The tension in each of two ropes holding up a box

Start with a guess. A box weighing $98\text{N}$ (that is a $10\text{kg}$ box, since $W=mg$ and $g=9.8{\text{m/s}}^2$) hangs from two ropes, each making a $30^{\circ }$ angle with the ceiling: the angles ${\theta }_1$ and ${\theta }_2$ marked in the first figure. With two ropes holding it up, is the tension in each rope half the weight, $49\text{N}$? Each rope does hold half the weight, $49\text{N}$, through the upward part of its tension, $T\sin \theta$. But the tension in each rope is $98\text{N}$: at $30^{\circ }$ only half of the tension points up. The other part points sideways and is canceled by the other rope. The tension can be larger than the weight itself. The cells below show why, give the tension for any pair of angles, and hang the same box a second way, with one rope steeper than the other.

The box's weight

A box's weight is the force gravity pulls it down with: its mass times $g=9.8{\text{m/s}}^2$, measured in newtons (N). The program measures angles in degrees, the way problems state them, and converts to radians only where it calls sin and cos.

A $10\text{kg}$ box weighs $98\text{N}$ (weight).

The setup, with its angles

The figures on this page are drawn by the program from the angles themselves. Each angle is measured at the ceiling, between the ceiling and the rope, and marked there with an arc. The function below draws one hanging box: the ceiling, the two ropes from where they are fixed down to the knot, the box under the knot, and the two angles.

What pulls on the knot

The knot is the point where the two ropes meet, just above the box. If we look at the knot, there are three forces acting on it, and they must cancel, because the knot is not moving. The box pulls straight down with its weight $W$. The left rope pulls up and to the left along its own length with some force $T_1$, and the right rope pulls up and to the right with $T_2$. A rope can only pull along itself; that pull is what tension means.

Splitting each pull into sideways and upward parts

The angle a rope makes with the ceiling is also the angle it makes with the horizontal line through the knot: the ceiling and that line are parallel, and the rope crosses both. So a rope at angle $\theta$ pulls on the knot partly sideways and partly up. With tension $T$, the sideways part is $T\cos \theta$ and the upward part is $T\sin \theta$: the rope's pull is the long side of a right triangle, $\cos$ gives the side along the horizontal, $\sin$ the side going up. The free-body diagram further down draws those triangles.

Now require that nothing moves.

Sideways: the left rope's sideways pull must cancel the right rope's: $$T_1\cos {\theta }_1=T_2\cos {\theta }_2.$$

Up and down: the two upward parts together must hold up the weight: $$T_1\sin {\theta }_1+T_2\sin {\theta }_2=W.$$

Two equations, two unknowns, $T_1$ and $T_2$. The function below is those two equations written as code. Given candidate tensions, it adds up the three forces and reports how much is left over. For the right tensions the answer is zero; the checks further down use it on every answer.

Solving the symmetric case first

When both ropes make the same angle $\theta$, the sideways equation says $T_1=T_2$. Call it $T$. The up-and-down equation becomes $2T\sin \theta =W$, so $$\frac{T}{W}=\frac{1}{2\sin \theta }.$$

Before running: what is $T/W$ with the ropes straight up ($90^{\circ }$), and at $30^{\circ }$?

Straight up, each rope carries half the weight, $0.5W$ (ratio_straight_up), as the guess at the top expected. At $30^{\circ }$ it is $1W$ (ratio_at_30_degrees): $\sin 30^{\circ }=\frac{1}{2}$, so $T=W$, the answer from the start, now explained. Each rope still holds half the weight through its upward part, $T\sin \theta$. At $30^{\circ }$ only half of the tension points up, so the tension is twice that half: $98\text{N}$, equal to the weight of the box. The rest of the tension, $T\cos \theta$, points sideways and is canceled by the other rope.

Solving for any two angles

From the sideways equation, $T_2=T_1\cos {\theta }_1/\cos {\theta }_2$. Put that into the up-and-down equation, multiply through by $\cos {\theta }_2$, and use $\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2=\sin ({\theta }_1+{\theta }_2)$: $$T_1=\frac{W\cos {\theta }_2}{\sin ({\theta }_1+{\theta }_2)},T_2=\frac{W\cos {\theta }_1}{\sin ({\theta }_1+{\theta }_2)}.$$

Each rope's tension has the other rope's angle on top.

At $30^{\circ }$ and $30^{\circ }$ the general formula gives $98\text{N}$ (tension_at_30_degrees), the same as the symmetric case. Of each rope's $98\text{N}$, $84.9\text{N}$ points sideways (sideways_part_at_30) and $49\text{N}$ points up (upward_part_at_30). The $49\text{N}$ up is the half of the weight each rope holds; the $84.9\text{N}$ sideways is canceled by the other rope's sideways part.

The free-body diagram, to scale

The next figure is the knot alone, with the three forces on it drawn as arrows, all to one scale. Each rope's pull is split into its sideways part and its upward part, drawn dashed as the two short sides of a triangle. The two sideways parts are equal and opposite; the two upward parts add up to the weight. The rope arrows come out as long as the weight arrow, because each tension is $98\text{N}$, while each upward part is $49\text{N}$.

Two ways to hang the same box

Now hang the same $98\text{N}$ box with the left rope steeper, at $60^{\circ }$, and the right rope still at $30^{\circ }$. Before running: which rope carries more tension? And the $30^{\circ }$ rope, which carried $98\text{N}$ in the symmetric case: more or less now?

With $60^{\circ }$ and $30^{\circ }$ the steep rope's tension is $84.9\text{N}$ and the tension in the rope at the smaller angle is $49\text{N}$ (steep_and_shallow): per unit of weight, $0.866W$ and $0.5W$ (steep_and_shallow_ratio). The steeper rope, the one closer to vertical, carries more tension. The $30^{\circ }$ rope is at the same angle as before, yet its tension fell from $98\text{N}$ to $49\text{N}$, because the steep rope now holds most of the weight: $73.5\text{N}$ of it, against $24.5\text{N}$. The right panel balances the same way the symmetric one does: sideways, $84.9\cos 60^{\circ }=49\cos 30^{\circ }=42.4\text{N}$, and up, $84.9\sin 60^{\circ }+49\sin 30^{\circ }=73.5+24.5=98\text{N}$. The two tensions add up to more than $W$, because neither rope points straight up.

Checking the algebra a second way

The formulas came from a few lines of algebra, and algebra can go wrong. So the program also finds the tensions without them: the two balance equations are a pair of linear equations in $T_1$ and $T_2$, $$\left(\begin{array}{cc}\cos {\theta }_1& -\cos {\theta }_2\\ \sin {\theta }_1& \sin {\theta }_2\end{array}\right)\left(\begin{array}{c}{T}_1\\ T_2\end{array}\right)=\left(\begin{array}{c}0\\ W\end{array}\right),$$ and lin_solve solves such a system directly. The tests compare the two answers over eight pairs of angles, and check with leftover_force that the formula's tensions cancel the weight.

Why a flatter rope carries more tension

Before running: $T/W=1/(2\sin \theta )$ was $1$ at $30^{\circ }$. At $10^{\circ }$, is it more or less than $2$? At $5^{\circ }$?

More: $2.9W$ at $10^{\circ }$ (ratio_at_10_degrees) and $5.7W$ at $5^{\circ }$ (ratio_at_5_degrees). The figure shows the whole curve. At $90^{\circ }$ each tension is $W/2$, and as the ropes flatten out the tension climbs, slowly at first and then steeply. A rope pulled perfectly flat would need infinite tension, because none of its pull would point up. That is why a clothesline sags: no rope can be pulled tight enough to be straight while holding anything.

The angle trap

Problems measure the angle in two ways, and mixing them up is a common mistake. If the angle $\phi$ is measured from the vertical instead of the ceiling, then $\phi =90^{\circ }-\theta$, and $\sin \theta$ becomes $\cos \phi$: $$T=\frac{W}{2\cos \phi }.$$ The ropes at $30^{\circ }$ above the ceiling are at $60^{\circ }$ from the vertical. Put into the new formula, that should give the same $98\text{N}$.

It does: $98\text{N}$ (tension_from_the_vertical). Putting $30^{\circ }$ into the cosine formula by mistake would give $W/(2\cos 30^{\circ })=0.58W$ instead (wrong_angle_ratio), which is less than the weight and wrong. Before using a formula, check which angle the problem gives you.

Checking an answer without the formula

Three quick checks find many mistakes. If both ropes hang straight down ($\theta =90^{\circ }$), each tension should be $W/2$. If the two angles are equal, $T_1$ and $T_2$ should be equal. And as a rope goes flat, its tension should grow without limit; at $0.2^{\circ }$ each tension should be more than $100W$. The test puts the general formula through all three.

Try one

A $50\text{N}$ sign hangs from two wires at $45^{\circ }$ and $60^{\circ }$ above the horizontal. Which wire carries more tension, and how much more? Work it out before running the cell: the $45^{\circ }$ wire's tension has $\cos 60^{\circ }$ on top, the $60^{\circ }$ wire's has $\cos 45^{\circ }$, and both have $\sin 105^{\circ }$ underneath.

The $60^{\circ }$ wire carries more tension: $T=50\cos 45^{\circ }/\sin 105^{\circ }=36.6\text{N}$, against $25.9\text{N}$ for the $45^{\circ }$ wire (sign, in that order: $45^{\circ }$ first). The steeper wire carries more, as before.

What was checked

The program computes each tension two independent ways, from the formulas and by solving the two balance equations directly, and checks that they agree and that the three forces cancel. The last test checks every number quoted on this page against a third computation outside the program: posts/refs/a-box-hanging-from-two-ropes.py solves the same balance equations with numpy.linalg.solve, and its output is beside it.