given force

from the principle of moments which states that the sum of clockwise moments ,must be equal to the sum of anticlockwise moments.

also, sum of upward forces must be equal to sum of downward forces

Going by the aforementioned,

taking moments about $T_{r}$

W*(l-x)=$T_{l}$*(L)

$T_{l}$=W*(L-x)/(L)..............1

at x=0

$T_{l}$=W*L/L

also

$T_{l}$+ $T_{r}$=W

$T_{l}$=W-$T_{r}$

Answer 2

The tension in the left rope is expressed by $T_{L} = W\cdot \left(1-\frac{x}{L} \right)$. For $x = 0$, $T_{L} = W$.

How to apply moment formulas to determine the tension in the left rope

A representation of the system is presented in the image attached below. Let suppose that the entire system represents a rigid body, that is, a body whose geometry cannot be neglected and does not experiment deformation due to the application of external loads.

The tension in the left rope ($T_{L}$) can be found by applying Newton's laws and D'Alembert's Principle applied on the rightmost point of the bar:

$\Sigma F = W\cdot (L-x)-T_{L}\cdot L = 0$

$T_{L} = \frac{W\cdot (L-x)}{L}$

$T_{L} = W\cdot \left(1-\frac{x}{L} \right)$   (1)

The tension in the left rope is expressed by $T_{L} = W\cdot \left(1-\frac{x}{L} \right)$. For $x = 0$, $T_{L} = W$. $\blacksquare$

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