Find the angle of two equal ropes supporting a weight

Key takeaway: This is a classic equilibrium problem. The two rope tensions have equal horizontal parts that cancel, while their vertical parts add up to the weight. Once you set that balance correctly, the angle drops out cleanly.

Let the angle each rope makes with the ceiling be $\theta$.

Because the setup is symmetric, the horizontal components of the two tensions cancel. So we only balance vertical forces.

1) Vertical components

Each rope has tension $40\,\text{N}$, and the angle is measured from the ceiling, so the vertical component of each tension is $40\sin\theta$

There are two ropes, so the total upward force is $2(40\sin\theta)=80\sin\theta$

2) Set equal to the weight

The weight is $60\,\text{N}$, so $80\sin\theta=60$ $\sin\theta=\frac{60}{80}=0.75$

3) Solve for the angle

$\theta=\sin^{-1}(0.75)\approx 48.6^\circ$

Answer

The ropes make an angle of about $\boxed{48.6^\circ}$ with the ceiling.

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Pitfalls the pros know 👇 A common mistake is using cosine instead of sine. Since the angle is measured from the ceiling, the vertical component is the opposite side, so it is $T\sin\theta$, not $T\cos\theta$.

What if the problem changes? If the angle were measured from the vertical wall instead of the ceiling, the component equation would change to $2T\cos\theta=60$. That would give a different numerical angle, but the force balance idea is the same.

Tags: equilibrium, tension, resolved components